Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been... | Ok, here we go.
Let $f(a,b,c,\lambda)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a^3+b^3+c^3}{3}+\lambda(a+b+c-3)$.
$\nabla f=0$ means:
*
*$\frac{\partial f}{\partial a}=-a^2-\frac{1}{a^2}+\lambda=0$
*$\frac{\partial f}{\partial b}=-b^2-\frac{1}{b^2}+\lambda=0$
*$\frac{\partial f}{\partial c}=-c^2-\frac{1}{c^2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How to prove: $\left(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}}}-1\right)^{4}=5$? Question:
show that: the beautiful ${\tt sqrt}$-identity:
$$
\left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\
+\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4}
=5
$$
Can you someone... | [This is a paraphrase on JimmyK4542's elegant answer earlier]
Let $$b=-5^{\frac 14}$$
Then the long expression under the long square root sign becomes an arithmetico-geometric series:
$$k=4+3b+2b^2+b^3$$
Multiplying by $b$:
$$\begin{align}
b\cdot k&=\quad \qquad 4b+3b^2+2b^3+b^4\\
&=\quad \qquad 4b+3b^2+2b^3+5 \end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
By plugging $p=1-q$, into the $3$ equations show that $x=y=z$ By plugging $p=1-q$, into the 3 equations:
$$\begin{cases} z=py+qx \\ x=pz+qy \\ y=px+qz \end{cases}$$ show that $\boxed{x=y=z}$
This is from the final part of question 7 in this STEP paper,
and is following the advice of another students solution , only i ... | The equations are equivalent to:
$$\begin{cases} z=q(x-y)+y \\ x=q(y-z)+z \\ y=q(z-x)+x \end{cases}$$
Substituting $z$ on the third equation we get:
$$y = -q^2(y-x) + q(y-x) + x \Rightarrow (y-x)(q^2-q+1) = 0$$
Similarly, $(x-z)(q^2-q+1) = 0$ and $(z-y)(q^2-q+1) = 0$.
So either $x = y = z$ or $q^2 -q +1 = 0$, but there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivati... | @Tunk-Fey makes a good point and the two expressions are not necessarily equal as they can differ by a constant. However, surprisingly, the two sides are equal in this case. Just ''rationalize'' the numerator of the expression inside the logarithm of the right side. To be more precise, perform the following computation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve $(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$
Solve $$(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$$
I have a very similar problem like this on my homework, and I have no clue how to set it up or even start. How could I set this up?
| $$(x-3) \frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=\frac{-y}{x-3} \Rightarrow -\frac{dy}{y}=\frac{dx}{x-3} \Rightarrow \int \left ( \frac{-1}{y} \right)dy=\int \frac{1}{x-3} dx \\ \Rightarrow -\ln |y| =\ln |x-3|+c \Rightarrow e^{-\ln |y|}=e^{\ln |x-3|+c} \Rightarrow \frac{1}{|y|}=C |x-3| \Rightarrow |y|=\frac{c'}{|x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$ Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.
$$
\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-... | Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following:
$$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot \sin{(x)}=\sin{(x)}$$
$$\cos{(2 \pi+x)}=\cos{(2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Sum of the series $\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$ How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$
I think the sum can be convert... | Using generalized Binomial Theorem, $$(1+z)^n=1+nz+\frac{n(n-1)}{2!}z^2+\frac{n(n-1)(n-2)}{3!}z^3+\cdots$$ for $|z|<1$
$\displaystyle S=\sum_{r=1}^\infty\dfrac{2\cdot6\cdots(4r-2)}{5\cdot10\cdots5(r+1)}$
Observe that there are $r$ terms in the numerator unlike the denominator which has $r+1$ terms
So, we write multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/900687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane. I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equa... | Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align}
x \cos\theta + y \sin\theta - p &= 0 \\
x \cos\phi + y \sin\phi - q &= 0 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
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How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with d... | By a tedious expansion of $P(x)=(x-r_1)(x-r_2)\ldots$ that other answers have covered or by using Vieta's formulas, you can find that
$$P(x)=x^4+\left[-\sin\left(\frac{\pi}{24}\right)-\sin\left(\frac{7\pi}{24}\right)-\sin\left(\frac{11\pi}{24}\right)-\sin\left(\frac{13\pi}{24}\right)\right]x^3$$
$$\ldots+\left[\sin\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$? Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail.
| Here is another Feynman's way to evaluate the integral. Set $x=\frac1t$, then
$$
\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\int_0^1\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Now consider
$$
\mathcal{I}(\alpha)=\int_0^1t^\alpha\cdot\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Hence
\begin{align}
\frac{d^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$ If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$.
| Hint: Just use $\sin^2\theta = 1 - \cos^2\theta$, get the eqn in $(a-b)^2 = 0$ form.
Then, simply use the fact that $\sec\theta = \frac1{\cos\theta}$
Spoiler :
!> $$\sin^2 \theta + 2\cos\theta - 2 = 0 \\
\implies 1 - \cos^2\theta + 2\cos\theta -2 = 0\\
\implies -(\cos^2\theta - 2 \cos\theta + 1^2) = 0\\
\implies (\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Algebra equation with functions, constraints and a graph. Consider the function $f:[1,3]\to\mathbf R$, $f(x)=-x^4+8x^3+ax^2+bx+d$, where $a$, $b$, $d$ are real constants. Find the values of $d$ for which $f$ has 3 stationary points between $x=1$ and $x=3$ and $f(1)=f(3)=0$.
| $f(1)=f(3)=0$, then $7+a+b+d=0$ and $135+9a+3b+d=0$. We know that $a=(d-57)/3$ and $b=12-4d/3$. Since there're three roots for the first derivative, by Rolle's theorem there are two roots for second derivative between $[1,3]$. Namely, $-12x^2+48x+2a=0$ has two roots in $[1,3]$, that is $2\pm \sqrt{4+\frac{a}{6}}\in[1,3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable... | We have
$$\begin{align}
\frac3{a+b+c+1} &- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\
&= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\
&= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\
&= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \\
&\le \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solve a limit with radicals I don't know how to solve this limit. What should I do?
$$
\lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
$$
Thank you!!
| As David H commented, rationalizing the denominator would be a good starting point.
Now, if you know Taylor series, the problem starts to be simple since, around $x=0$, you have $$\sqrt{x^3+2x+1}=1+x+O\left(x^2\right)$$ $$\sqrt{x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{4x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
| $\sin{A}+\sin{B}+\sin{C}=2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+\sin{C}\le 2\sin{\frac{A+B}{2}}+\sin{(\frac{5\pi}{4}-(A+B))}.$ Equality holds when $A=B$. Thus we consider the function $$f(x)=2\sin{x}+\sin{\left(\frac{5\pi}{4}-2x\right)}=2\sin{x}+\sin{\left(2x-\frac{\pi}{4}\right)}.$$
defined on $(0,\frac{5\pi}{8})$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to find the sum of sequence $ 1+4+4^2+\cdots+4^{X+Y} $? I see the following sequence and it's:
$$h=1+4+4^2+\cdots+4^{X+Y}=\frac{4^{X+Y+1}-1}{4-1}$$
how we get this sequence?
I know this is a primary question but I confused :)
| This is an example of a geometric series. Let's say we try to sum:
$$S=1+r+r^2+r^3+\dotsb+r^n$$
In your example, $r=4$, and $n=X+Y$.
