Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution ... | You have $(x^2+1)^2=3x(x^2+1)$ This means that $(x^2+1)^2-3x(x^2+1)=0$ or $(x^2+1)(x^2+1-3x)=0$ Can you solve it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Some exact values of $\cos \theta$ using de Moivre's theorem Presently I am faced with the following question:
By showing that $$\cos5\theta = \cos\theta(16\cos^4\theta - 20\cos^2\theta + 5)$$ and then solving the equation $\cos5\theta = 0$, deduce that
$$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt{5}}{8}$$
and... | Since $\cos^2 \theta$ is strictly decreasing on $\left[0, \frac{\pi}{2}\right]$, we must have that
$$\cos^2 \left(\frac{\pi}{10}\right) > \cos^2 \left(\frac{3 \pi}{10}\right),$$ and so
$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5 + \sqrt{5}}{8} \qquad \text{and} \qquad \cos^2 \left(\frac{3\pi}{10}\right) = \frac{5 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Cannot understand how does he find the minimum value of this funtion My teacher has find the minimum value of $y = x + \frac{2}{x}$ when $x \ge 0$ without using
differential:
$y = x + \frac{2}{x}$
Multiplied both side by $x$ and moved $y$ to other side
$x^2 - yx + 2 = 0$
Then he wrote that having an answer in this equ... | If the discriminant is smaller than zero($b^2-4ac<0$) than $x_{1,2}\in\mathbb{C}$ so clearly $b^2-4ac\geq0$ now if $y\in(-2\sqrt{2},2\sqrt{2})$ it means that there is no solution in reals,and since $x>0$ than $y>0$ so we have that $y\geq2\sqrt{2}$ clearly since we are looking for the minimum we pick the smallest $y$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probablity of rolling the same number twice after n times Im trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice n times is?
The chance to get some number is 1/6 and getting that same number two times in a row must be 1/36, but with this calcul... |
I'm trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice $n$ times is?
Exactly two of a specified number, or exactly two of any number?
Case 1 : The probability of getting exactly two sixes. $n\in \{2,3,....\}$
$$\mathsf P(T_6) = {n\choose 2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is
$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}... | You would be better off using Heron's formula. But there is an even better way: what is the maximum value that $\frac12ab\sin \theta$ can attain, if $a \le 1$, $b \le 2$, and $\theta$ can be any angle? Is this value attainable given the additional constraint $2 \le c \le 3$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Evaluating $\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$
Evaluate $$\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$$
This is a geometric series and since $a = \dfrac{1}{2}$ Then the infinite sum is jsut $S = \dfrac{1}{1-\frac{1}{2}} = 2$ Then I multiply by $2$ to get $4$ right? But the actual answer should just be $2$. Am I missing... | The first term in $\sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2}+\frac{1}{4}+\dots$ is $\frac{1}{2}$, so the sum of the geometric series is $\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.
Another way to do it is to absorb the $2$ into the denominator and then change the index from $n$ to $k = n-1$:
$$\sum_{n=1}^{\infty}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,... | Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 0
} |
Differentiate $f(x)=\dfrac{\sin^2(3x)}{2x}$ Consider following function
$$ f(x) =
\begin{cases}
\dfrac{\sin^2(3x)}{2x}, & x\neq0 \\
0, & x=0
\end{cases}$$
Evaluate $f'(0)$.
Is this function differentiable at $x=0$ ?
| Take the derivative first, using the quotient rule:
$$\frac{d}{dx}\left(\frac{\sin^2(3x)}{2x}\right)=\frac{2x\cdot2\sin(3x)\cdot\cos(3x)\cdot3-\sin^2(3x)\cdot2}{4x^2}$$
Then we need to take the limit as x approaches 0 of this expression:
$$\lim_{x\rightarrow 0} \frac{2x\cdot2\sin(3x)\cdot\cos(3x)\cdot3-\sin^2(3x)\cdot2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Improper Integral of $\int\frac{dx}{(2x-1)^3}$ Improper Integral of
$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$
from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$
The integral, substituting $u= (2x-1)$:
$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$
Definite solution to use ... | Rather than multiplying by $-2,$ you should be dividing, since $$\int u^{n}\,du=\frac1{n+1}u^{n+1}$$ for all integers $n\ne-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove reduction formula for $\int \cos^n (x)\sin^m (x) \, dx$ $$\displaystyle\int \:\sin^n\left(x\right)\cos^m\left(x\right)\mathrm dx=\frac{\sin^{n+1}x\cos^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \:\sin^nx\cos^{m-2}x\,\mathrm dx$$
I have been trying to solve for over a week now can someone please help me.
| To start, we rewrite:
$$I=\int\sin^m\left(x\right)\cos^n\left(x\right)dx=\frac{1}{m+1}\int\big[\sin^{m+1}(x)\big]'\cos^{n-1}(x)dx$$
Partial integrating:
\begin{align}
I&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)-\frac{1}{m+1}\int\sin^{m+1}(x)\big[\cos^{n-1}(x)\big]'dx\\
&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)+\frac{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$ I have a function:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do i... | It is important to remember that
$$(A+B)^2=A^2+\mathbf{2AB}+B^2$$
This what I'd do:
$$y(1+\sqrt x)=1-\sqrt x$$
$$y+y\sqrt x=1-\sqrt x$$
$$(y+1)\sqrt x=1-y$$
$$(y+1)^2x=(1-y)^2$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find Cartesian coordinates of polar curve $r =5\sin(\theta) + 5\cos(\theta)$ Polar equation of the form $r = 5\sin(\theta) + 5\cos(\theta)$
The Cartesian equation is of the form $(x-A)^2+(y-B)^2 = R^2$
Find $A,B$, and $R$.
