Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How many squares in base $9$ consist of only ones How many perfect squares exist of the following form: $(1111111....11)_9$?
Let the required number equal be $(111111...1\space (n \space ones))_9$. This can written as:
$9^{n-1} + 9^{n-2} + \dots + 9^2 + 9 + 1 = \frac{9^{n} - 1}{8} = k^2$ for some k.
Since $\frac14$ is... | The following is to show there are not solutions with $n>1$ of the equation.
The equation $(9^n-1)/8=k^2$ may, after doubling the sides, be factored as
$$\frac{3^n-1}{2}\cdot \frac{3^n+1}{2}=2k^2.$$
The two factors on the left here are adjacent integers and so are coprime. This implies, from the right side of the equat... | {
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"url": "https://math.stackexchange.com/questions/522412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$ I know I have to use long division first, but I don't really know how to do it in this case
$$\int \frac{x^3 + 36}{x^2 + 36}dx$$
| Using http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)- 36x +36 }{x^2+ 36} $$
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)}{x^2+36} - \frac {36x}{x^2+36} + \frac{36}{x^2+36} $$
$$ \int \frac{x^3+36}{x^2+36}dx = \int \frac{x(x^2+36)}{x^2+36} dx -\int \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523209",
"timestamp": "2023-03-29T00:00:00",
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Finding minimum $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}$ I would appreciate if somebody could help me with the following problem
Q. Finding maximum minimum
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
| $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
Consider the following :
$a , \frac{1}{a}$ we know that $A.M. \geq G.M.$
$\therefore \frac{a+ \frac{1}{a}}{2} \geq \sqrt{a . \frac{1}{a}}$
$\Rightarrow \frac{a^2+1}{2a} \geq 1 $
$\Rightarrow a^2 + 1 \geq 2a $
$\Rightarrow (a-1)^2 \geq 0$
$\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Derivative of $\lceil 1/x \rceil$ I'm looking for the derivative $\frac{d}{dx}\lceil 1/x \rceil$.
I would like to find a real number $1<x \le y$ satisfying the minimum of $\left\lceil \frac{y}{x} \right\rceil x$, when $y$ is a fixed value $>0$.
Is the ceiling function a problem ?
| $\large x > 0:$
\begin{align}
\left\lceil x\right\rceil
&=
\Theta\left(x\right)\Theta\left(1 - x\right)
+
2\Theta\left(x - 1\right)\Theta\left(2 - x\right)
+
3\Theta\left(x - 2\right)\Theta\left(3 - x\right) + \cdots
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\Theta\left(x - n\right)
\Theta\left(n + 1 - x\right)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limit -> Infinity with radicals First guess to multiply by $x^{-1.4}$ so the radical in numerator with $x^7$ becomes 1 and other stuff becomes 0. But then denominator becomes $-\infty$. What is the right approach?
$$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} $... | For starters: Note, that the denominator has as highest power of $x$ the $x^1$, so we multiply by $1 = \frac {1/x}{1/x}$, giving
$$\frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} =
\frac{\sqrt[5]{x^2 + 3x^{-5}} - \sqrt[4]{2x^{-1} - x^{-4}}}{\sqrt[8]{x^{-1} + x^{-6} + x^{-8}} - 1}
$$
Now th... | {
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"timestamp": "2023-03-29T00:00:00",
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Is my calculation of $\frac{\partial}{\partial x}\frac{x+y}{\sqrt{y^2-x^2}}$ correct? Is my calculation of the partial derivative (with respect to $x$) of the function $$f(x,y)=\frac{x+y}{\sqrt{y^2-x^2}}$$ correct?
$$f'_x=\frac{\sqrt{y^2-x^2}-(x+y)(\frac{-x}{\sqrt{y^2-x^2}})}{|y^2-x^2|}\\=\frac{\sqrt{y^2-x^2}+\frac{x^... | All correct except the last line
$$f'_x=\frac{\sqrt{y^2-x^2}-(x+y)(\frac{-x}{\sqrt{y^2-x^2}})}{|y^2-x^2|}\\=\frac{\sqrt{y^2-x^2}+\frac{x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=\frac{\frac{|y^2-x^2|+x^2+xy}{\sqrt{y^2-x^2}}}{|y^2-x^2|}\\=\frac{|y^2-x^2|+x^2+xy}{(y^2-x^2)^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$ How do I prove
$$_4F_3\left(\frac18,\frac38,\frac58,\frac78;\ \frac14,\frac12,\frac34;\ \frac12\right)\stackrel?=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2\phantom{|}}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2... | Some related values, based on Kirill's answer, given by radicals:
$$\begin{align}
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac59\right) &= \frac{1}{4}\sqrt{9+\sqrt{30}+2\sqrt{15+3\sqrt{30}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Checking the equality of an equation(Trigonometery) The equation
$\displaystyle\cos^2\theta=\frac{(x+y)^2}{4xy}$ is only possible when
What i need to compare is x and y and derive a relation between them
i.)x
ii.)x=-y
iii.)x>y
iv.)x=y
Pls Answer
| $$\cos^2\theta=\frac{(x+y)^2}{4xy}\iff \sec^2\theta=\frac{4xy}{(x+y)^2}$$
$$\implies \tan^2\theta=\frac{4xy}{(x+y)^2}-1=-\frac{(x+y)^2-4xy}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
$$\implies \tan^2\theta+\left(\frac{x-y}{x+y}\right)^2=0$$
As $\theta$ is real, $\tan^2\theta\ge0$ and $x,y$ are real $\displaystyle\lef... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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probability regarding three people throwing a die There are 3 players, A, B, C, taking turns to roll a die in the order ABCABC....
What's the probability of A is the first to throw a 6, B is the second, and C is the third?
The answer said it's 216/1001, but I always got 125/1001
The way I did it was:
Let $X_1$= num... | Since they go in the order A,B,C, you want $\sum_{i\le j\le k}$.
Detail: We want $$\sum_{0\le i\le j\le k}(5/6)^i(5/6)^j(5/6)^k(1/6)^3.$$
Summing from $k=j$ to $\infty$, we get
$$\sum_{0\le i\le j}(1/6)^2(5/6)^i (5/6)^{2j}.$$
The summation with respect to $j$ yields
$$\sum_{0\le i} (5/6)^{3i},$$
which is $\frac{6^3}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/536420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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gaussian and mean curvatures I am trying to review, and learn about how to compute and gaussian and mean curvature. Given $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, how can I compute the gaussian and mean curvatures?
This is what I have so far,
$$K(u, v) = \frac{a^2 b^2 c^2}{[c^2 \sin^2(v) (a^2 \sin^2(u... | The parametric equations of an ellipsoid can be written as
\begin{align}
x &=a \cos u \sin v\\
y &=b \sin u \sin v\\
z &=c \cos v
\end{align} for $u \in [0,2\pi)$ and $v \in [0,\pi]$.