There is a well known "trick" for solving this. Multiply by $r$:
$$Sr=r+r^2+r^3+\dotsb+r^n+r^{n+1}$$
Notice how this is the same thing as $S$, except without the $1$ at front and with an ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Hard Definite integral involving the Zeta function Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to... | Note that $\frac{1-x}{1-x^6}=\sum_{k=0}^\infty (x^{6k}-x^{6k+1})$. And the integration $\int_0^1 x^n \ln{x}^4=\partial_n^4 \int_0^1 x^n dx=\frac{24}{(n+1)^5}$. We have
$$LHS = 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
Use the discrete Fourier, and denote $\xi=\exp(i\frac{\pi}{3})$, $\xi_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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How many ways to generate unique multiplication result from given set? From set {2, 2, 3, 5}, I can have 8 ways to generate unique multiplication result, which:
- two number multiplication: 2*2, 2*3, 2*5, 3*5
- three number multiplication: 2*2*3, 2*2*5, 2*3*5
- four number multiplication: 2*2*3*5
Then, how many uniqu... | Every positive integer has a unique prime factorization. So your products will all be of the form
$$ 11^a\cdot 7^b\cdot 5^c\cdot 3^d\cdot 2^e $$
where $0\le a,b,c,d,e$ and $a,b\le 5;\; c\le 4;\; d,e\le 3$. Which gives you
$$ 6\cdot 6\cdot 5\cdot 4\cdot 4 = 2880 $$
possible numbers of this form. But that includes the ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the equation of normal line to the graph of given function
Give the equation of the normal line to the graph of
$$y = 2x \sqrt{x^2+8} + 2$$
at the point $(0,2)$
What I've done so far is:
Taken the derivative and got
$$(2x^2)/\sqrt{ x^2+8} + 2\sqrt{x^2+8}$$
I have no idea if this is right, it was pretty har... | $$f(x)=2x \sqrt{x^2+8}+2 \Rightarrow f'(x)=2\sqrt{x^2+8}+\frac{x}{\sqrt{x^2+8}}2x=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{2(x^2+8)+2x^2}{\sqrt{x^2}8}=\frac{4x^2+16}{\sqrt{x^2+8}}$$
At the point $(0,2)$, $f'(0)=\frac{16}{\sqrt{8}}=\frac{16}{2 \sqrt{2}}=\frac{8}{\sqrt{2}}$.
The slope of the normal line is $\frac{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ if $a^2+b^2+c^2=1$
Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$
Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM be... | Write $x=a^2$, $y=b^2$, and $z=c^2$. Then, $x+y+z=1$. A first consequence of this is
$$
(1+z)\geq 2\sqrt{xy+z}.\tag{*}
$$
This is true because
$$
(1+z)^2-(2\sqrt{xy+z})^2=(1+z)^2-4z-4xy\\
=(1-z)^2-4xy=(x+y)^2-4xy=(x-y)^2\geq 0.
$$
Analogous to (*), we also have
$$
(1+y)\geq2\sqrt{xz+y},\quad (1+x)\geq 2\sqrt{yz+x}.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4... | $$f(x,y,z)=\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
$$\nabla f= \frac{(3 (2+5 y) (5+2 z) (x^2-y z))}{(x^2 y z)} \hat e_x+\frac{(-((2 x-15 y^2) (3 x+z) (5+2 z))}{(x y^2 z))}\hat e_y+\frac{(-((x+3 y) (2+5 y) (15 x-2 z^2))}{(x y z^2))}\hat e_z$$
We get 4 real solutions:
$$(5/6,-5/18,-5/2),(1,-\sqrt{2/15},\sqrt{15/2}),(1,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How did Ulam and Neumann find this solution? In the book "Chaos, Fractals and Noise - Stochastic Aspects of
Dynamics" from Lasota and Mackey the operator $P: L^1[0,1] \to L^1[0,1]$
$$ (Pf)(x) = \frac{1}{4\sqrt{1-x}} \left[ f\left(\frac{1}{2}\left(1-\sqrt{1-x}\right)\right) + f\left(\frac{1}{2}\left(1+\sqrt{1-x}\right)... | I don't have the reference for the derivation, but I can proof, that $f^*$ is indeed an eigenfunction of $P$ with eigenvalue $1$, i.e. $Pf^*=f^*$.
Proof
\begin{align}
(Pf^*)(x)&= \frac{1}{4\sqrt{1-x}} \Bigl( f\Bigl(\frac12 (1-\sqrt{1-x})\Bigr)+
f\Bigl(\frac12 (1+\sqrt{1-x})\Bigr)\Bigr)
\\
&=\frac{1}{4\sqrt{1-x}} \Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$ I'm trying to find a closed form of this sum:
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}.\tag{1}$$
WolframAlpha gives a large expressions containing multiple generalized hypergeomet... | Another possible closed form of $S$ is the following. It containts also a generalized hypergeometric function, but just one.
$$S = \frac{\sqrt{\pi}}{648} {_6F_5}\left(\begin{array}c\ 1,\frac32,\frac32,\frac32,\frac32,\frac32\\2,\frac52,\frac52,\frac52,\frac52\end{array}\middle|\,\frac14\right).$$
WolframAlpha's simplif... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
| $$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
Start by simplifying the numerator. Specifically, add the two fractions.
$$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}=\frac{\frac{3x}{6}+\frac{2y}{6}}{6x+4y}=\frac{\frac{3x+2y}{6}}{6x+4y}$$
Then, since the fraction bar means division, you have:
$$\frac{\frac{3x+2y}{6}}{6x+4y}=\frac{... | {
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"question_score": "3",
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equat... | It might help to go back to the basic definition of multiplication as repeated addition:
$$2 \cdot 5 = 2 + 2 + 2 + 2 + 2 = 10$$
and that division is the opposite of multiplication:
$$\frac{10}{2} \mapsto 10 \underbrace{{}- 2 - 2 - 2 - 2 - 2}_{\text{5 times}} = 0.$$
So $\frac{10}{2}=5$, remainder $0$
Also, for $\frac{11... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of $\psi\left(\frac{1}{2}\right)$ I apologise if this is a dumb question, but I have trouble deriving $\displaystyle\psi\left(\frac{1}{2}\right)=-\gamma-2\ln{2}$. I have tried the following.