Guess: Let $x = R\cos(\theta) + A$ and $y = R\sin(\theta)+B$. Plug them in and get on the left ha... | I would say
$r = 5\sin \theta + 5\cos \theta,\quad x = r\cos \theta, \quad y = r\sin \theta$
$\Rightarrow r^2 = 5x + 5y\Rightarrow x^2+y^2=5x + 5y\Rightarrow (x-\frac{5}{2})^2+(y-\frac{5}{2})^2=25/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Strange double integral What is wrong with this computation of $\int_0^1\int_{-y}^y \sqrt[3]{x} \, dx \, dy$?
I'm considering real functions only.
Since $x^{4/3}$ is an antiderivative of the integrand, we will get $\frac{3}{4}[x^{4/3}]_{-y}^y =\frac{3}{4}(y^{4/3}-(-y)^{4/3})=\frac{3}{4}(y^{4/3}-y^{4/3})=0$. Thus $\int_... | As others have already stated, Maple is interpreting $\sqrt[3]{x}$ as a complex function. Here's how it's being calculated
$$ \int_0^1\int_{-y}^y\sqrt[3]{x}\ dxdy= \int_0^1\int_{-y}^y x^{\frac{1}{3}}\ dxdy $$
$$ = \int_0^1\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{-y}^y dy = \int_0^1\left[\frac{x^{\frac{4}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\r... | I would say
$\displaystyle \lim_{x\to \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor=\lim_{x\to \infty} \lfloor \sqrt{x^2+x+1 }\rfloor+\{ \sqrt{x^2+x+1 } \}-\lfloor \sqrt{x^2+x+1 }\rfloor=$
$\displaystyle=\lim_{x\to \infty} \{ \sqrt{x^2+x+1 } \}= (0 \sim 1) \Rightarrow\, $limit not exist
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $\lfloor \sqrt x \rfloor = \lfloor x/2 \rfloor$ for real $x$ I'm trying to solve $$\lfloor \sqrt x \rfloor = \left\lfloor \frac{x}{2} \right\rfloor$$
for real $x$.
Obviously this can't be true for any negative reals, since the root isn't defined for such.
My approach is the following: Let $x=:n+r$, $n \in \math... | It is completely obvious that $x$ must be a non-negative real number. We decompose it as the following format:
$$x=k^2+n+r$$
in which $k^2$ is the largest integer less than or equal to $x$ in which is a squre of an integer number. So it is completely apparent that
$$n<(k+1)^2-k^2\rightarrow n<2k+1$$
$r$ is a real numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Number of real solutions of the equation $1+8^x+27^x = 2^x+12^x+9^x$
Find the number of real solutions $x\in\mathbb{R}$ of the equation
$$
1+8^x+27^x = 2^x+12^x+9^x
$$
My Attempt:
Let $2^x=a>0$ and $3^x=b>0$ where $x\in \mathbb{R}$. This allows us to change the equation to
$$
1+a^3+b^3 = a+a^2b+b^2
$$
This can be r... | A different method. Define:
$$f(a,b)= 1 + a^3 + b^3 - (a + a^2 b + b^2)$$
Show by solving $f'_a = f'_b=0$ that the minimum of $f$ is at $a=b=1$, and that $f(1,1)=0$, proving there is a single solution, $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | Nice Work
Just don't start by assuming the equality or inequality
Indeed there is another way, There's an elementary way
Since we are only concerned about acute angles
$$\left(a-\frac1a\right)^2\ge0$$
$$a^2-2+\frac1{a^2}\ge0$$
$$a^2+\frac1{a^2}\ge2$$
put $a^2=\tan x$
$$\tan x+\frac1{\tan x}\ge2$$
$$\tan x+\cot x\ge2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Lagrange Multipliers: Find $\min$ of $f(x,y)=3(x+1) +2(y-1)$ subject to the constraint $x^2+y^2=4$
Find the minumum value of the function $f(x,y)=3(x+1) +2(y-1)$, subject to the constraint that $x^2+y^2=4$.
The problem states to use Lagrange Multipliers. In doing so I obtained the point $(\frac 6{\sqrt{13}},\frac 4{\... | $$minimize \hspace{3mm} 3(x+1) + 2(y-1)$$
$$s.t. \hspace{3mm} x^2+y^2=4.$$
The lagragian function is $\mathcal{L} = 3(x+1) + 2(y-1) + \lambda (x^2+y^2-4)$. Thus,
$$\frac{\partial\mathcal{L} }{\partial x} = 3 + 2\lambda x = 0 \Rightarrow \lambda = \frac{-3}{2x},$$ and
$$\frac{\partial\mathcal{L} }{\partial y} = 2 + 2\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Intersection multiplicity of the curves I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.
That`s what I have tried:
$f$ and $g$ have a common tangent, the $y=0$.
So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg... | From the system of equations,
$$x^5+x^4+y^2=x^6-x^5+y^2=0,$$
you can eliminate $y$, and
$$-x^6+2x^5+x^4=0.$$
This polynomial has a quadruple root at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $\lim_{x\rightarrow 1}x^{\frac{1}{n-x-x^2-x^3-\ldots -x^n}}=\frac{1}{\sqrt[k]{e}}$ Proving
$$\lim_{x\rightarrow 1}x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\frac{1}{\sqrt[k]{e}}$$
when $$k=1+2+3+\ldots+n$$
| Hint: $$x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\exp\left(\ln x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}\right)=\exp\left(\frac{\ln x}{n-x-x^2-\ldots-x^n}\right)$$ and now due to continuity of $\exp$ it suffices to show with L' Hopitals rule that $$\frac{\ln x}{n-x-x^2-\ldots-x^n} \to -\frac{1}{1+2+3+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove using De Moivre's formula,that $\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}$ I've been asked to prove that:
$$
\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}
$$
When $0<x<2\pi$.