In this parametrization, the coefficients of the first fundamental form are
\begin{align}
E &= (b^2\cos^2u+a^2\sin^2u)\sin^2v\\
F &= (b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$ Prove this
$$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$
My try: let
$$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$
| Allow me to present an approach that uses dilogarithms. Split the integral up into 2 and substitute $x \mapsto \dfrac{1}{x}$ for the second integral. This yields
\begin{align}
\int^\infty_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx
&=\int^1_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx+\int^\infty_1\frac{1}{x}\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Are these ideals the same? I have already proved that $(X^3-Y^3,X^2Y-X)\subseteq(X^2-Y,X-Y^2)$ since the elements $X^3-Y^3$ and $X^2Y-X $ can be written as a linear combination of $(X^2-Y,X-Y^2)$.
However, I can't write $X^2-Y,X-Y^2$ as a combination of $(X^3-Y^3,X^2Y-X)$. How can I prove that $(X^2-Y,X-Y^2) \subseteq... | If $(X^2-Y,X-Y^2) \subseteq(X^3-Y^3,X^2Y-X)$, then $X^2-Y\in (X^3-Y^3,X^2Y-X)$. There exist two polynomials $f,g$ such that $X^2-Y=(X^3-Y^3)f(X,Y)+(X^2Y-X)g(X,Y)$. Send $X$ to $0$ in the equation and find $Y=Y^3f(0,Y)$ $\Leftrightarrow$ $1=Y^2f(0,Y)$. Now send $Y$ to $0$ and obtain $1=0$, a contradiction.
Conclusion: ... | {
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"timestamp": "2023-03-29T00:00:00",
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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?
| Use Newton's identities:
$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.
Here
$p_1= x+y+z = e_1$
$p_2= x^2+y^2+z^2$
$p_3= x^3+y^3+z^3$
$e_2 = xy + xz + yz$
$e_3 = xyz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/543991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
| Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, by AM-GM and Schur we obtain:
$$a^2+b^2+c^2+2abc+1\geq a^2+b^2+c^2+3\sqrt{a^2b^2c^2}=\sum_{cyc}(x^6+x^2y^2z^2)\geq$$
$$\geq\sum_{cyc}(x^4y^2+y^4x^2)\geq2\sum_{cyc}x^3y^3=2(ab+ac+bc)$$
and we are done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$
| We first remark the following identity:
\begin{align*}
\int_{0}^{\infty} \frac{x^{s-1}}{ze^{x} - 1} \, dx
&= \int_{0}^{\infty} \frac{z^{-1}x^{s-1}e^{-x}}{1 - z^{-1}e^{-x}} \, dx \\
&= \sum_{n=1}^{\infty} z^{-n} \int_{0}^{\infty} x^{s-1} e^{-nx} \, dx \\
&= \Gamma(s) \sum_{n=1}^{\infty} \frac{z^{-n}}{n^{s}}
= \Gamma(s)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.
| Here is a non-calculus method which works in these cases:
Put
$$S = \frac{1}{3} + \frac{4}{9} + \frac{9}{27} + \frac{16}{81} + \dots$$
then $$\frac{1}{3}S = \frac{1}{9} + \frac{4}{27} + \frac{9}{81} + \dots$$
subtract:
$$\frac{2}{3}S = \frac{1}{3} + \frac{3}{9} + \frac{5}{27} + \dots$$
This has simplified the problem (... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$ I'm having trouble proving that:
$$\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$$
Edit: The next step is actually a mistake. I've put up a comment to the accepted answer explaining why it's a mi... | For any $\;1\le k\le 2^n\;$ we have that
$$\frac1{2^n+k}\ge\frac1{2^n+2^n}=\frac1{2^{n+1}}\;,\;\;\text{so:}$$
$$\underbrace{\frac1{2^n+1}+\ldots+\frac1{2^n+2^n}}_{2^n\;\text{summands}}\ge\frac{2^n}{2^{n+1}}=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of series of ratios I need to calculate
$$\lim_{n \to \infty}\frac{1}{n}\left(\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n}+\cdots+\frac{3n}{4n}\right).$$
It can be written as
$$\lim_{n \to \infty}\frac{1}{n^2}\left(\frac{1}{\frac{1}{n}+1} + \frac{1}{\frac{2}{n}+1} + \frac{1}{\frac{3}{n}+1}+\cdots+\frac{1}{\frac{4... | Hint:
$$
\frac{1}{n}\sum_{k=1}^{mn} \frac{1}{\frac{k}{n} + 1} \approx \int_0^m \frac{1}{x+1}dx
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ 7^{50} \cdot 4^{102} ≡ x \pmod {110} $ The way I would solve this would be: $$ (7^3)^{15} \cdot 7^5 \cdot (4^4)^{25} \cdot 4^2 $$ and take it from there, but I know that this is most likely in an inefficient way. Does anyone have more efficient methods?
| Do prime factorization on $110$. It's $110 = 5 \cdot 2 \cdot 11$.
Now work with the prime factors as moduli.
$$7^{50} \cdot 4^{102} \equiv 0 \pmod 2 \text{ trivial, as 2 \mid 4}$$
Then apply Fermat's Little Theorem so we have:
$$7^4 \equiv 1 \pmod 5 \implies 7^{50} \equiv 49 \equiv 4 \pmod 5$$
$$4^4 \equiv 1 \pmod 5 \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Combinatorics error correcting code (56) * (36)^4 * (-55) + (35)(67)(-14)^2 mod 17. Find the least non-negative residue of the expression module the given n.
First I just want to make sure I understand what the question wants. To do that, I give a simple case 15 mod 7. 1 is what we looking for right?
Then get back to ... | The calculation can be done without a calculator. Since $56=3\cdot 17+5$, we have $56\equiv 5\pmod{17}$. We have $36\equiv 2\pmod{17}$, so $36^4\equiv 2^4\equiv -1\pmod{17}$. But $(-1)(-55)=55\equiv 4\pmod{17}$. Thus the first part of our expression is $\equiv 20\equiv 3\pmod{17}$.
A similar calculation shows that the... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12$ How can I prove that
$$\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}$$
| The answer can be achieved using simple limit theorems as follows
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\left(\frac{1}{\log(x + \sqrt{1 + x^{2}})} - \frac{1}{\log(1 + x)}\right)\\
&= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(x + \sqrt{1 + x^{2}})\log(1 + x)}\\
&= \lim_{x \to 0}\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How do you factor $(10x+24)^2-x^4$? I tried expanding then decomposition but couldn't find a common factor between two terms
| Whoa there, Ladies and Gentlemen! We ain't done yet! As pointed out by yanbo and Hugus in their answers, a good first step is to observe that the identitiy
$a^2 - b^2 = (a + b)(a - b) \tag{1}$
may be directly applied to
$(10x+24)^2-x^4 \tag{2}$
and we obtain
$(10x+24)^2-x^4 = (10x + 24 + x^2)(10x + 24 - x^2); \tag{3}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| let $$E(a,b,c)=a^3b^2+b^3c^2+c^3a^2-abc(a^2+b^2+c^2)$$
then
$$2E(a,b,c)=\sum a^3(b-c)^2-\sum a^2(b^3-c^3)=\sum a^2(b-c)^2(a+c-b)\ge 0$$
because
$$\sum a^2(b^3-c^3)=\sum a^2(b-c)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integration of $\int\frac{x^2-1}{\sqrt{x^4+1}} \, dx$ Integration of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+1}} \,dx$
$\bf{My\; Try}$:: Let $x^2=\tan \theta$ and $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$
$$
\begin{align}
& = \int\frac{\tan \t... | For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\dfrac{x^2-1}{\sqrt{x^4+1}}dx$
$=\int(x^2-1)\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+2}}{4^n(n!)^2}dx-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
| use this
since
$$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz\Longrightarrow xy+yz+xz=2$$
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$
so
$$7-3xyz=27-18\Longrightarrow xyz=-\dfrac{2}{3}$$
use
$$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the... |
Proposition : $$ f(z,a) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}$$
Proof : Note that,
$$ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} $$
where $ U_{n} (x) $ is the Chebyshev Polynomial of the second ki... | {
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"url": "https://math.stackexchange.com/questions/557230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
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Imaginary numbers and polynomials question I have a task which I do not understand:
Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.