\begin{align}
\psi\left(\frac{1}{2}\right)
&=\lim_{N\to\infty}\left[-\gamma-2+\sum^N_{k=1}\left(\frac{1}{k}-\frac{1}{k+1/2}... | One could evaluate the limit as follows. Set $N=2 n$:
$$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1}=\lim_{n\to \infty} \sum_{k=n}^{2n} \frac{2}{2k+1}=\lim_{n\to \infty} 2\cdot\left(\sum_{k=0}^{2n} \frac{1}{2k+1}-\sum_{k=0}^{n-1} \frac{1}{2k+1}\right)=2\cdot\lim_{n\to \infty} \left(\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-... | $$M=1+3+5+7+...+(2n-3)+(2n-1)\\M=(2n-1)+(2n-3)+...+7+5+3+1\\M+M=(1+2n-1)+(3+2n-3)+(5+2n-5)+....(2n-3+3)+(2n-1+1)\\n-term\\M+M=n(2n)\\2M=2n^2\\M=n^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a solution to Laplace's equation that satisfies polar coordinates and show that any solution produces perpendicular lines. 7a. Find a solution of Laplace's equations $u_{xx}+u_{yy}=0$ of the form $u(x,y)=Ax^2+Bxy+Cy^2 (A^2+B^2+C^2 \neq 0)$ which satisfies the boundary condition $u(cos ( \theta),sin( \theta))=cos(2... | The graph of the function $u$ is the subset $\{(x,y,u(x,y)):x,y\in\Bbb R\}\subset\Bbb R^3$. The points on the $xy$-plane are precisely those with $z$-coordinate $0$. Hence the intersection with our graph with the $xy$-plane is the set of all points $(x,y)$ with $u(x,y)=0$. You have $u(x,y)=x^2+2xy-y^2=0$. Can you solve... | {
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"source": "stackexchange",
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Inequality $\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$ with weird condition I want to prove the following inequality:
$$\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$$
Where $a,b,c$ are positive reals and with the horrible condi... | The condition $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2 \iff abc = a+b+c+2$, so as you deduced through two successive substitutions, we can have
$$a = \frac{u+v}w, \, b = \frac{v+w}u, \, c = \frac{w+u}v$$ to get
$$\sum_{cyc} \sqrt{\frac{u+v}w} \ge 2 \sum_{cyc} \sqrt{\frac{w}{u+v}} \iff \sum_{cyc} \frac{u+v-2w}{\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor ... | If $|x|<1$ :
The series converges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\frac{x+1}{(1-x)^3}+\frac{x^2+4x+1}{(1-x)^2(x^2-2x+1)}=\frac{4 x + 2}{x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1}$
If $|x|>1$ :
The series diverges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\sum_{n=1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$
x^3 + 1 = 2\sqrt[3]{{2x - 1}}
$$
I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go ... | We have $$x^3+1=2(2x-1)^{1/3}\iff x^3=2(2x-1)^{1/3} -1.$$
Here, setting $y=(2x-1)^{1/3}$ gives us
$$y^3=2x-1 \ \ \text{and}\ \ x^3=2y-1.$$
Hence, we have
$$\begin{align}y^3-x^3=(2x-1)-(2y-1)&\Rightarrow (y-x)(y^2+yx+x^2)=2(x-y)\\&\Rightarrow (y-x)(y^2+yx+x^2+2)=0\\&\Rightarrow (y-x)\{(x+(y/2))^2 + (3/4)y^2+2\}=0\\&\R... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus Limits Problem L'Hopital's Rule is not allowed.
Question 1:
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \ ?$$
I tried to cross multiply $\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}$with $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ and I got $x+2$ on both LHS and RHS thus I conclude
$\frac{\sqrt{3+x}+1}{\sqrt{6+x}+... | Answer 1:
Set $x=-2+y$
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \lim_{y\to 0} \frac{\sqrt{4+y}-2}{\sqrt{1+y}-1}$$
$$=\lim_{y\to 0} \frac{(y/4)+O(y^2)}{(y/2)+O(y^2)}=\lim_{y\to 0} \frac{y/4}{y/2}=(1/2)$$
Answer 2:
Set $x=\pi+y$
$$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \lim_{y\to 0} ... | {
"language": "en",
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Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $... | Try expanding $2(k+1)^2 + 3(k+1)$ in order to compare it with $6k^2 + 9k$ (I.e., what you need to ensure that $6k^2 + 9k \geq 2(k+1)^2 + 3(k+1))$:
$$\begin{align} 2(k+1)^2 + 3(k+1) & = 2(k^2 + 2k + 1) + 3k + 3 \\ &= 2k^2 + 4k + 2 + 3k + 3 \\ &= 2k^2 + 7k +5\end{align}$$
Now, all that remains (while keeping in mind that... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ a... | $$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2$$
$$\implies 2 < x^2 < 6$$
$$ \implies \sqrt{2} < |x| < \sqrt{6}$$
So
$\sqrt{2} < -x < \sqrt{6} \implies -\sqrt{2} > x > -\sqrt{6}$
or
$ \implies \sqrt{2} < x < \sqrt{6}$ and its done...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation with two unknowns I have this equation to solve and I solved it but don't know if the result is correct.
$\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$
Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$
Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\p... | Let's consider the system
\begin{cases}
x+y=a\\
x^2+y^2=b
\end{cases}
which is the same as yours with $a=12/\pi$ and $b=20/\pi$. We can rewrite the second equation as
$$
(x+y)^2-2xy=b
$$
so, taking into account the first equation, it becomes
$$
2xy=a^2-b
$$
Now the problem is reduced to finding two numbers of which we ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{... | Why not directly? Since
$$\int\frac1{\sqrt x}dx=\int x^{-1/2}dx=\frac{x^{1/2}}{1/2}+C=2\sqrt x+C$$
we get that for any differentiable (and positive) function $\;f\;$:
$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
In our case,
$$f(x)=1+2x^2\;,\;\;f'(x)=4x\implies\int\frac x{\sqrt{1+2x^2}}dx=\frac14\int\frac{(1+2x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^... | Since $ \dbinom{n}{r} = 0 $ for $ r > n $, we can rewrite the sum as
$$ \text{S} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r} $$
From the Binomial Theorem, we see that the sum is the coefficient of $x^n$ in
$$f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots $$
$$ = \dfrac{(1-x)^{n+1}}{x^2-x+1} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
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"source": "stackexchange",
"question_score": "9",
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Does the series converge or diverge? I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$
converges or diverges.
$a$ is a constant number
Ratio test
$$\begin{align}
& \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n... | I don't know much about Gamma Function and summation under the integral so here is a more elementary proof.
*
*If $a>0$ we have $\ln a_n=-(\ln (1+\frac{a}{1})+\ln (1+\frac{a}{2})+\dots+\ln (1+\frac{a}{n}))$ and using
$$x\ge\ln(1+x)\ge x-\frac{x^2}{2}$$
(for $x>0$ proven by using monotonic function) we have
$$-a\su... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal. Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.
Part A:
$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0... | Transformation T(x,y,z) is defined as in the question posted $$
T(x,y,z) =
\left[ {\begin{array}{cc}
1 & 1 & 0\\
1 &-1 & 0\\
1 & 0 & 1\\
\end{array} } \right]\left[ {\begin{array}{cc}
v_1\\
v_2\\
v_3\\
\end{array} } \right]
$$
which is of the form $T(x,y,z)=Ax.$ Such that $A=\left[ {\begin... | {
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Square in Interval of Primes Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.
The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. ... | Let $p$ be the $(n+1)$st prime. Then $a_n\le (-8)+(1+3+5+7+9+\ldots +p-2)=(\frac{p-1}{2})^2-8$ provided $p\ge 11$ (the smaller cases can be dealt with by checking manually). So with $b_n:=\lfloor \sqrt {a_n}\rfloor $ we have $b_n< \frac{p-1}{2}$. Then $a_n<(b_n+1)^2=b_n^2+2b_n+1< a_n+p=a_{n+1}$ as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the... | $$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} = \lim_{x \to -1} \frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)} \\ \\ = \lim_{x \to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\\ \\ \lim_{x \to -1} \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| Notice that
$$(a+b+c+d)^2=\underbrace{a^2+b^2+c^2+d^2}_{\text{the sum of square of all terms}}+\underbrace{2ab+2ac+2ad+2bc+2bd+2cd}_{\text{the sum of the double products of the terms taken 2 by 2 }}$$
so we find
$$\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 ..\right)^2=\underbrace{ 1^2+(2\times 1\times... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving the general formula [nx] where [.] is the floor function. I've been trying to solve a exercise that asks me to prove the following generalization for the floor function:
$$\lfloor nx\rfloor = \sum_{k=0}^{n-1} {\lfloor x + \frac kn \rfloor}$$
I've already proven the special cases where $n = 1,2,3$. I proved it a... | Here is a classical neat solution:
Let
$$f(x)= \lfloor nx\rfloor - \sum_{k=0}^{n-1} \lfloor x + \frac kn \rfloor$$
Then the following are immediate:
*
*$f(x+\frac{1}{n})=f(x)$.