I know there are many similar posts on this site, but usi... | You asked to prove this by DeMovire's formula, which states $$e^{ikx} = (\cos x + \sin x)^k = \cos kx + \sin kx.$$
Everything you did until the final line is correct.
The final move is to do the following, $$\sin \left ( \frac{nx}{2} \right ) \frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}} = \frac{\cos(x/2) - \cos(n + 1/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt... | This gets easier if you compute $\sin 15^\circ$ in the same way that you computed $\sin 75^\circ$:
$$
\sin 15^\circ = \sin(45^\circ-30^\circ)=\sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ = \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \, .
$$
For an alternat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using ... |
We can also split the triangle into three smaller triangles using $\frac{1}{2} ab \sin C$. Therefore $\Delta ABC$ equals:
$$\frac{1}{2} r^2 \big(\sin \theta + \sin \theta + \sin(360º - 2 \theta ) \big)$$
$$\frac{1}{2} r^2 \big(2 \sin \theta + \sin(- 2 \theta ) \big)$$
Since the area is only dependent on $\theta$, we c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| Square both sides, and you get
$$4\cdot 3\overset{n\ times}{8\cdots8}9 = 4(9+8\sum_{k=1}^n 10^k+3\cdot 10^{n+1})=
36+32\sum_{k=1}^n 10^k +12\cdot 10^{n+1}= $$
$$=6+3\cdot 10+ 2\sum_{k=1}^n 10^k+3\sum_{k=1}^n 10^{k+1}+2\cdot 10^{n+1}+10^{n+2}=
$$
$$=
6+ 2\sum_{k=1}^{n+1} 10^k+ 3\sum_{k=1}^{n+1} 10^k +10^{n+2}=
6+5\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an ... | Note that
$$\left|\frac{1}{x_n}-\frac{1}{2}\right|= \left|\frac{2-x_n}{2x_n} \right|$$
Choose $N$ large enough that $x_n>1$ and $|x_n-2|<\epsilon$ for all $n>N$. Then
$$ \left|\frac{2-x_n}{2x_n}\right|\leq \epsilon/2< \epsilon $$
for all $n>N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $y'(x)\left(4-3y(x)x^2\right)=4x$
Solve the differential equation
$$y'(x)\left(4-3y(x)x^2\right)=4x$$
I would appreciate some help with this problem.
| $$\frac{dy}{dx}\left(4-3y(x)x^2\right)=4x$$
$$4-3yx^2=4x\frac{dx}{dy}$$
You can use the sustitution $u=x^2$, so $\frac{du}{dy}=2x\frac{dx}{dy}$
In the equation:
$$ 4-3yu=2\frac{du}{dy} $$
Rewriting:
$$ 2\frac{du}{dy}+(3y)u=4 $$
$$ \frac{du}{dy}+\frac{3}{2}yu=2 $$
That is a first order linear ordinary differential equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the determinant of a matrix definition Let $A$ be a matrix that is defined like this:
$$A_{ij}=\begin{cases}
\alpha, & \text{if i=j} \\
\beta , & \text{if i $\ne$ j}
\end{cases}
$$
So I realized this matrix looks somehow like this
$$
\begin{pmatrix}
\alpha & \beta & \beta \\
\beta & \alpha... | We know that each determinant should consist of $n!$ products of $n$ entries each if we have an $n\times n$ matrix. Considering each $\alpha$ on the diagonal, such an $\alpha$ is multiplied by the determinant of the $(n-1)\times(n-1)$ version of the matrix. So if we take the previous determinant, multiply it by $n\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
A Sum that came up while solving a integral While evaluating $I$, I did the following-
$$\begin{align}I= \int_{0}^{1} \log \left(\dfrac{1+x}{1-x}\right) \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x &= 2 \int_{0}^{1}\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1} \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x\\ &=2\sum_{n=0}^{\infty} \int... | Instead of series expansion, you can do this:
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx=\frac{1}{2}\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx$$
$$=\int_{-1}^1 \frac{\ln(1+x)}{x\sqrt{1-x^2}}\,dx=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+\sin\theta)}{\sin\theta}\,d\theta$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Why is this intuitive method valid? Problem. There are $2$ white and $3$ black balls in the urn. A person randomly picked $2$ balls and put $1$ white ball. What is the probability of the event that the next randomly-picked ball would be white.
To solve this problem formally you have to consider conditional probabilitie... | Grinding balls into powder is mathematically the same as calculating expected value.
In fact, the reason why mathematicians often speak of expected value calculations as the sum of the possible values, weighted by their probabilities, is exactly why your intuition of weights of grinded balls is relevant.
The expecte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why sometimes we get only one root of quadratic equations? What is logic behind getting (sometimes) only one root of a quadratic equation which satisfies the equation?
| The logic is that every positive number has two square roots, but zero only has one.