Given that Im $w = 0$, show that $| z | = 1$.
Partial solution (thanks to @ABC and @aranya):
If I substitute $z$ with $x + iy$ then we have $w = \frac... | You're basically done, since $x^2+y^2=\left|z\right|^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3... | Let $n$=0. The sum is then $\sum_{k=1}^{3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{-k}= \frac{1}{2}\cdot4+\frac{1}{4}\cdot16+\frac{1}{8}\cdot64=2+4+8=14$. This equals $2\left(\frac{1}{4} \right)^{0}\left(2^{0+3}-1\right)=2\cdot 7$, and it does not equal $\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation $\sqrt{x}+\sqrt{y}=\sqrt{2013}$ in rationals Can we find all rational numbers $x,y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{2013}$?
Certainly possible answers are $(2013,0)$ and $(0,2013)$.
If we square the equation, we get $x+y+2\sqrt{xy}=2013$, so $\sqrt{xy}$ must be rational.
| if $x\neq 0$ then $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational.
$x+y+2\sqrt{xy} = 2013$
$x + xq^2+2xq = x(q+1)^2=2013$
$x= \frac{2013}{(q+1)^2}, y=xq^2, q\in \mathbb Q^+\cup\{0\}$
if $x=0$ then $y = 2013$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/564422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$ Regarding this problem, I conjectured that
$$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \ma... | Just for references, I remark that the following proposition was proved in my answer:
Proposition. If $0 < r < 1$ and $r < s$, then
$$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 2,
"answer_id": 1
} |
Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \... | How do you get from $$\lim\limits_{c \rightarrow 0^+} \frac{\sqrt{1+c} - 1 + \sqrt{1-c} - 1}{c^2}$$ to $$\frac{1}{2}c - \frac{1}{2}c?$$
You cannot just evaluate part of the limit and leave another part. Compare your mistake with "evaluating" $\lim_{x \rightarrow 0} \frac{x^2}{x^2}$ by noting that $\lim_{x \rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the radius of convergence and the interval of convergence of the series $\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$ Series is:
$$\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$$
So, I understand that I use the ratio test to find r, but I can't simplify the equation to the point where I can do this. Here's where I am so f... | $$
\lim_{n \to \infty} \left|\frac{(n+1)(x-4)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right|\lim_{n \to \infty} \left|\frac{(n+1)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right| < 1 \\
R = 1 \ \text{and} \ x \in (3, 5)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate the limit of the function Please help me to calculate the limit of the function. I do not know where to start.
Thank you.
| To get started, Assuming limit L exists then
$$
\begin{align}
\\ \lim_{x\to0} f(x)^{g(x)} &={\rm L}
\\ \lim_{x\to0} {g(x)} \ln f(x) &=\ln {\rm L}
\end{align}
$$
Where
$$
\begin{align}
f(x)&= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
\\ g(x) &=-\frac{1}{\left(1-\ope... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
| Let $f(y) = \frac12 (y^3+1)$, we have
$$2\sqrt[3]{2y-1} = y^3+1\quad\iff\quad f^{-1}(y) = f(y) \quad\implies\quad y = f(f(y))$$
Since $f(y)$ is a strictly increasing function in $y$, we have
*
*If $f(x) > x$, then $f(f(x)) > x$.
*If $f(x) < x$, then $f(f(x)) < x$.
This means
$$\begin{align}
y = f(f(y)) \implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$ Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example?
$$\begin{align}S&=\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx\\\vphantom... | Incomplete solution:
In fact, we have $${}_2F_1(-\frac14,\frac54;1;\frac{x}{2})=\frac{8\sqrt{2}}{\pi\sqrt{2+\sqrt{2x}}}((2+\sqrt{2x})E(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})-K(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})).$$
Here $E(m)=\int^1_0\sqrt{\frac{1-mt^2}{1-t^2}}~dt$ and $K(m)=\int^1_0\frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 3,
"answer_id": 0
} |
Generalizing a series of numbers I know that I studied this long ago but I can't seem to bring the information to mind. I am looking to construct a general formula for the following sequence of numbers ->
when $x=4$ : $n=(3+x)+2(x-1)$;
when $x=5$ : $n=(3+x)+3(x-1)+2(x-2)+1(x-3)$;
when $x=6$ : $n=(3+x)+4(x-1)+3(x-2)+2(... | Note that we can rewrite your statements:
when $x=4:n=(3+x)+2(x-1)$. Obviously, if $x=4$, then we have $n=(3+4)+2(4-1) = 13$.
$\begin{align} x=5:n&=(3+x)+3(x-1)+2(x-2)+1(x-3)\\ &= 3+4+3(4-1)+2(4-2)+1(4-3)\\ &= 7+9+4+1 \\ &= 21\end{align}$
$\begin{align} x=6:n&=(3+x)+4(x-1)+3(x-2)+2(x-3)+1(x-4) \\ &= 3+6+4(6-1)+3(6-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| By QM-AM for the first inequality and AM-GM for the second,
$$\frac{a^2+b^2+c^2}{3}\geq\left( \frac{a+b+c}{3} \right)^2\geq\frac{a+b+c}{3}\left( \sqrt[3]{abc} \right)$$
so since $abc=1, a^2+b^2+c^2\geq a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Prove $13|19^n-6^n$ by congruences I am trying to prove $13|19^n-6^n$. With induction its not so bad but by congruences its quite difficult to know how to get started.
Any hints?
| Modular Arithmetic is an extremely useful way of thinking about numbers. It can be used to calculate the last three digits of $2011^{2011}$ by hand. The day of the week, the time of the day, and the flipping of a light switch are all based upon modular arithmetic. The day of the week is in $\mod 7$ (cycles in a period ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Intuitive explanation for formula of maximum length of a pipe moving around a corner? For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):
My ... | Here's a derivation without trigonometry.
There is also a short discussion of the geometric significance
of the exponents in the answer after the derivation.
Let $L$ be the length of a line touching the inner corner of the
two corridors and extending to the outer wall of each corridor.