*$f(x)=0$ for all $x \in [0, \frac{1}{n})$.
Moreover, any function satisfying these two must be identically 0.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck her... | Use Substitution instead. Let $u^2=x^2+4$. Then $u\,du=x\,dx$ and we end up integrating $(u^2-4)(u^2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer wo... | You correctly started with $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)$$ Now consider that, for small values of $\epsilon $, $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+\frac{\epsilon ^3}{16}+O\left(\epsilon
^4\right)$$ and replace first $\... | {
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Finding the cube root of a complex number $z$ $\text{Let }z = -2-2i \text{ where }i \text{ is imaginary. Find in Modulus-Argument form the cube roots of }z$
So far I've done this
$$r = \sqrt8 = 2 \sqrt2 \\
\alpha = \frac{-\pi + \frac{\pi}{4}}{3} = \frac{-\pi}{4}
$$
Which then leads me to believe that the first answer i... | Seems like you forgot to take the cube root of $2\sqrt{2}$, after which, you get the answer.
If you look take a complex number $re^{i\theta}$ to the $n$th power, you get $r^n e^{in\theta}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about $\lim_{x \to -\infty}\frac{\sqrt{10+11x^2}}{12+13x}$
$\lim_{x \to -\infty}\dfrac{\sqrt{10+11x^2}}{12+13x}$
= multiply top and bottom by $\dfrac{1}{x}=-\dfrac{1}{\sqrt{x^2}}$
My question is, why is the negative sign in front so crucial, I don't get it:
$\lim_{x \to -\infty}-\dfrac{\sqrt{10/x^2+11x^2... | Maybe this is easier to visualize:
\begin{align*}
\lim_{x \to -\infty} \frac{\sqrt{10+11x^2}}{12+13x} & = \lim_{x \to -\infty} \frac{\sqrt{x^2\big(\frac{10}{x^2}+11\big)}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} \frac{\sqrt{x^2} \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\cos(x+y)$ and $\sin(x+y)$ given that $\cos x + \cos y = a$ and $\sin x + \sin y = b$ If $\cos x + \cos y = a$ and $\sin x + \sin y = b$.
Find $\cos(x+y)$ and $\sin(x+y)$.
I only need some hints to start as I am not able to get any way to go forward to.
| Using Prosthaphaeresis Formulas
$$2\sin\frac{x+y}2\cos\frac{x-y}2=a$$
and $$2\cos\frac{x+y}2\cos\frac{x-y}2=b$$
Divide to find $\tan\dfrac{x+y}2$ assuming $ab\cos\dfrac{x-y}2\ne0$
Now apply Weierstrass substitution
Alternatively find $a^2+b^2,a^2-b^2,ab$ and use Prosthaphaeresis Formulas
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
... | Classic mistake occurs at the line where you go to cancel a $\sin x$ from each side...that only holds if $\sin x$ isn't 0. You also get answers whenever $\sin x=0$, which occurs at increments of $n\pi$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \... | PS:
Thomas Andrews says the values of $A,B,C$ are in error. I haven't checked those, but the technique outlined below still works if different numbers are involved.
end of PS
$$
\int\frac{8x+31}{x^2+2x+5}\,dx
$$
First let $w=x^2+2x+5$ so that $dw=(2x+2)\,dx$. Then we have
$$
\int\frac{8x+31}{x^2+2x+5}\,dx = 4\int\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Inverting a Characteristic Function for half-cubic Student's T entailing a Modified Bessel of 2nd kind The Characteristic function of the Student's T with $\alpha$ degrees of freedom,
$C(t)=\frac{2^{1-\frac{\alpha }{2}} \alpha ^{\alpha /4} \left| t\right| ^{\alpha /2}
K_{\frac{\alpha }{2}}\left(\sqrt{\alpha } \left... | Well, according to Mathematica
FourierTransform[
Abs[t]^(9/4)*BesselK[3/4, (\[Sqrt](3/2)*Abs[t])^3], t, w],
which translates to
$$\mathcal{F}_t\left[\left| t\right| ^{9/4} K_{3/4}\left(\left(\sqrt{\frac{3}{2}} \left| t\right| \right)^3\right)\right](w)$$
is equal to
$$\frac{4\ 2^{3/8} \pi \, _1F_4\left(\frac{11}{12};... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Partial Fraction Decomposition Problem I am having trouble with this problem.
I need to integrate: $$\frac1{T^4}\times \frac1{K-T}$$ with respect to $T$.
If I do PFD:
$$\frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T}$$
Does this look right?
From here, I am having trouble solving for the coeff... | \begin{align*} \frac1{T^4(K-T)} &= \frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T} \\ 1 &= (K-T)A + T(K-T)B + T^2(K-T)C + T^3(K-T)D + T^4E \\ \text{Now equate coefficients} \\ \text{(Unity)} \qquad 1 &= KA \implies A=\tfrac1K \\ (T) \qquad 0 &=-A+KB \implies B = \tfrac1{K^2} \\ (T^2) \qquad ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Confusion about a certain series expansion While reading some old notes on contour integration, I noticed the author uses series expansion:
$$\frac{\sinh sx}{\sinh sa}=\frac{sx(1+\frac{1}{6}s^2x^2+\cdots)}{sa(1+\frac{1}{6}s^2a^2+\cdots)}=\frac{x}{a}(1+\frac{1}{6}s^2(x^2-a^2)+\cdots)$$
for small $s$
This may be a si... | The first part is an application of the Taylor series of $\sinh(z)$, which is
$$\sinh(z)=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\frac{z^7}{7!}...$$
Collecting $z$ we get
$$\sinh(z)=z(1+\frac{z^2}{3!}+\frac{z^4}{5!}+\frac{z^6}{7!}...)$$
which corresponds to the first expression provided in the question.
The second expression, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Discrete math. Finding a perfect square. The problem is: Find all natural numbers $n$ for which $2^n + 1$ is a perfect square?
I am having a bit of trouble finding a generic way of finding these numbers. Of course the first obvious solution is $n = 3.$ For which we have $8 + 1 = 3^2.$
Anyone has any smart ideas?
| Method $\#1:$
If $2^n+1=m^2\iff2^n=(m+1)(m-1)$
We can easily test for $n\le2$
For $n>2,$ clearly $m$ is odd
and we have $$2^{n-2}=\frac{m-1}2\cdot\frac{m+1}2$$
But as $\displaystyle\frac{m+1}2-\frac{m-1}2=1,\left(\frac{m-1}2,\frac{m+1}2\right)=1,$ at least one of them is odd
But each divides $2^{n-2},$ the odd must b... | {
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"timestamp": "2023-03-29T00:00:00",
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Difference in derivative between $\frac{6}{3x^2+1}$ and $\frac{6}{3x^2}$ Is there a difference in derivative between $\frac{6}{3x^2+1}$ and $\frac{6}{3x^2}$ I thought there wouldn't be and I've asked several people all with different results?