The quadratic equation comes from the algebraic process known as completing the square. Given an equation $ax^2 + bx + c = 0$, divide by $a$:
$$
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
$$
Then add $\left(\frac{b}{2a}\right)^2$ and to bot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
A faster way of calculating this determinant? I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3
\end{vmatrix}
So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal... | Determinants are not changed by adding multiples of a column to another. So
$$
\begin{vmatrix}1&1&1\\ a&b&c\\ a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}1&0&0\\ a&b-a&c-a\\ a^3&b^3-a^3&c^3-a^3\end{vmatrix}=(b-a)(c^3-a^3)-(b^3-a^3)(c-a)
$$
Edit: since in the comments to the other answer you were asking how to factor, here... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx$ I have tried to evaluate $$∫\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm d x$$ using the following identity:
$$\frac{d(\sin^{-1}{u})}{du} = \frac{du}{1+u^2}$$
So I then reformed the integral to this:
$$1/9\int\frac{\sin(8x)}{1+\sin^4(4x)/9}\,\mathrm dx = 1/9\int\frac{\... | $\sin^4 (4x) = \dfrac{\left(1- \cos (8x)\right)^2}{4} \Rightarrow u = \cos (8x) \Rightarrow du = -8\sin (8x)dx \Rightarrow \displaystyle \int \dfrac{\sin (8x)}{9+ \sin^4(4x)} dx = -\dfrac{1}{8}\cdot \displaystyle \int \dfrac{1}{9+\dfrac{(1-u)^2}{4}} du = -\dfrac{1}{2}\cdot \displaystyle \int \dfrac{1}{6^2 + (1-u)^2} du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding number of integral solutions I am really getting confused in this question.
Number of integral solutions of the equation.
$x_1x_2x_3x_4=770$
options-
*
*$2^{11}$
*$2^{10}$
*$4^4$
*$5^5$
I attemtemted it by saying that $7*11*5*2=770$
so solutions should be $4!$
then we can say $77*5*2*1=770$
so $4!$
and $7... | Case 1: $2\cdot 5\cdot 7\cdot 11 = 770 \to 4! = 24$
Case 2: $10\cdot 7\cdot 11\cdot 1 = 770 \to 4!$
Case 3: $14\cdot 5 \cdot 11\cdot 1 = 770 \to 4!$
Case 4: $22 \cdot 5\cdot 7\cdot 1 = 770 \to 4!$
Case 5: $35 \cdot 2\cdot 11\cdot 1 = 770 \to 4!$
Case 6: $55 \cdot 2\cdot 7\cdot 1 = 770 \to 4!$
Case 7: $77\cdot 2\cdot 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove two identities relating to series Show that:
$$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$
$$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$
Thank you in advance.
NOTE:
The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2... | For (1) Consider the product $P_N = \displaystyle\prod_{n = 1}^{N}\cos\dfrac{x}{2^n}$. Multiply by $\sin\dfrac{x}{2^N}$ and use the identity $\cos\theta\sin\theta = \dfrac{1}{2}\sin 2\theta$ repeatedly to get $P_N\sin\dfrac{x}{2^N} = \dfrac{1}{2^N}\sin x$.
Therefore, $P_N = \dfrac{\sin x}{2^N\sin\frac{x}{2^N}} \to \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Drawing the two associated curves, we see that there are $6$ simple solutions (the maximum according to Bezout theorem). Let $s=x+y,p=xy$. If $(x,y)$ is a solution, then $(-y,-x)$ is also a solution. Thus the solutions in $p$ have multiplicity $2$ and $p$ is any root of a polynomial of degree $3$. In the same way, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater... | Hint:
$$(a-b+1)^2=a^2+b^2+1-2ab-2b+2a$$
Solution:
$$3a^2+3b^2-2b+2a+1=2a^2+2b^2+(a-b+1)^2+2ab=a^2+b^2+(a+b)^2+(a-b+1)^2>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{-2}^{-1}\frac{\text{d}x}{\sqrt{-x^2-6x}}$. Problem statement [from Charlie Marshak's Math GRE Prep Problems]:
Evaluate $\displaystyle \int\limits_{-2}^{-1}\dfrac{\text{d}x}{\sqrt{-x^2-6x}}$.
My work: notice that
$$\begin{align}
-x^2-6x &= -(x^2 - 6x) \\&= -\left[x^2-6x+\left(\dfrac{6}{2}\right)^2-\le... | Where you have $x-3$ you should have $x+3$. The exact point you went wrong is in the first line, where you put $-6x$ rather than $+6x$. There is already a minus sign. This is the error:
$$
-x^2-6x \neq -(x^2 - 6x)
$$
$$
-x^2-6x = -(x^2 + 6x)
$$
Note also that $\sin^{-1}(x)$ is meaningless if $|x|>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmap... | As you have found $Aut_{\mathbb{Q}}L \simeq S_3$ then the intermediate fields of $Gal (X^3-2, \mathbb{Q})$ are in correspondence to the subgroups of $Aut_{\mathbb{Q}}L$, according to the Fundamental Theorem of Galois. Now,
$$S_3 = \{Id, \sigma, \sigma^2, \tau, \sigma\tau, \sigma^2\tau\}$$
Notice that there is a elemen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
solving for sin of the sum of two angles of a triangle In triangle $ABC$,
$3\sin B+4\cos C=6$ and $4\sin C+3\cos B=1$.
Show that $\sin(B+C)=0.5$.