Where this line meets the outer ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
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inequality $\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$ Given real numbers $a, b, c, d$ with this condition: $$abcd=1$$
Let's prove this inequality: $$\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$$
Thank everybody!
| The function $f(x ) = \frac{1}{1+3 e^x}$ is convex on $[\ln(\frac{1}{3}),\infty)$ look: convexity proof.
Whenever $y\leq 1/3$ then we have that $\dfrac{1}{1+3y} \geq \dfrac{1}{2}$, therefore whenever more than two of the $a,b,c,d$ are less than $1/3$ then the inequality holds. So, it remains to deal with the case tha... | {
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"url": "https://math.stackexchange.com/questions/584227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Particular solution of a non-homogenous recurrence relation I need some help with the following non-homogenous recurrence relation.
$$a_n-2a_{n-1}+a_{n-2}=n+1$$
$$a_0=0, a_1=1$$
When I solve the associated homogenous equation I use the auxiliary equation $x^2-2x+1=0$ and obtain the root $x=1$. Hence, I obtain the equa... | The form of the equation allows us to solve it by using only first-order difference equations (see Remark 1).
Arrange the equation as
$$\begin{cases}[a_{n}-a_{n-1}]-[a_{n-1}-a_{n-2}]=n+1,{\quad}n=3,4,\cdots\\ a_{1}=1,\ a_{0}=0,\end{cases}$$
which is a first-order difference equation in $b_{n}:=[a_{n-1}-a_{n-2}]$ for $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| Alternatively, let $a_1,\,a_2,\,a_3,\,\cdots,\,a_n$ be the following sequence
$$\sqrt{7},\,\sqrt{7\sqrt{7}},\,\sqrt{7\sqrt{7\sqrt{7}}},\,\cdots,\underbrace{\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots\sqrt{7}}}}}}_{\large n\,\text{times}}$$
respectively.
Notice that
$$\large a_n=7^{\Large 1-2^{-n}}$$
Hence
$$\sqrt{7\sqrt{7\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 6
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How find this integral $\int\frac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$ Find this integral
$$\int\dfrac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$$
My try: let
$$\arctan{(x\sqrt{x-1})}=t$$
and that's very ugly,Thank you
| $\because$ according to http://integrals.wolfram.com/index.jsp?expr=%28x-1%29%5E%281%2F2%29%2Fx&random=false, $\int\dfrac{\sqrt{x-1}}{x}dx=2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1}+C$
$\therefore\int\dfrac{\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})}{x}dx$
$=\int\tan^{-1}(x\sqrt{x-1})~d(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})$
$=(2\sqrt{x-1}-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x ten... | You need to evaluate $$\lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^{2} - 4}}$$ and you are not supposed to replace $\sqrt{x} - \sqrt{2}$ by $\sqrt{x - 2}$. This is not allowed through any rule of algebra/calculus. What we do here is not difficult, but a bit of careful and simple algebraical mani... | {
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"url": "https://math.stackexchange.com/questions/589470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $... | Note that if $x\lt0$, then $\lfloor x^2\rfloor-3\lfloor x\rfloor+2\ge5$, so any solution $x=n+f$ with $0\le f\lt1$ must have $n\ge0$. Picking up at the OP's equation $n^2+\lfloor f^2+2nf\rfloor-3n+2=0$ and letting $r$ denote the non-negative integer $\lfloor f^2+2nf\rfloor$, we find
$$(2n-3)^2=1-4r$$
Thus $r$ must be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| $$\begin{eqnarray}
\\ \text{Let } 2^{(n-1)} + 2^{(n-2)} + \cdots + 2 + 1 &=& S
\\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 &=& 2S \tag {multiply both sides by 2}
\\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 +1 &=& 2S +1\tag {add 1 to both sides }
\\ 2^{n} + S &=& 2S +1\tag {replace the term on LHS by S }
\\ 2^{n} &=& S +1\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Let a function $f(x)= \sqrt{(x^2-2x+2)(-x^2+3x+10)}-\sqrt{(-x^{2}+x-1)(x^{2}+x-6)}$ Let a function $f(x)= \sqrt{(x^2-2x+2)(-x^2+3x+10)}-\sqrt{(-x^{2}+x-1)(x^{2}+x-6)}$ and its domain is $D=[a,b+c]$. What's the value of $a^3+b^3+c^3 \over abc$ ?
Any ideas ?
The answer is $\boxed{3}$ .
| in the first term $x^2-2x+2$ is always greater than $0$. hence $-x^2+3x+10$ should be greater than or equal to $0$.
similarly for the second term $-x^2+x-1$ is always less than $0$. hence $x^2+x-6$ should be less than or equal to $0$.
now find the common interval in which both inequality satisfies . the domain will be... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Fast way to prove that $(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$ What is the most simplest way to prove that $$(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$$
Please!! Thanxx
| For brevity's sake, let $x = a - b$, $y = b - c$, and $z = c - a$. We compute a few products:
$$-xz = (a - b)(a - c) = a^2 - ab - ac + bc$$
$$-xy = (b - a)(b - c) = b^2 - ab - bc + ac$$
$$-yz = (c - a)(c - b) = c^2 - ac - bc + ab$$
Adding these all together gives:
$$ - (xy + yz + zx) = a^2 + b^2 + c^2 - ab - ac - bc $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know... | First we compute the sum $S_1=\sum_{n=1}^\infty \frac{n}{2^n}$ using the identity $n=\sum_{k=1}^n1$ $$S_1=\sum_{n=1}^\infty \frac{n}{2^n}=\sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{2^n}=\sum_{ k \leq n }\frac{1}{2^n}=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{2^n}=\sum_{k=1}^\infty \frac{1/2^k}{1-1/2}=2 \sum_{k=1}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 1
} |
Solid Revolution question I'm trying to do the following problem:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the $y=3$ axis specified by means of the circular arrow drawn :
$$ y = 3 \sin x, y = 3 \cos x, 0 \le x \le π/4 $$
First off, am I drawing the picture correct... | Yahoo!Answers is notoriously unreliable on math and technical matters; my informal count is that the answer is alright only about half the time. (I suspect the poster overlooked the rotation axis being $ \ y = 3 \ , $ rather than the $ \ x-$ axis.) Your integrand looks fine and reduces to
$$ \ (9 - 18 \sin x + 9 \sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$a, b \in\Bbb N$, $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$. If $a$ and $b$ are positive integers such that $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$.
By contradiction assuming that $a <b$ then it follows that $2 (a +1) <A <2 (b +1)$.
This means that $A = 2a + r,$ $... | First, if $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, then $\sqrt{n} \in \Bbb N$.
To see why, notice first that if $n$ is a perfect square, there is nothing to prove.