| Let $$f(x) = \frac{6}{3x^2+1}$$
Then we have
$$\tag{1}f'(x) = \frac{d/dx(6)\cdot(3x^2+1)-6\cdot d/dx(3x^2+1)}{(3x^2+1)^2} = \frac{-6\cdot6x}{(3x^2+1)^2}$$
Since $\frac{d}{dx} ~3x^2+1 = \frac{d}{dx}~ 3x^2$ there is no difference upstairs in $(1)$, but only in the denominator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What am I doing wrong? Finding a limit as $x$ approaches $0$ $$\lim_{x\to 0} {a-\sqrt{a^2-x^2}\over x^2} =$$
$${a-\sqrt{a^2-x^2}\over x^2}\cdot{a+\sqrt{a^2-x^2}\over a+\sqrt{a^2-x^2}} = $$
$$a^{2} - a^{2} - x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}} $$
$$-x^{2}\over ax^{2}+x^{2}\sqrt{a^{2}-x^{2}}$$
$$-1\over a+\sqrt{a... | Third equality $$\left (a - \sqrt{a^2-x^2}\right )\left (a+\sqrt{a^2+x^2}\right ) \color{blue}{=x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find and sketch the line $x=1$ under the mapping $f(z)=1/z$ Find and sketch the image of the vertical line $x=1$ under the mapping $f(z)=\frac 1z$
I started by using $u(x,y) + iv(x,y)=f(z)=\frac 1z = \frac 1{x+iy}$ From here I multiplied $f(z)$ by the conjugate to get $\frac x {x^2 + y^2} - \frac {iy} {x^2 + y^2}$ Then... | Using $u_y=u(1,y)$ and $v_y=v(1,y)$, you obtain the equation
$(u_y-\frac{1}{2})^{2} +v_y^{2} =(\frac{1}{2})^{2}$ for any $y\in\mathbb{R}$.
Conversely you have for any point $z=x+iy$ in the circle with center $(\frac{1}{2},0)$ and radius $\frac{1}{2}$
\begin{equation}
(x-\frac{1}{2})^{2} +y^{2} =(\frac{1}{2})^{2}\\
\Rig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of an identity involving binomial coefficients I have found numerically that the following identity holds:
\begin{equation}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{x^2+t^2-t}{2t-1},
\end{equation}
where $n$, $t$, and $x$ are positive integers ($x... | The right-hand side is not correct, but we can show for non-negative integers $0\leq x\leq t$:
\begin{align*}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{1}{2}\left(\frac{x^2+t^2-t}{2t-1}-x\right)\tag{1}
\end{align*}
It is convenient to use the coeffic... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.
| Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then:
$$ n^4+n^2+1 = (n^2-\omega)(n^2-\omega^2), $$
so:
$$\frac{1}{1+n^2+n^4}=\frac{1}{i\sqrt{3}}\left(\frac{1}{n^2-\omega}-\frac{1}{n^2-\omega^2}\right)$$
and:
$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{\sqrt{3}}\Im\sum_{n=1}^{+\infty}\frac{1}{n^2-\omega}$$
can b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $ $$ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $$
If I pick $ x = 0$ I get:
$$ \lim_{(x,y) \to (0,0)} \frac{0}{2y^2} = 0$$
So if the limit exists it must be $0$
Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$
So I can use the Taylor series of $sin(t)... | We have $\;3x^2+2y^2\geq 2\sqrt{6}|xy|\;$ by AM-GM, thus we have
$\dfrac{\sin^2(xy)}{2\sqrt{6}|xy|}=\left(\dfrac{\sin(xy)}{xy}\right)^2\dfrac{|xy|}{2\sqrt{6}}\to0\;$ for $\;(x,y)\to(0,0)\;.$
I think your proof is ok. You can read also my proof if you want, it is a little bit shorter.
| {
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Use row reduction to show that the determinant is equal to this variable. Show
determinant of:
\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}
is equal to $(b - a)(c - a)(c - b)$
I'm not sure if you can use squares or square roots hmmm.. please help me. I'm sure it's a simple question. Much appreciated.
| Subtracting $\DeclareMathOperator{Col}{Col}\Col_1$ from $\Col_2$ and $\Col_3$ gives
$$
\begin{pmatrix}
1 & 0 & 0 \\
a & b-a & c-a \\
a^2 & b^2-a^2 & c^2-a^2 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
a & b-a & c-a \\
a^2 & (b+a)(b-a) & (c+a)(c-a)
\end{pmatrix}
$$
The determinant is then
\begin{align*}
(b-a)(c+a)(c... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction that $f(n) = 1-2^{2^n}$, where $f(0) = 3$ and $f(n) = 2 f(n-1) - (f(n-1))^2$ I am doing a textbook question which state that a function $f:\mathbb{N}\to\mathbb{Z}$ is a recursively defined as shown bellow
$f(0) =3$,
$f(n) = 2\cdot f(n-1) -(f(n-1))^2 $ if $n\ge1$.
Prove that $f(n) = 1-2^{2^n}$ for all... | $$\begin{align}f(k)&=2f(k-1)-(f(k-1))^2\\&=2 \left(1-2^{2^{k-1}}\right)-\left({1-2^{2^{k-1}}}\right)^2\\&=2-2^{2^{k-1}+1}-\left(1-2^{2^{k-1}+1}+2^{2^k}\right)\\&=1-2^{2^{k}}.\end{align}$$
P.S.
$$2\left(1-2^{2^{k-1}}\right)=2-2^1\cdot 2^{2^{k-1}}=2-2^{1+2^{k-1}}=2-2^{2^{k-1}+1}.$$
$$\left(1-2^{2^{{k-1}}}\right)^2=1^2-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/969923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When simplifying $\sin(\arctan(x))$, why is negative $x$ not considered? Let $u = \arctan(x)$, hence $x = \tan(u)$ for $u$ belongs in $(-\frac\pi2, \frac\pi2)$. Since $u$ belongs in $(-\frac\pi2, \frac\pi2)$, we consider $\sin(u)$ where $u$ belongs in $(-\frac\pi2, \frac\pi2)$.
I used the unit circle to determine that ... | To add some additional information regarding user46234's comment, we can also understand the omission of a $\pm$ sign as a simple consequence of algebra.
Here are the relevant pieces of information to conclude this:
*
*By definition, $\sqrt{x^2}=|x|$
*$\arctan$ is a strictly increasing function bounded between $(-\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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The common tangents to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle Problem :
Show that the common tangents to circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle.
Solution :
Let $C_1 : x^2+y^2+2x=0$
here centre of the circle is $(-1,0) $ and radius 1 unit.
$C_2:x^2+... | More systematic and direct method!
Tangent line to $y=f(x)$ at the point $(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0).$$
Let $(x_1,y_1)$ and $(x_2,y_2)$ be the tangent points of the common increasing tangent line to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$, respectively.
Equate the slopes and intercepts of the tange... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Using only addition, subtraction and multiplication I have the numbers 6, 30, 8, 8, 3, 7, 1, 2, and 5. Using only addition, subtraction, and multiplication, can you use those numbers to make 60, 54, and 52?
| $$6+30+8+8+3+7+1+2\color{red}{-}5=60.$$
$$6+30+8+8\color{red}{-}3+7+1+2\color{red}{-}5=54.$$
$$6+30+8+8\color{red}{-}3+7\color{red}{-}1+2\color{red}{-}5=52.$$
P.S.
$$30\times 2=60.$$
$$6\times (8+1)=54.$$
$$(8+5)\times (7-3)=52.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving AM-GM for the special case $n=3$ I know the AM-GM inequality and its proof which is relatively complex, though the case for $n=2$ is quite simple. However, I don't know of any special easier proof for the case $n=3$, specifically:
$$\frac{a+b+c}3\ge \sqrt[3]{abc}$$
What is the most elegant proof for this? :)
| The case for $n=3$ can be proved by using the cases for $n=2,4$.