Can we assume $\angle A = 180 - ( B + C)$ and use sum formula.
| Hint: By squaring both sides of the equation:
$$9\sin^2 B + 24 \sin B\cos C + 16 \cos^2 C = 36$$
$$9\cos^2 B + 24 \cos B\sin C + 16 \sin^2 C = 1$$
By adding the equations:
$$9(\sin^2B+\cos^2B)+24(\sin B\cos C+\cos B\sin C)+16(\cos^2 C+\sin^2C)=37$$
Using $\sin^2x+\cos^2x=1$
$$9(1)+24(\sin B\cos C+\cos B\sin C)+16(1)=37... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding $\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$ I am working on this double integral
$$\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$$
so far, I don't know how to start.
Can someone give a hint?
Thanks
| You can employ the following formula
$\displaystyle\bbox[#EFF,15px,border:3px solid blue]{\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin^3(y)f\big[\sin\left(x\right)\sin\left(y\right)\big]{\mathrm{d}}x{\mathrm{d}}y=\frac{\pi}{4}\int_{0}^{1}\left(t^{2}+1\right)f\left(t\right){\mathrm{d}}t}$
Then
\begin{align*} &\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ I was just playing around with a calculator, and came to the conclusion that:
$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$
Now I'm curious. Is it possible to evaluate the exact value of ... | It is easy to show, assuming the limit exists:
$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$
It is also easy to see the following set of inequalities:
$$\sqrt{\frac{1}{2}+1} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}$$
$$\sqrt{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 1
} |
Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| i will try $x = {2t \over 1 + t^2}$
$$dx = 2{ (1+t^2)dt - t \cdot2tdt \over (1+t^2)^2} = {2(1-t^2)dt \over (1+t^2)^2}
\\ 1 - x = {(1-t)^2 \over 1 + t^2}, 1 + x = {(1+t)^2 \over 1 + t^2}, \sqrt{{1-x \over 1 + x}}= {1 - t \over 1 + t} $$
putting all these together,
$$\int{1 \over x} \sqrt{{1-x \over 1 + x}} dx = \int{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Algebra on a Louvre tablet Problem: On a Louvre tablet of about 300 B.C. are four problems concerning rectangles of unit area and given semiperimeter. Let the sides and semiperimeter be $x,y$ and $a$. Then we have
\begin{equation}
xy=1, \qquad x+y=a. \tag{1}
\end{equation}
Solve this system by using the identity
$$
\bi... | I think the intended solution was as follows:
$$
\left(\frac{x-y}{2}\right)^2=\left(\frac{a}{2}\right)^2-1.
$$
Thus $x-y$ can be solved for:
$$
x-y=\pm2\sqrt{\left(\frac{a}{2}\right)^2-1}=\pm\sqrt{a^2-4}
$$ Adding and subtracting with $x+y$ gives us:
$$
x=\frac{(x+y)+(x-y)}{2}=\frac{a\pm \sqrt{a^2-4}}{2},
$$
and likewi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 ... | There are many great answers already written up, so I'm not sure if this is going to add anything, but for what it's worth, here's how I did it.
Recognise that $\underbrace{555\cdots555}_{1000\ \text{times}} = 5\cdot \underbrace{111\cdots111}_{1000\ \text{times}} = 5 \cdot 9^{-1} \cdot \underbrace{999\cdots999}_{1000\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 8
} |
Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection? This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, n... | Model of the projection
Let the light come from above, along the $z$-axis.
The shadow of an triangle above the $x-y$-plane is then the projection onto the $x-y$-plane ($z = 0$).
The triangle has a unit normal vector $n$ to describe its orientation ($n^2 = 1$).
The shadow projection $P$ is
$$
P =
\left(
\begin{matrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Evaluating the sum : $\;\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$ How to evaluate this sum?
$$\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$$
Please give some technique. Binomial not working.
| Note
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{(n+2)n!}&=\sum_{n=1}^{\infty}\dfrac{n+1}{(n+2)!}=\sum_{n=1}^{\infty}\dfrac{(n+2)-1}{(n+2)!}=\sum_{n=1}^{\infty}\left(\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}\right)\\
&=\dfrac{1}{2}-\lim_{n\to\infty}\dfrac{1}{(n+2)!}\\
&=\dfrac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Quadratic equation with parameter and conditions on the roots For which values of $m$ the equation
$$(2m+1)x^2-(4m+2)x+m-1=0$$
has two real roots $x_1$ and $x_2$ that satisfy the condition: $x_1<x_2<2$?
I found that $$m\in\left(-\frac{1}{2}; -\frac{1}{3}\right).$$ Is it correct?
| The discriminant is $$\Delta=b^2-4ac=(4m+2)^2-4(2m+1)(m+1)=8m^2+20m+8$$ To get real roots, $m$ must satisfy $$\Delta>0$$
which reduces as $$2m^2+5m+2>0$$
and has for solution $m>-1/2$ or $m<-2$. The roots are then $$x_{2,1}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{4m+2\pm\sqrt{8m^2+20m+8}}{4m+2}$$
with $x_1=1-\frac{\sqrt{8m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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An infinite matrix series Playing around with my CAS, I found that apparently
$$\sum_{n=1}^\infty\begin{bmatrix}1/2 & 1/3\\1/4 & 1/5\end{bmatrix}^n = \begin{bmatrix}29/19 & 20/19\\15/19 & 11/19\end{bmatrix}$$
If that is indeed true, it's pretty cool.
How can I prove this, and what is the general formula if the coeffici... | Given any $n \times n$ square matrix $A = (a_{ij})$ over a field $K = \mathbb{R}$ or $\mathbb{C}$.