So, suppose $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, and $n$ is not a perfect square. Then there is a prime number $\alpha$ such that $\alpha$ factors in $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{\log(2x+1)-\log 4}{1-\log(3x+2)}=1$ My attempt:$$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$
$$\frac{\log(2x+1)-\log4}{\log10-\log(3x+2)}=\log10$$
$$\frac{\log\frac{(2x+1)}{4}}{\log\frac{10}{(3x+2)}}=\log10$$
$$\frac{\frac{(2x+1)}{4}}{\frac{10}{(3x+2)}}=10$$
$$\frac{(2x+1)}{4}=10\cdot\frac{10}{(3x+2)}$$(is ... | $$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$
$$\frac{\log(2x+1)/4}{\log 10/(3x+2)}=1$$
$$\log(2x+1)/4=\log 10/(3x+2)$$
$$(2x+1)/4=10/(3x+2)$$
$$(2x+1)(3x+2)=40$$
$$6x^2+7x-38=0$$
$$x_{1,2}=\frac{-7\pm31}{12}:x_1=2,x_2=-\frac{19}{6}$$
in real field only $x_1=2$ is a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/597162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$ Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\r... | Let $a_k = (-1)^k \left(\frac{1}{4k+1} + \frac{1}{4k+3}\right)$ and $b_k = \frac{1}{(4k+1)^2} + \frac{1}{(4k+3)^2}$. The goal is to show that: $$ \left(\sum_{i=0}^\infty a_i\right)^2 = \sum_{i=0}^\infty b_i $$
The key observation that I missed on my previous attempt is that: $$ \sum_{i=0}^n a_i = \sum_{i=-n-1}^n \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 2,
"answer_id": 1
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Closed form for $\int_0^\alpha \int_0^{\alpha-x} \frac{1}{[(x+y)(\beta-x)(1-\beta-y)]^{3/2}} \,dy \,dx$ Let $0\leq \alpha \leq \frac{1}{2},$ and let $\beta$ be such that $\alpha \leq \beta \leq 1-\alpha$.
Define $\varphi(\alpha,\beta)$ to be the integral
$$\varphi(\alpha,\beta) = \int_0^\alpha \int_0^{\alpha-x} \frac 1... | Let $z = x+y$, we have
$$\varphi(\alpha,\beta) = \int_0^\alpha \frac{dz}{z^{3/2}}\left\{\int_0^z \frac{dy}{((\beta-z+y)(1-\beta-y))^{3/2}}\right\}$$
For fixed $z$, introduce variables $u, v$ such that $u = \beta -z + y$, $v = 1 - \beta-y$. It is easy to check
$$d\left(\frac{u-v}{\sqrt{uv}}\right)
= \left(\frac{2}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/599454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that for every $k\geq 1$, $1\cdot 3\cdot(1!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$ Prove that for every $k\ge1,$
$\displaystyle1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$
I start off with considering induction, but I can't quite get the left side to match the right side.... | Let
$\displaystyle f(k): 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$ holds true for $k=m$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(m+2)\cdot(m!)^2=\{(m+1)!\}^2-1 $
For $k=m+1,$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/599886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.
| Observe that the Right Hand side $\displaystyle\frac{\sin x+1}{\cos x}=\tan x+\sec x$
So, I want to utilize $\displaystyle\sec^2x-\tan^2x=1$
Dividing the numerator & the denominator by $\cos x,$
$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\tan x-1+\sec x}{\tan x+1-\sec x}$$
$$=\frac{\tan x+\sec x-(\sec^2x-\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What is $ \lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2} $? I have limit:
$$
\lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2}
$$
Why is the result $8$ ?
| $$\lim_{x,y\to{2,2}}\frac{x^4-y^4}{x^2-y^2}$$ $$= \lim_{x,y\to{2,2}}\frac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}$$
$$\lim_{x,y\to{2,2}} x^2+y^2 =...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Inverse Trig & Trig Sub Can someone explain to me how to solve this using inverse trig and trig sub?
$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$
Thank you.
| You can also use integration by part: let $u=x^2$ and $dv=\frac{x}{\sqrt{1+x^2}}$ then you will have:
\begin{align*}
\int udv&=uv-\int vdu\\
&=x^2\sqrt{1+x^2}-\int2x\sqrt{1+x^2}\,dx\\
&=x^2\sqrt{1+x^2}-\frac{2}{3}(1+x^2)^{\frac{3}{2}}+C
\end{align*}
where the last integral was solved by substitution $\ u=1+x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality: if $\cos^2a+\cos^2b+\cos^2c=1$, prove $\tan a+\tan b+\tan c\geq 2(\cot a+\cot b+\cot c)$ Let $a,b,c$ be in $\left(0;\dfrac{\pi}2\right)$ such that $\cos^2a+\cos^2b+\cos^2c=1$. I am trying to prove the following inequality: $\tan a+\tan b+\tan c\geq 2\left(\cot a+\cot b+\cot c\right)$, but I do not know how.... | Lemma:
let $x,y,z>0$,then
$$\dfrac{3}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\le\dfrac{x+y+z}{3}\le\sqrt{\dfrac{x^2+y^2+z^2}{3}}$$
let $$\cos{a}=x,\cos{b}=y,\cos{c}=z,x,y,z>0$$
Note
$$\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\sqrt{1-\cos^2{a}}}{x}=\dfrac{\sqrt{y^2+z^2}}{x}\ge\dfrac{1}{\sqrt{2}}\dfrac{y+z}{x}$$
so
$$\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after lo... | Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$
Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$
and
$$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/612181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 1
} |
Elevator stop, other approach There is a $10$-floor building, $10$ people get in the elevator in the ground floor and each gets off at one of the floors randomly and independently. What is the probability that the elevator stops at floor $5$?
$$1 - (9/10)^{10}$$
I understand the approach of using the probability that n... | Yes, there is an alternative way to solve this, although it's more difficult than the method you mentioned.
Start by considering the case where there are 2 floors and 2 people. The probability of at least one person exiting on our chosen floor, say floor 2, is
$$P(\geq\text{1 person on floor; 2 people and 2 floors}) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the ratio of areas produced by perpendiculars from the $3$ sides of an equilateral triangle.
A point O is inside an equilateral triangle $PQR$ and the perpendiculars $OL,OM,\text{and } ON$ are drawn to the sides $PQ,QR,\text{and } RP$ respectively. The ratios of lengths of the perpendiculars $OL:OM:ON \text{ i... | Here's a different approach.
Well-known Fact. The sum of the distances from an interior point to the sides of an equilateral triangle is equal to the height of that triangle.
Proof (in case you've never seen it). Using our current triangle, writing $s$ for its side-length, and $h$ for its height:
$$\frac{1}{2} s h =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \... | Your last expression is completely wrong. This question is not solved in your way.
HINT:
$$\frac{x^{1/3}-x^{1/5}}{x^{1/3}+x^{1/5}}=\frac{1-x^{(1/5)-(1/3)}}{1+x^{(1/5)-(1/3)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Limits of square root $$\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x + \sqrt x} }}-\sqrt x\right) $$
(original screenshot)
Compute the limit
Can you please help me out with this limit problem
| Here's a comparatively clean way to do it: $$ \sqrt{x+\sqrt x}-\sqrt x\le\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}\le\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}-\sqrt{x} $$
Now, let $u=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$. Then $$u^2=x+u\implies u=\frac{1+\sqrt{1+4x}}{2}=\frac12+\sqrt{\frac14+x}$$
(Note that $u$ is strictly po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the ma... | Using parametric form, $\displaystyle x-2=\cos\phi,3\left(y+\frac13\right)=\sin\phi$
$\displaystyle\implies x=2+\cos\phi, 3y=\sin\phi-1 $
$\displaystyle\implies\frac{4x-9y}2=\frac{4(2+\cos\phi)-3(\sin\phi-1)}2=\frac{11+4\cos\phi-3\sin\phi}2$
Can you prove $$-\sqrt{a^2+b^2}\le a\cos \theta+b\sin\theta\le \sqrt{a^2+b^2}?... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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So close yet so far Finding $\int \frac {\sec x \tan x}{3x+5} dx$ Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts. below is his work.