For $p,q\gt 0$, we have$$(\sqrt p-\sqrt q)^2\ge0\iff \frac{p+q}{2}\ge\sqrt{pq}.$$
So, we have for $s,t,u,v\gt 0,$$$s+t\ge 2\sqrt{st},\ \ \ u+v\ge 2\sqrt{uv}.$$
Hence, we have
$$s+t+u+v\ge 2\sqrt{st}+2\sqrt{uv}\ge 2\sqrt{2\sqrt{st}\cdot 2\sqrt{uv}}=4(stuv)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\s... | We have
$$x(5+2\sqrt{6})=289$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$
$$\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}$$
$$\Rightarrow x=289(5-2\sqrt{6})$$
| {
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"url": "https://math.stackexchange.com/questions/977429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | $$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Done!
| {
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Asymptotic approximation of the arctangent? That is, I am looking for an algebraic function $f(x)$ that approximates $\arctan x$ for large values of $x$.
The approximation could be reasonably modest -- perhaps something like
$$\tan (f(x)) = \frac{\pi}{4} + O\left(\frac{1}{x^2}\right).$$
| Use the following
$$\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} \Leftrightarrow$$
$$\arctan(x) = \frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$$
The series for $\arctan(x)$ is
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$
Now substitute $\frac1x$ to get the result
$$\arctan(x)=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$ How can we find the integral:
$$\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$$
I tried to find and got it to be $\cfrac{\pi}{\sqrt2}$. Am I correct? Please help me with an appropriate method. I tried to use sum of resi... | First, use the change of variable $x=\frac{1}{u} => dx = -\frac{1}{u^2}du $
$I = \int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{u^2} {1 + {u^4}}} \;{\mathrm{d}}u $
=> $2*I = 2*\int_{ 0 }^{ + \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
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Continued fraction of the golden ratio It is known, that the continued fraction of $\phi = \frac{1+\sqrt{5}}{2}$ is $[\bar{1}]$. This can be shown via the equation $x^2-x-1=0$:
$$ x^2-x-1=0 \Rightarrow x = 1+\frac{1}{x} = 1+ \frac{1}{1+\frac{1}{x}} = \cdots $$
As far as I can see, the only thing that has been used here... | The golden ratio
$$
x : 1 = 1 + x : x
$$
leads to the equation
$$
x^2 - x - 1 = 0 \quad (\#)
$$
It can be transformed to two different equations of the form
$$
x = F(x)
$$
which then can be used to substitute the $x$ on the right hand side by $F(x)$
$$
x = F(F(x)) = F(F(F(x))) = \cdots
$$
Your transformed version of $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find out the primitive polynomial GF(3) 1.) $x^2 + 2x$
2.) $x^2 + 1$
3.) $x^2 + 2$
4.) $x^2 + 2x$
5.) $x^2 + 2x + 1$
6.) $x^2 + 2x + 2$
7.) $x^2 $
8.) $x^2 + x + 2$
9.) $x^2 + x + 1$
Can any one help me in listing out primitive polynomials and tell me why is it a primitive polynomial please.
| $$x^2 + 1,
x^2 + 2x + 2,
x^2 + x + 2,$$
are primitive polynomials over $gf(3)$.
Because they are irreducible and monic polynomials
Monic means coefficient of highest exponent of $x$ is $1$.
And irreducible means can not be further factorise into small polynomials.
Let's take first case $x^ 2 + 2x$ is further factorise ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Consider the lists of length six made with the symbols $P, R, O, F, S$ where repetition is allowed. Consider the lists of length six made with the symbols $P, R, O, F, S$ where
repetition is allowed. (For example, the following is such a list: $(P,R,O,O,F,S)$.)
How many such lists can be made if the list must end in an... | We want to count $5$ letter words with at least $2$ O letters.
*
*The number of arbitrary $5$ letter words is $5^5$.
*The number of $5$ letter words with exactly one "O" is $5 \cdot 4^4$ (fix the location of the O, and the rest can consist of the four remaining letters).
*The number of $5$ letter words without the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the roots of $(\frac{z-1}{z})^5=1$
Write down the fifth roots of unity in the form $\cos \theta + i \sin \theta$ where $ 0 \leq \theta \leq 2\pi$
Hence, or otherwise, find the fifth roots of i in a similar form
By writing the equation $(z-1)^5=z^5$ in the form :
$${\left(\frac{z-1}{z}\right)}^{5}=1$$
show ... |
Solve$$\displaystyle \left(\frac{z-1}{z}\right)^5=1\rightarrow \frac{z-1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}$$ for $k=0,1,2,3,4$.
Hence $$\displaystyle 1-\frac{1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}\rightarrow 1-e^{\Large \frac{i\cdot{2k}\pi}{5}}=\frac{1}{z}\rightarrow z=\frac{1}{1-e^{\Large \frac{i\cdot{2k}\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality relating coefficients and roots of a complex polynomial While going through some olympiad handouts I stumbled upon a problem related to an upper bound for the Mahler measure, which stated that
Given a polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0 \in \mathbb{C}[x]$ that has roots $z_1, z_2, ... , z_... | Suppose the roots of polynomial $f(z)$ are $z_1,z_2,\cdots,z_n$,
where, $|z_1| \ge |z_2| \ge \cdots \ge |z_m| > 1 \ge |z_{m+1}| \ge \cdots \ge |z_n|$
Let, $g(z) = z^nf\left(\frac{1}{z}\right) = 1 + a_{n-1}z + \dots + a_0z^n$
Then, $\{1/z_k\}_{1\le k \le m}$ are the zeros of $g$ in the disk $|z| \le r = 1-\epsilon < 1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Prove this equality $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
This is how I did it:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4 \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=64$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=64$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11... | You can begin with
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x$$
and find $x$.
Alternatively, let $z=2+11i$. You have to show that $\sqrt[3]z+\sqrt[3]{\bar z}=4$. Since $\sqrt[3]z$ and $\sqrt[3]{\bar z}$ are conjugate, it suffices to show that $\Re\sqrt[3] z=2$.
If you try to solve for $b$
$$(2+ib)^3=2+11i$$
you get
$$8-6b^2=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
| From $a^2+b^2=c^2$ we get $a^2=c^2-b^2=(c+b)(c-b)$, i.e. a factorization of $a^2$ into two distinct factors $c+b>c-b$. The only such factorizations for the square of a prime is $a^2\cdot 1$, i.e. we conclude $c-b=1$, hence $b=2$, $c=3$. But then $a^2=5$, qea.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 6
} |
Closed form of $\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$ What real analysis tools would you recommend me for getting the closed form of the integral below?
$$\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$$
| The place I would start is the nifty result, proven here, that
$$\frac{\sin{x}}{\cosh{t} - \cos{x}} = 2 \sum_{k=1}^{\infty} e^{-k t} \sin{k x} $$
Of course, the integral actually looks like
$$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} - \sin{x}} \log{x} $$
so we need to map $x \mapsto \pi/2 - x$ and we have that the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
A triangle ABC with vertex $C(4,3)$. The bisector and the median line equation drawn from the same vertex are given. Find the vertices A & B. A triangle $\triangle ABC$ one of his vertex is the point $C(4,3)$.