If its operator norm $\|A\|_{op}$:
$$\|A\|_{op} \stackrel{def}{=} \sup\big\{\; \|Av\| : v \in K^n \text{ with } \|v\| = 1 \;\big\}$$
or any sub-multiplicative matrix norm, e.g. the
Frobenius norm
$\|A\|_F$:
$$\|A\|_{F} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Hint for Estimating an Infinite Series I am currently working through E. V. Shchepin's book Uppsala Lectures on Calculus: On Euler's Footsteps, and I've encountered a problem with estimating an infinite sum.
I need a hint estimating: S = $\sum_{k=1}^\infty{1 \over2^kk}$. This problem appears early in the book in the s... | Write
$$
\frac{1}{k} = 1 - \frac{1}{1\cdot 2} -\frac{1}{2\cdot 3} -\frac{1}{3\cdot 4} \cdots \frac{1}{(k-1)k}
$$
Then group terms to get
$$S = 1 \sum \frac{1}{2^k} - \frac{1}{1 \cdot 2} \sum \frac{1}{2^{k+1}} + \cdots
$$
So for example your fifthapproximation will be
$$
1 - \frac{1}{2} \frac{1}{2} - \frac{1}{4} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability of rolling a sum of N with up to infinite rolls of a die I'm trying to figure out if there is some (relatively simple) formula for calculating the probability of rolling a sum of N with as many rolls as needed with a single regular six-sided die.
For example:
$N=1$ is $0.16666 = 1/6 = 1/6$ (1)
$N=2$ is $0.1... | Let $q(x)=\frac{1}{6}\left(x^1+x^2+x^3+x^4+x^5+x^6\right)$. Write:
$$\begin{align}G(x)&=\sum_{n=0}^\infty q(x)^n\\&=\frac{1}{1-q(x)}\\&=\frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}
\end{align}$$
Then the coefficient of $x^N$ in $G(x)$ is your probability.
So you need to know something about the roots of the denominator. $1$ is on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
| Let $$f(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}.$$
Then taking the derivatives wrt $x$, the factors in the denominator disappear:
$$f''''(x)=\sum_{n=0}^{\infty }(-x^2)^n=\frac1{1+x^2}.$$
Integrate four times from $0$ to $x$ (factors added for convenience).
$$\begin{align}f'''(x)&=\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Investigate convergence of the series Investigate convergence of the series:
$$\left( \frac{n^2+3n+10}{n^2+5n+17} \right)^{n^2 (\sqrt{n+1}-\sqrt{n-1})}$$
It should be solvable with simple manipulations with the formula, i guess, but how to do that?
| Let $a_n$ be the above term. Note first that
$$\frac{n^2+3 n+10}{n^2+5 n+17} = \left (1+\frac{3}{n}+\frac{10}{n^2} \right ) \left (1+\frac{5}{n}+\frac{17}{n^2} \right )^{-1} = 1-\frac{2}{n}+\frac{3}{n^2}+O\left (\frac1{n^3}\right )$$
Also,
$$n^2 \left ( \sqrt{n+1}-\sqrt{n-1}\right ) = n^{3/2} \left (1+\frac{3}{8 n^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pma... | Let's look at your first identity. Express everything in terms of $F_{n-1}$ and $F_n$.
$$ \eqalign{F_{n-2} F_{n-1} F_{n+3} - F_{n}^3 &= 3 F_{n}^2 F_{n-1} - F_n F_{n-1}^2 - 2 F_{n-1}^3 - F_{n}^3 \cr
&= (F_n - 2 F_{n-1})(-F_n^2 + F_n F_{n-1} + F_{n-1}^2)\cr
(-1)^n F_{n-3} &= -(-1)^n (F_n - 2 F_{n-1})\cr}$$
so the ident... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$
Any method for solving this problem? I've not achieved anything so far. Thanks in advance!
| For $k=0$, the answer is $n$.
For $k>0$, the answer is ${{n+1}\choose{k+1}}$.
This can be seen by considering how many ways there are to choose $k+1$ integers from the set $\{1,2,\dots,n+1\}$ and conditioning on the greatest integer chosen, because there have to be $k$ choices less than the greatest integer chosen.
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Finding the derivative $f(x)=\sqrt{x^2 -9}$, I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,
$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$
The answer book says the slope is ${1\over 4}$
Here's what I did,
$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)... | In step 6, the denominator should be $2 \sqrt{x^2-9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$ What is the value of $\displaystyle\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$?
This is a question I came up with myself. It is not homework.
I constructed this example to make the following technique work:
Integrate $\... | \begin{align}
\int^\infty_0\frac{x\ln^2{x}}{x^4+x^2+1}dx
&=\frac{1}{8}\int^\infty_0\frac{\ln^2{x}}{x^2+x+1}dx\\
&=\frac{1}{4}\int^1_0\frac{(1-x)\ln^2{x}}{1-x^3}dx\\
&=\frac{1}{4}\sum^\infty_{n=0}\int^1_0\left(x^{3n}-x^{3n+1}\right)\ln^2{x}dx\\
&=\frac{1}{2}\sum^\infty_{n=0}\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Writing a piecewise function for $f(x) = \mid x+3\mid -\mid x-1\mid $
I am wanting to write a piecewise function for the following:
$$f(x)= \mid x+3\mid -\mid x-1\mid $$
I know how to write piecewise functions for functions that have a single set of absolute value brackets, but I don't know how to deal with two sets ... | The idea is the following:
The absolute value of $y$ is given by
$$|y| = \begin{cases} y&\text{if } y\geq 0\\-y&\text{if } y<0\end{cases}$$
The point is now to check when $x+3$ and $x-1$ are positive, and when they are negative.