How can I calculate
$$
\int {\sec\left(x... | Finally figured this one out.
${let: 3x+5= \sec \theta, x=\frac {\sec\theta+5}{3}, dx=\frac{1}{3}\sec\theta\tan\theta d\theta}$
then we have $${\frac {1}{3}\int\frac{\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}{\sec\theta}\sec\theta\tan\theta d\theta=}$$ $${\frac{1}{3}\int\cos\theta {\sec\frac{\sec\theta+5}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\l... | It is not true that
$\lfloor x \rfloor
\to x
$.
What is true is that
$\frac{\lfloor x \rfloor}{x}
\to 1
$.
All that you need for
(1) is that
$\frac{\ln x}{x}
\to 0
$.
For (2),
note that
$x^2+x
\le
\lfloor x^2+x \rfloor
<
x^2+x+1
$.
Therefore
$\lfloor x^2+x \rfloor
= x^2+x+c
$
where
$0 \le c < 1
$.
You can then write
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers. Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers
and
$$(1+1/x)(1+1/y)(1+1/z) = 3.$$
I know
$(x+1)(y+1)(z+1) = 3xyz$ which is no big deal. I can't move forward ... | Let $x$ be the smallest of the three.
Case 1: $x=1$.
Then
$$ y=\frac{2\,z+2}{z-2} = 2 + \frac{6}{z-2}$$
So $z-2$ should divide $6$
So the possibilities are $z=3,4,5,8$ If we want $x \le y \le z$, we can eliminate two of them.
case 2: $x=2$
Then
$$ y = \frac{z+1}{z-1} = 1 + \frac{2}{z-1}$$
So $z-1$ should divide 2, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$ Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$.
I already proved that it converges but I can't find the sum, Although I think it should be $1+\frac{1}{\sqrt3 +1}$, from calculations.
| Start with following infinite product expansion which is valid for $|q| < 1$,
$$\prod_{n=0}^\infty \left(1 + q^{2^n}\right) = \sum_{n=0}^\infty q^n = \frac{1}{1-q}$$
Taking logarithm and apply $q\frac{\partial}{\partial q}$ on both sides, one get
$$\sum_{n=0}^\infty \frac{2^n q^{2^n}}{1 + q^{2^n}} = \frac{q}{1-q}$$
Sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
How to find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$ Using which way can we find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$? I have used binomial theorem but failed to find an answer for it.
| You can decompose
$$
\frac{1}{(x+3)(x-2)^2}=\frac{A}{1+x/3}+\frac{B}{1-x/2}+\frac{C}{(1-x/2)^2}
$$
and it's just a matter of finding $A$, $B$ and $C$. For $x=0$, $x=1$ and $x=-1$, we get
$$
\begin{cases}
\frac{1}{12}=A+B+C\\
\frac{1}{4}=\frac{3}{4}A+2B+4C\\
\frac{1}{18}=\frac{3}{2}A+\frac{2}{3}B+\frac{4}{9}C
\end{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$ Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$
$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$
Now Let $r(x) = ax^2+bx+c$
So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c.................... | Your method is correct. You are essentially computing the first few terms of the Taylor power series expansion of $\,(x+1)^n$ at $\,x=1$ (or, equivalenly, of $\,(z+2)^n$ at $\,z =0,\,$ where $\,z = x-1).\,$ Alternatively, one may compute the expansion by using the Binomial Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How prove this $f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$ Let $x,y>0$,and prove or disprove
$$f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$$
I know
$$f(1,1)=(4-7)(1+1+1)+9=0$$http://www.wolframalpha.com/input/?i=%28%283x%2By-7%29%281%2Bx%5E2%2Bxy%29%2B9%29
since I have use nice methods solve follow inequality
let $a,b>0$,show that
$$9a^2... | Here is something strange, for which I don't really have an entirely rigorous justification, but which works and for which the calcluations seem simpler.
Firstly, (as Christian Blatter's answer does) note that for $x\to\infty$ or $y \to \infty$, we have $f(x,y)\to\infty$ as well. So $f(x, y)$ attains its minimum either... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An equivalent of : $f(x)=\int_0^{+\infty}\frac{e^{-xt}}{(1+t^3)^{1/3}} dt$ $\forall\ x\ \in\ \left]0,+\infty\right[\ $ we put:
$$
{\rm f}\left(x\right)
=
\int_{0}^{\infty}{{\rm e}^{-xt} \over \left(1 + t^{3}\right)^{1/3}}\,{\rm d}t
$$
The question is the question is to find an equivalent of $\,\,{\rm f}\left(x\right)$ ... | Mathematica gives
$$
\frac{G_{1,4}^{4,1}\left(\frac{s^3}{27}|
\begin{array}{c}
\frac{2}{3} \\
0,0,\frac{1}{3},\frac{2}{3} \\
\end{array}
\right)}{2 \sqrt{3} \pi \Gamma \left(\frac{1}{3}\right)}$$ for the integral, and
expanding this in a power series at $0,$ gets
$$
\frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is $\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$? I am stuck on this one; Its a sum and don't know how to calculate the denominator.
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$$
| Stolz-Cesaro lemma is applied twice successively:
$$L=\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}} =\lim_{n\to\infty} \frac{1}{n+2-(n+1)} \frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\lim_{n\to\infty}\frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = ... | A standard result (http://en.wikipedia.org/wiki/Matrix_determinant_lemma) is $\det(I+AB) = \det(I+BA)$.
Since the matrix above can be written as $I+ e e^T$, where $e$ is a vector of ones, we have $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove or disprove the implication: Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$
$ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain.
Does anyone have a suggestion?
Thank you very much!
| I will follow the method suggested by Blue in his comment to show that the implication does not hold.