The bisector line equation is $x+2y-5=0$ and the median line equation is $4x+3y-10=0$ drawn from the same ver... | The segment connecting $A$ in $(1,2)$ with point $C$ in $(4,3)$ lies on the line $y=\frac{x}{3}+\frac{5}{3}$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$.
Note that the bisector crosses the $x$-axis in point $(5,0)$. Let us call $D$ this point. The segment $CD$ lies on the line $y=-3x+15$ and has length $\sqrt{3^2+1^2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions Have been working on this for years. Need a system which proves that there exists a number $C$ which has certain properties. I will give a specific example, but am looking for a system which could possibly be generalized.
Find (or prove th... | Say
$$
x^2 -47x - c = (x-a)(x-b)
$$
for integers $a$ and $b$.
Then
$$
a + b = 47 \\
-ab = C.
$$
So $a$ and $b$ must have only $2,3,5$ in their prime factorization.
Say $b$ is divisible by $5$ (one of them must be). So say $b = 5n$. So then $47 - a$ is divisible by $5$. And $a$ must be $-13 , -8 , -3 , 2, 7, 12, 17, \do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int\frac{\ln x\,dx}{x^2+2x+4}$
$$K=\int\frac{\ln x\,dx}{x^2+2x+4}$$
I did this $x^2+2x+4=(x+\alpha)(x+\beta)$, then used partial fraction, I am then unsure how to integrate $\int\frac{\ln x}{x+c}\,dx$.
I tried Integration by parts also taking first function as both of them which ended nowhere
Also... | One way to attack improper integrals of the form
$$\int_0^{\infty} dx \, f(x) $$
is to use the residue theorem, i.e. contour integration in the complex plane. To do this, one considers the contour integral
$$\oint_C dz \, f(z) \log{z} $$
where $C$ is a keyhole contour about the positive real axis, of inner radius $\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Converting Summation to Expression How does the summation break down from
$$\displaystyle\sum_{n \geq 0} (x + x^2) ^ n$$
to
$$\frac1{1 - x - x^2} $$
per this answer?
| Assuming $$|x+x^2|\lt 1$$
$$s=\displaystyle\sum_{n \geq 0} (x + x^2) ^ n$$
Notice that
$$\quad \\\\\\\ \ \ \quad s=1+(x+x^2)+(x+x^2)^2+(x+x^2)^3+\cdots$$
$$s(x+x^2)=(x+x^2)+(x+x^2)^2+(x+x^2)^3+\cdots$$
$$s-s(x+x^2)=s(1-(x+x^2))=1$$
$$\Rightarrow s=\frac{1}{1-(x+x^2)}=\frac{1}{1-x-x^2}$$
$$\Rightarrow \displaystyle\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find sufficient and necessary conditions in which $f(x)$ is a natural number Let us consider the function $$f(x)=(-√3+2)^{2^{x-2}}+(√3+2)^{2^{x-2}}$$ where $x≥3$.
We know for example that if $x$ is an integer, then $f(x)$ is also an integer.
My question is: Find sufficient and necessary conditions in which $f(x)$ is a... | 1st claim :
Let $k \in \mathbb{N}$ with $k \geq 2$. Then :
$$ \Big( 2\cosh(x) = k \Big) \, \Leftrightarrow \, x = \ln \Big( \frac{k+\sqrt{k^2-4}}{2} \Big). $$
Let $k \in \mathbb{N}$ with $k \geq 2$. We have :
$$ \begin{align*}
2\cosh(x) = k &\Leftrightarrow {} e^{2x} - ke^{x} + 1 = 0 \\
\end{align*}
$$
Let $u=e^{x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with epsilon delta proof $x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.
So I have
$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$
$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$
$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$
$$=|(x-3)|<\epsilon/28$$
How do I prove $x^2+... | There's a safer way to think. Let $\epsilon > 0$, and suppose that $0 < |x - 3| < \delta$. let's find $\delta$. Notice: $$|x| - 3 < |x-3| < \delta \implies |x| < \delta + 3.$$
Once we find $\delta$, any other value $\delta'$ less than this $\delta$ will also work, so we impose that $\delta \leq 1$. If the value of $\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
| Notice that in this case by doing some rearrangement
$$\frac{p(x)}{q(x)}=A\frac{q(x)}{q(x)}+B\frac{q'(x)}{q(x)}$$
$$\begin{align}
\int\frac{2\sin x + \cos x}{\sin x + 2\cos x}dx&=-\frac{3}{5}\int\frac{\cos x-2\sin x}{\sin x + 2\cos x}dx+\frac{4}{5}\int\frac{\sin x+2\cos x}{\sin x + 2\cos x}dx\\
\end{align}$$
$$\int\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$ $$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$$
What I did
I multiplied by the denominator's conjugate and got the following $$2x+\sqrt{4x^2-2x}$$
My question is, what would I now do to evaluate the limit? The positive infinity I would do as fol... | When $x \to +\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 +\sqrt{1- \dfrac{1}{2x}}\right) \sim 4x\to +\infty$$
When $x \to -\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 -\sqrt{1- \dfrac{1}{2x}}\right) = \dfrac{1}{1 +\sqrt{1- \dfrac{1}{2x}}} \to \dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
| $$ 3(2x-1)^2=147 $$
$$ (2x-1)^2=\frac{147}{3} $$
$$ (2x-1)^2=49 $$
$$ (2x-1)^2=(\pm 7)^2 $$
$$ \sqrt{(2x-1)^2}=\sqrt{(\pm 7)^2} $$
$$ 2x-1=\pm 7$$
$$ 2x=1\pm 7$$
$$ x=\frac{1\pm 7}{2}$$
Therefore
$$ x_1=\frac{1+ 7}{2}=\frac82=4$$
$$ x_2=\frac{1- 7}{2}=-\frac62=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Specific question on proving continuity of this function If $c\in\mathbb{R}$, I want to show that $f(x)=\dfrac{cx}{x^2+c^2}$ is continuous using $\varepsilon-\delta$. Unfortunately from an algebraic standpoint I seem to be missing the strategy. I let $p\in\mathbb{R}$ ($p\neq 0)$, and I want $\forall \varepsilon>0$ $\ex... | Choose $\delta < |p|/2$ such that when $|x-p| < \delta$,
$$||x|-|p|| \leq |x-p|< |p|/2$$
and
$$|p|/2 <|x| < 3|p|/2$$
Note that
$$|f(x) - f(p)| < \frac{|c||p||x|}{|x^2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|x^2+c^2||p^2+c^2|}|x-p| \\ < \frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}|x-p|\\ = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Better substitution calculating integral? I'm calculating
$$ \iint\limits_S \, \left(\frac{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}{1+\frac{x^2}{a^2}+\frac{y^2}{b^2}} \right)^\frac{1}{2} \, dA$$ with $$S =\left\{ (x, \, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\right\}.$$
I take
$$x = ar\cos \theta$$
$$... | One way to do this is to substitute $u^2 = 1 + r^2$, so $2u \, du = 2r\, dr$, or $r\, dr = u\, du$
$$\int_0^1 \sqrt{\frac{1-r^2}{1+r^2}}\,r\,dr = \int_0^1 \sqrt{\frac{2-(1+r^2)}{1+r^2}}\,r\,dr\\
= \int_1^\sqrt{2} \sqrt{\frac{2-u^2}{u^2}}\,u\,du = \int_1^\sqrt{2} \sqrt{2- u^2} \, du $$
Now substitute $u = \sqrt{2} \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
| To avoid that nasty quotient rule
(and throwing away that "$2$"),
$\begin{array}\\
\left(\frac{1+4\cos x}{ \sqrt{x+4\sin x}}\right)'
&=\left((1+4\cos x)(x+4\sin x)^{-1/2}\right)'\\
&=(1+4\cos x)\left((x+4\sin x)^{-1/2}\right)'
+(1+4\cos x)'(x+4\sin x)^{-1/2}\\
&=(1+4\cos x)\left((-1/2)((x+4\sin x)'(x+4\sin x)^{-3/2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Determine all values of n such that this quadratic Determine all values of $n^2 + 19n + 99$ is a perfect square. I tried setting some square $b^2$ equal to the following, and then factoring as a Diophantine equation with $2$ variables... Didn't work.