Let us look at the case where $x-1\geq 0$ first. Then $x\geq 1$, and it trivially follows t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of $\sum\limits_{n=1}^\infty \bigl[ \frac{(3 - \sqrt 5)^n}{2^nn^3}\bigr]$ Please help me to find the sum of
$\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$
Is there any special technique to solve this one ?
| According to Maple,
$$ \text{polylog}\left(3, \dfrac{3-\sqrt{5}}{2}\right) = \dfrac{4}{5} \zeta(3) + \dfrac{\pi^2}{15} \ln \left(\dfrac{3-\sqrt{5}}{2}\right) - \dfrac{1}{12} \ln\left(\dfrac{3-\sqrt{5}}{2}\right)^3 $$
I don't know where it gets this rather remarkable identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.... | Use the Brahmagupta-Fibonacci Identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$
This identity can be verified by multiplying out each side, and in nicer ways.
From the Identity, we get
$$65^2=(8^2+1^2)(7^2+4^2)=(8\cdot 7-1\cdot 4)^2 +(8\cdot 4+1\cdot 7)^2.$$
We can get another representation of $65^2$ as the sum of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Example of Stirling Numbers of the First Kind I am trying to calculate the stirling numbers of the first kind. I am not very good in math and there is not a single example somewhere in the internet.
So I would really appreciate it if you could help me there.
The stirling number which I want to calculate is s1(4,2). I k... | You could use the rising factorial to find $\left[\begin{array}{c} 4 \\ 2 \end{array}\right]$. It goes as follows:
$$(x)^4 = x(x+1)(x+2)(x+3) = x^4 + 6x^3 + 11x^2 + 6x$$
$\left[\begin{array}{c} 4 \\ k \end{array}\right]$ is now the coefficient with $x^k$. Thus $\left[\begin{array}{c} 4 \\ 2 \end{array}\right]$ = 11.
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
System of congruences and Chinese remainder theorem Find all the integers satisfying this system of congruences
$$\begin{cases}
x \equiv 2 \pmod 5\\
x \equiv 1 \pmod {10}\\
x \equiv 0 \pmod 3
\end{cases}
$$
I think you use Chinese remainder theorem but I'm not sure how to.
| Basically the two equivalences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
have a common solution if and only if $$a \equiv b \pmod{\gcd(A,B)}$$
In the case of \begin{align}
x &\equiv 2 \pmod 5\\
x &\equiv 1 \pmod {10}\\
x &\equiv 0 \pmod 3
\end{align}
We notice that $1 \not \equiv 2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a... | Dividing $a$ by $3$ allows to write $a=3q+r$. Now
$$
(3q+r)^3=27q^3+9q^2r+3qr^3+r^3=3s+r^3
$$
collecting together all multiples of $3$ in the expansion. Dividing again the result by $3$ we have
$$
(3q+r)^3=3Q+R
$$
where $R$ is the remainder of the division by 3 of $r^3$ (because $3s$ is a multiple of $3$).
But there ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $ Calculate $S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!
| $$u_k = \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2k(k+1)} - \dfrac{1}{2(k+1)(k+2)} =
f_k - f_{k+1}$$
so $$u_1 + u_2 + \cdots u_n = (f_1 - f_2) + (f_2 - f_3) + \cdots + (f_n - f_{n+1}) = f_1 - f_{n+1} =
\dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Check: Find and plot all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ . Find all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ .
For each natural number $n$ there are exactly $n$ $n$-th roots of unity, which can be expressed as: $$z_k=e^{i\frac{2\pi}{n}k},k=0,1,\cdot, n-1$$ In our c... | First, start with finding the polar notation of $7 - 3i$ which is :
$$7-3i = \sqrt{58}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Then go even further to get only exponential :
$$7-3i = e^{\frac{1}{2}\ln{(58)}}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Now, it's pretty simple : the $n$th roots of $7-3i$ are :
$$e^{\frac{1}{2\cdot 6}\ln{(58... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$ Proving $$\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$$
By using the numerical calculation I saw that the convergence of product series is slow, so I need the proving.
thanks.
| Use the infinite product representation: $\displaystyle \dfrac{\sin \pi x}{\pi x} = \prod\limits_{n=1}^{\infty} \left(1-\frac{x^2}{n^2}\right)$ to rewrite the product $$\displaystyle \prod_{n=1}^{\infty }\left(1+\frac{3/4}{4n^2-1}\right) = \prod_{n=1}^{\infty }\frac{\left(1-\frac{1}{16n^2}\right)}{\left(1-\frac{1}{4n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Help me find the following limit : $\lim_{{n}\to{\infty}} (\frac{2^x+3^x+\cdots+n^x}{n-1})^\frac{1}{x} = ?$ I have no idea where to start.$$\begin{align}\lim_{{n}\to{\infty}} \left(\dfrac{2^x+3^x+\cdots+n^x}{n-1}\right)^{1/x} = ?, n >1\\\end{align}$$
| Another approach:
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\le\left(\frac{n-1}{(n-1)2^x}\right)^{1/x}=\frac12\implies \sum_{n=1}^\infty\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}$$
converges, and then
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\xrightarrow[n\to\infty]{}0\implies\left(\frac{2^x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Volume using triple Integrals, cylindrical coordinates I want to calculate the volume of a solid with $z+1\ge x^2+y^2$ and $3\left(z-1\right)\le -\left(x^2+y^2\right)$. After cylindrical coordinates
x=rcosϕ, y=rsinϕ
I got $r^2-1\le z$, and $z\le \frac{\left(3-r^2\right)}{3}$ , so $r^2-1\le z\le \frac{\left(3-r^2\right... | Since $r^2-1=1-\frac{r^2}{3}\implies r^2=\frac{3}{2}\implies r=\sqrt{\frac{3}{2}}$,
$V=\displaystyle\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\int_{r^2-1}^{1-\frac{r^2}{3}} r\;dz dr d\theta=\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\left(2r-\frac{4}{3}r^3\right)\;dr d\theta$
$\displaystyle\;\;=\int_0^{2\pi}\left[r^2-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
|
Sketch indicates the truth of the proposition. Blue boxes : $x+y+z > \sqrt{3}$.