The sides $a$, $b$, $c$ are proportional to $\sin A$, $\sin B$ and $\sin C$, respectively, by the sine law, so the equation is equivalent to $S = 0$, where
$$ S = \sin^2 A \tan(B-C) + \sin^2 B \tan(C-A) + \sin^2 C \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
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Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1... | Using the root test, we can prove easily,
note that,
$$\limsup_{n\to \infty} \sqrt[n]a_n=\lim_{n\to \infty}\sqrt[2n]{\frac 1 {2^n}}=\frac 1 {\sqrt 2}<1 $$
Hence the convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/634558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating a trigonometric integral by means of contour $\int_0^{\pi} \frac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta$ I am studying for a qualifying exam, and this contour integral is getting pretty messy:
$\displaystyle I = \int_0^{\pi} \dfrac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta $
I first notice that the integran... | Note that $\sin 4\theta$ is an odd function, so we can simplify to
$$\begin{align}
I &= \frac12 \int_{-\pi}^\pi \frac{e^{4i\theta}}{1 + \cos^2\theta}\,d\theta\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^4}{1 + \left(\frac{z+z^{-1}}{2}\right)^2}\,\frac{dz}{z}\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
| Maybe this will help you to see things clearer, but it is the same concept as @NasuSama..
$$\underbrace{\frac14x^2y(x-1)^3}_\text{one expression}-\underbrace{\frac54xy(x-1)^2}_\text{another expression}$$
Let $z=(x-1)$, your expression now becomes:
$$\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2$$
You can easily see now that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Supremum and infimum of a set of numbers $\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$ Let $n \geq 3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_1,...,a_n)$ of positive numbers. Find the supremum and infimum of the set of numbers $$\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$$ where... | (Write $S(a)=\sum \dots$ for any such sum)
Plug in $q,q^2,\dots,q^n$ for a $q > 0$ and get
$$
S(a) = (n-2)\frac{1}{1+q+q^2}+\frac{q^{n-1}}{q^{n-1}+q^n+q^1}+\frac{q^{n}}{q^{n}+q^1+q^2}
$$
let $q \rightarrow 0,\infty$ yield inf $\leq 1$ and $\sup\geq n-2$
Not let $a_1,\dots,a_n$ be any such sequence, let $A:=\sum a_i$
1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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In how many ways can you arrange a circle of partners so that no partners are touching? There are a lot of similar questions to this but none that is quite the same so I figured I would ask a new question. The problem is you have a group of people that came in pairs, in how many ways can the $N$ people be arranged in a... | Here is the calculation for $8:$
Seat one person of the first couple. This fixes the rotation of the circle. Seat the second person of the first couple. The remaining seats can be in groups of $1+5 (2$ ways), $2+4 (2$ ways), or $3+3, (1$ way).
From $1+5$ we can go to $4$ in $4$ ways, $1+3$ in $6$ ways, $2+2$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Calculate a multiple sum of inverse integers. The question is to calculate a following sum:
\begin{equation}
{\mathcal S}_p(n) :=\sum\limits_{1\le j_1 < j_2 < \dots <j_p \le n-1} \prod\limits_{q=1}^p \frac{1}{n-j_q}
\end{equation}
for $p=1,2,..$ and $n\ge 1$.
From purely combinatorial reasoning we have:
\begin{eqnarray... | Expanding the right hand side of the identity given by achille hui we get ``a compact'' expression for the sum:
\begin{eqnarray}
{\mathcal S}_p(n) &=& \sum\limits_{m=1}^p \frac{(-1)^{-m+p} }{m!} \sum\limits_{p_1+p_2+\dots+p_m=p} \prod\limits_{q=1}^m \frac{H_{n-1}^{(p_q)}}{p_q} \\
&=& \frac{1}{1!} \frac{(-1)^{p-1}}{p} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/641667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to get the theta in this projectile equation? The equation for the Parabolic Trajectory is
$$y = x\tan{\theta}- \frac {gx^2}{2V_0^2\cos^2{\theta}}$$
It will be easy to get any variables if there's an angle, but how I can solve if the missing one is the angle? What will be the formula to get the angle? In quadratic ... | Note that due to the Pythagorean Trigonometric Identity : $1 + \tan^2{\theta} = \sec^2{\theta}$,
$$\begin{align}12.5 &= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2\cos^2\theta}\\
&= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2}\sec^2\theta \\
&= 6.7\tan{\theta} - \frac{(9.81)(6.7)^2}{(2)(24)^2}(1 + \tan^2\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc$ How to prove the following equality?
$$(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc$$
I did it
$$\begin{aligned}
a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc &=a^2(b + c) + bc(b + c) + a(b^2 + c^2 +2bc)\\&=a^2(b + c) + bc(b+ c) +a(b + c)^2\\&=(b + c)(a^2 + bc + ab + ac)\\&... | Set $\displaystyle a+b+c=x$
So, we have $\displaystyle (x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+(ab+bc+ca)x-abc$
Now, $\displaystyle x^3-x^2(a+b+c)=x^2(x-\overline{a+b+c})=x^2\cdot0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as sayi... | $$10^x = (2\cdot 5)^x = 2^x\cdot 5^x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
The value of (x-y) of quadrilateral inscribed in circle A quadrilateral that has consecutive sides of length $70,90,110,130$ is inscribed in a circle and also has a circle inscribed in it . The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of $x$ and $y$ . If $y\g... | Call the quadrilateral $ABCD$, with sides $AB=70, BC=90, CD=130, DA=110$, i.e. in the order
$$\begin{array}{ccc}
A&\leftarrow70\rightarrow&B\\
\begin{array}{c}\uparrow\\110\\\downarrow\end{array}&O&\begin{array}{c}\uparrow\\90\\\downarrow\end{array}\\
D&E&C\\
&\leftarrow130\rightarrow&\\
\end{array}$$
For $ABCD$ to be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A moving point has its distance from (1,3) always one-third of its distance from (8,2). Find the equation of its Locus. A moving point has its distance from (1,3) always one-third of its distance from (8,2).
Find the equation of its Locus.
My equation displays a circle formed by the loci, I don't know if it's right. P... | $$\begin{align*}3^2\left[(x-1)^2+(y-3)^2\right] =& (x-8)^2 + (y-2)^2\\
9\left(x^2-2x+y^2-6y+10\right) =& x^2-16x+y^2-4x+68\\
8x^2-2x+8y^2-50y+22 =& 0\\
x^2-\frac14x+y^2-\frac{25}4y+\frac{11}4=&0\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\frac{225}{32}\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation for limits equality. $$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{1}{x}}} = \exp \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{a^x} + {b^x}}}{2} - 1}}{x}} \right)$$
I am familiar with the "taking the exponent trick" which bring us to:
$$\\exp \left( {\mat... | \begin{align*}
\left(\frac{a^x+b^x}{2}\right)^{\frac 1 x}=\exp\left(\frac 1 x{\color{\red}\ln}\left(\frac{a^x+b^x}{2}\right)\right).
\end{align*}
Now, as $x\to0$, $(a^x+b^x)/2\to (a^0+b^0)/2=(1+1)/2=1$ (assuming that $a,b\neq0$). Also, $\ln(1)=0=1-1$, so that $\lim_{y\to1}\ln y=\lim_{y\to1}(y-1)$, since both functions ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this outcome change depending on infinity Why does the outcome of the limit as x approaches infinity of $$\sqrt{x^2+2x}- \sqrt{x^2-2x}$$
which simplifies to
$$\dfrac {4x}{x \left(\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}\right) }= \dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$
change depending on wh... | $$\sqrt{x^2} = |x|$$
So when $x$ tends to $- \infty$, $\sqrt{x^2} = -x$.