| $$
\begin{array}{c}
n^2 + 19n+99 = r^2 \\
(4n^2+76n) = 4r^2 - 396 \\
(2n+19)^2 -361 = 4r^2 - 396 \\
r^2 - (2n+19)^2 = 35 \\
(r+ (2n+19)) (r-(2n+19)) = 35
\end{array}
$$
Then if: $$
\begin{array}{ccc}
(r+ (2n+19)) =35, & (r-(2n+19)) = 1 & \longrightarrow n = -1 \\
(r+ (2n+19)) =7, & (r-(2n+19)) = 5 & \longrightarrow n =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proving that $(n-1)^n>n^{n-1}$ I want to prove that $(n-1)^n>n^{n-1}$, for $n>4$, $n$ is an integer.
So I divided by $n^n$ and got:
$(1-\frac{1}{n})^{n}>\frac{1}{n}$
I know that $(1-\frac{1}{n+1})^{n+1}$>$(1-\frac{1}{n})^{n+1}=(1-\frac{1}{n})^n(\frac{n-1}{n})>\frac{1}{n}(\frac{n-1}{n})$.
How can I continue?
Thanks.
| Solution using induction:
Base case - $n=4$, so $(4-1)^4>4^{4-1}$ which is true.
Inductive step - assume that the statement holds for some k and show that it holds for k+1, hence we want to show that $k^{k+1}>(k+1)^k$ for $k \ge 3$.
Our assumption is $(k-1)^k>k^{k-1}$ or alternatively $\displaystyle k-1>\left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof $(1+1/n)^n$ is an increasing sequence I need help proving $a_n=\left(\dfrac{n+1}{n}\right)^n$ is increasing sequence on the positive integers.
An exercise in the analysis book by Mattuck asks to prove $a_n=\left(\dfrac{2^n+1}{2^n}\right)^{2^n}$ is increasing.
But this is easy since $\left(\dfrac{2^n+1}{2^n}\right... | Use the Bernoulli inequality,
$$1+(n+1)(a-1)\leq a^{n+1}$$ with
$$a=\frac{n(n+1)}{(n+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show sequance is monotonic Let $x>0$ (fixed) and $n$ be natural. Show that $$\displaystyle (x^n+x^{n-1}+...+1)^{\frac{1}{n}}$$ is monotonic.
I tried by induction but didn't work but intuition tells me it's decreasing.
| The first couple are decreasing at least:
$(x+1)^2-(x^2+x+1)^1=x$
$(x^2+x+1)^3-(x^3+x^2+x+1)^2=x^5+3x^4+3x^3+3x^2+x\\=x(x+1)(x^3+x^2+x+1)+x^2(x^2+x+1)$
Next is $(x^3+x^2+x+1)^4-(x^4+x^3+x^2+x+1)^3$, which is
$$x^{11}+4x^{10}+10x^9+16x^8+22x^7+25x^6+22x^5+16x^4+10x^3+4x^2+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Another quick induction question for a recursively defined sequence (with closed form formula given) I was given:
A sequence is defined recursively by $a_0 = 1$, $a_1 = 4$, and for $n\ge2$, $a_n = 5a_{n-1} - 6a_{n-2}$.
Use induction to prove that the closed form formula for $a_n$ is $a_n = 2\cdot3^n-2^n, n\ge0$.
So far... | Substitute $a_{n-1}=2\cdot3^{n-1}-2^{n-1}$:
$$\begin{align}
a_{n+1}&=5(2\cdot3^n-2^n)-6a_{n-1}\\
&=5(2\cdot3^n-2^n)-6(2\cdot3^{n-1}-2^{n-1})\\
&=30\cdot3^{n-1}-10\cdot2^{n-1}-12\cdot3^{n-1}+6\cdot2^{n-1}\\
\end{align}$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is... | Here is a full proof.
Let us start the discussion for general $n$. Denote $S = \sum_{i=1}^n a_i$.
Since by AM-GM, $S \geq n \sqrt[n]{a_1a_2...a_n}$, we have
$$1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} \geq \frac{S}{S - (n-2)\sqrt[n]{a_1a_2...a_n}}$$
Hence a tighter claim is (simultaneously defining $L$ and $R$):
$$L ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Minimum without using of differential calculus Find minimum of $$x + y^5$$ where $x>0,y>0 $ $xy=1$ without using of differential calculus.
| $x+y^5=\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+y^5$. By inequality between arithmetic mean and geometric mean we get $$\frac{x+y^5}{6}\geq \sqrt[6]{\frac{x^5}{5^5}y^5}=5^{-5/6}\Rightarrow x+y^5\geq 6\cdot 5^{-5/6}$$ and equality holds iff $\frac{x}{5}=y^5$, so corresponding values of arguments are $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 +... | $$\sum^n_{k=0} (n-k)^2=n^2+(n-1)^2+\cdots + 0^2= 0^2+1^2+\cdots +n^2 =\sum^n_{k=0} k^2=\frac16n(n+1)(2n+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Inhomogeneus recurrence relation $a_{n+1} = 2a_n+3^n+4^n$ So this was given in class and the teacher weren't able to solve it, and I was wondering how a solution can be given?
$a_{n+1} = 2a_n+3^n+4^n, \enspace a_0 = 1$
Usually we'd consider the solution $a_n$ to be of the form $a_n = a_n^{(h)}+a_n^{(p)}$, where $a_n^... | $$\dfrac{a_{n+1}}{2^{n+1}} = \dfrac{a_n}{2^n} + \dfrac{1}{2}(\dfrac{3}{2})^n + \dfrac{1}{2}2^n$$
Let $b_{n+1} = \dfrac{a_{n+1}}{2^{n+1}}$, we have $$b_{n+1} = b_n + \dfrac{1}{2}(\dfrac{3}{2})^n + \dfrac{1}{2}2^n $$
so $b_n = b_0 + \sum_{k=1}^n\left(\dfrac{1}{2}(\dfrac{3}{2})^k + \dfrac{1}{2}2^k\right)$
Can you go from... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exact closed form expression of $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$ Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$
I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?
| Your series is
$1+2+\cdots+2^n$
$\,\,\,\,\,\,\,\,\,2+\cdots+2^n+2^{n+1}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2^n+2^{n+1}+\cdots+2^{2n}$
And Sum can be written as
$$\begin{align}s&=1+2(2+4+\cdots+2^{n})+2^{n+1}+2^n+2^{n+1}+2^{n+2}+\cdots2^{2n}\\
&=1+2(2+4+\cdots+2^{n})+2^{n+1}+(2^n+2^{n+1}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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