Hence then answer would be:
$$\frac{4}{-2} = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/659725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Cubic spline that interpolates $f$ Given the following cubic spline that interpolates $f$:
$$\left\{\begin{matrix}
(x+3)^3-9(x+3)^2+22(x+3)-10 & ,-3 \leq x < -1\\
(x+1)^3-3(x+1)^2-2(x+1)+6 & , -1\leq x <0\\
ax^3+bx^2+cx+d & ,0\leq x <2\\
(x-2)^3+6(x-2)^2+7(x-2) & ,2 \leq x \leq 3
\end{matrix}\right.$$
It's also giv... | HINT: Use the fact that $f(x)$ and $f'(x)$ must be continuous at $x=0$ and $x=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine for what value of $x$ the series converges $∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $ Determine for what value of $x$ the series converges
$∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $
Here is what I got
Using the ratiotest, I got
$D_n =\frac{\frac{(3^{n+1}+(-2)^{n+1} )}{n+1}(x+1)^{n+1}}{\frac {(3^n+(-2)^n )... | $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}$
$=lim_{n->\infty} \frac{3^{n+1}n(1+(-2/3)^{n+1} )}{3^nn(1+1/n)(1+(-2/3)^n)}$
$=lim_{n->\infty} \frac{3(1+(-2/3)^{n+1} )}{(1+1/n)(1+(-2/3)^n)}=\frac{3(1+0)}{1+0)(1+0)}=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/668404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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High degree polynom divisibility The problem is:
Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$
I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.
| It is not hard to see that $x^5-1$ divides $x^{55}-1$.
Thus $(x-1)(x^4+x^3+x^2+x+1)$ divides $(x^{11}-1)(x^{44}+x^{33}+x^{22}+x^{11}+1)$ and therefore
$x^4+x^3+x^2+x+1$ divides $(x^{10}+x^9+\cdots +x+1)( x^{44}+x^{33}+x^{22}+x^{11}+1)$.
But $x^4+x^3+x^2+x+1$ and $x^{10}+x^9+\cdots +x+1$ are relatively prime (use the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/670555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Showing that $\lim\limits_{n \to\infty} z_n = A$ implies $\lim\limits_{n \to\infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A$ In what follows let all values be in $\mathbb{C}$. I'm trying to show that if
$$\lim z_n = A,$$
that then
$$
\lim_{n \to \infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A.
$$
For ease of no... | In step 1., "through repeated applications of the triangle inequality we have", more interestingly, that
$$
\left| A - s_n \right|=\left|\frac1n\sum_{k=1}^n(A - z_k) \right| \leqslant\frac1n\sum_{k=1}^n\left| A - z_k \right| .
$$
Then one can proceed.
Later on, in step 4., you fall prey to the common fallacy that sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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L.C.M. and H.C.F. problem Let $a$ and $b$ be natural numbers with $ab>2$. Suppose that the sum of their highest common factor and least common multiple is divisible by $a+b$. Prove that the quotient is at most $\frac{a+b}4$. When is this quotient exactly equal to $\frac{a+b}4$.
| Let $g=\gcd(a,b)$ and $a=gA$, $b=gB$. Then, $\mathrm{lcm}(a,b)=gAB$ and we have
$$\frac{\mathrm{lcm}(a,b)+\gcd(a,b)}{a+b} = \frac{AB+1}{A+B}$$
Without loss of generality, we can assume $A\geq B$. Let's start with two simple cases:
*
*$A=B$; which implies $A=B=1$ (and $a=b=g$) due to $\gcd(A,B)=1$. The quotient is eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Agai... | $$\Longleftrightarrow (a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \le 27a^2b^2c^2$$
Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $$(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$$.
If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to prove the convergence of a series of prime numbers I have a bit of a problem proving that the series:
$$
\sum_{p\leq x} \frac{p\ln\left(p\right)}{x^2}
$$
where the sum is extended over all prime numbers, converges to 0.5.
Any ideas?
Thanks in advance,
Kijn
| A standard method is to write the sum as a sum over all positive integers $\leqslant x$, multiplying the term with $\pi(n) - \pi(n-1)$, where $\pi$ is the prime counting function, to annihilate the terms for composite $n$, and then rearrange:
$$\begin{align}
\sum_{p\leqslant x} p\log p &= \sum_{n\leqslant x} \bigl(\pi(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this simple trignometric problem? So this is the question that was given in a textbook and i attempted to win from the book which was saying i was wrong?
If $$\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{4}$$ then what is te value of$$\frac{\tan^2\theta + 1}{\tan^2\theta - 1}$$
so ... | Observe that $\dfrac a b=\dfrac c d$ doesn't imply $a=c$ and $b=d$.
Now you may proceed as follows,
\begin{align}
&\dfrac{\sin \theta +\cos\theta}{\sin \theta -\cos\theta}=\dfrac54
\\\implies&4\sin \theta +4\cos\theta=5\sin \theta-5\cos\theta
\\\implies&\sin\theta=9\cos\theta.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove this inequality $\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}\;?$
Let $a,b,c\ge 0$, and assume $a+b+c=3$.
I'd like to show that
$$\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$
My try: Use Hölder inequality
$$\left(\sum_{cy... | I'm going to use two facts. The first one can be obtained as following:
$$\begin{align}
ab + bc + ca &\leqslant a^2 + b^2 + c^2, \\
3(ab + bc + ca) &\leqslant (a + b + c)^2 = 9, \\
ab + bc + ca &\leqslant 3.
\end{align}$$
And the second one:
$$\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/678619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Roots of a certain family of polynomials Let $\xi \in (0,1)$ and $p$ be a positive non-zero integer.Show that in the limit $p \rightarrow \infty$ the following algebraic equation:
\begin{equation}
\frac{x^{p+2} }{\xi} + x^{p+1} + x^p = (-\xi)^p
\end{equation}
has roots:
\begin{equation}
\left\{ x_\xi, \bar{x}_\xi, \lef... | I do not see why a polynomial of degree $p+2$ would have $p+3$ roots...
This is a perturbation problem, essentially you have
$q(x)x^p-(-ξ)^p=0$ with the quadratic polynomial $q(x)=\frac1ξx^2+x+1$.
The quadratic polynomial $0=4ξq(x)=(2x+ξ)^2+4ξ-ξ^2$ has roots
$$x=\frac12\left(-ξ\pm\sqrt{ξ^2-4ξ}\right)=-\frac12ξ\pm i\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Triangle with sides a,b,c and area There is a triangle with sides $a,b,c$, then his area is:
a) $A$ not greater than $\frac{1}{2}(a^2-ab+b^2)$
b) $B$ greater than $\frac{1}{2}(a^2-ab+b^2)$
c) $C$ greater than $\frac{1}{2}(a^2-\frac{1}{2}ab+b^2)$
d) $D$ less than $\frac{1}{2}(a^2+ab+b^2)$
I am thinking about applying t... | HINT:
The area is $\displaystyle\triangle= \frac12ab\sin\gamma$
As $\displaystyle 0<\sin\gamma\le1, ab\sin\gamma\le ab\iff -ab\sin\gamma\ge -ab$
$\displaystyle\implies A-\triangle=\frac{a^2-ab+b^2-ab\sin\gamma}2\ge\frac{a^2-ab+b^2-ab}2$ which is $\displaystyle\frac{(a-b)^2}2\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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