Blue with purple centers $$xy + yz + zx > \frac{1}{\sqrt{x^2+y^2+z^2}}$$
The surfaces indicate where equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
The formula of the order of multistep methods How can I derive this $$(1+\xi) \left(1+\frac{1}{2}\xi-\frac{1}{12}\xi^{2}\right)+O(\xi^3)$$ from
$$\frac{1+\xi}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^{2}}+O(\xi^3)$$
?
The whole formula is below. This is from "A first course in the numerical analysis of differential equations by... | Use the series $\frac1{1-x}=1+x+x^2+O(x^3)$ to get
$$
\begin{align}
\frac1{1-\frac12\xi+\frac13\xi^2}
&=1+\left(\frac12\xi-\frac13\xi^2\right)+\left(\frac12\xi-\frac13\xi^2\right)^2+O(\xi^3)\\
&=1+\frac12\xi-\frac13\xi^2+\frac14\xi^2+O(\xi^3)\\
&=1+\frac12\xi-\frac1{12}\xi^2+O(\xi^3)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor series expansion of the function $f(x)=x \arctan x-0.5 \log(1+x^2)$ about the origin int the region {$|x|\le1$} Find the Taylor series expansion of the function
$\color {green}{f(x)=x \tan^{-1} x-0.5 \log(1+x^2)}$ about the origin int the region {$|x|\le1$}
My effort: I know $\displaystyle \log (1+x)=x-x^2/2+x... | it is easier to find the maclaurin series for
$$f^\prime(x) = \tan^{-1}(x) = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} + \cdots$$ and integrate this series alsin rthe fact that $f(0) = 0$ which gives you
$$f(x) = x\tan^{-1}x - 0.5 \ln{(1+x^2)} = \frac{x^2}{2} - \dfrac{x^4}{3.4} + \dfrac{x^6}{5.6} + \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Three mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circles Three circles of different radii are tangent to each other externally. The distance between their centers are $9\ cm$, $8\ cm$, and $11\ cm$.
*
*Find the radius of each circle.
*Fi... | Solving for the radii, we get $\{3,5,6\}$.
For the area we get
$$
\begin{align}
&\sqrt{14(14-8)(14-9)(14-11)}\\[9pt]
&-\frac{(8+9-11)^2}8\arccos\left(\frac{8^2+9^2-11^2}{2\cdot8\cdot9}\right)\\
&-\frac{(11+8-9)^2}8\arccos\left(\frac{11^2+8^2-9^2}{2\cdot11\cdot8}\right)\\
&-\frac{(9+11-8)^2}8\arccos\left(\frac{9^2+11^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Case $a+b\mid a^a+b^b$
Suppose $a$ is odd. Because $a+b\mid a^a+b^a$ we have $a+b\mid b^b-b^a$, hence $a+b\mid b^{|b-a|}-1$ because $\gcd(a,b)=1$, hence $a-b\mid a+b-1$ by Fermat. By symmetry, we get the same if $b$ is odd.
Case $a+b\mid a^b+b^a$
Suppose $a$ is odd. Because $a+b\mid a^a+b^a$ we have $a+b\mid a^b-a^a$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Integrating: $ \;\int \frac{1}{x^2+3x+2} dx $ How can I solve the following integral:
$$ \int \frac{1}{x^2+3x+2} dx $$
Should I proceed by changing the variable (substitution)? or should I use integration by parts? Or another method altogether?
Thank you!
| Just for fun, I tried it another way.
Complete the square
$$
\frac{1}{x^2+3x+2} = \frac{1}{(x+\frac{3}{2})^2-\frac{1}{4}}
$$
Change variables $y=x+\frac{3}{2}$
$$
\int\frac{dx}{x^2+3x+2} = \int\frac{4}{4y^2-1} = {-2\;\mathrm{atanh}(2y)+C}
$$
Substitute back
$$
\int\frac{dx}{x^2+3x+2}=-2\;\mathrm{atanh}(2x+3) + C
$$
Che... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (... | Use only your first identity. You should check where the series is absolutelly convergent. In that interval you can change the integral with the sum. That will lead to a known series you shall be able to evaluate. (If you can't solve with the hint I edit this answer later).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Ring with many one-sided zero-divisors Does there exist a ring all of whose elements are left zero-divisors but only one element is a right zero-divisor?
The motivation for asking this question is that if there exists atleast one left zero-divisor there exist atleast one-right zero divisor. Now, it seems a natural ques... | It seems the following.
Such a non-trivial ring is unique. Indeed, assume that $R$ is a ring all of whose elements are left zero-divisors but only one element $r\ne 0$ is a right zero-divisor and $R$ contains at least three distinct elements. Let $x\in R$ be an arbitrary non-zero element. Since $x$ is a left zero-divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Triangle in parabola I have a problem. In my triangle one vertex is in the vertex of the parabola and two others are in parabola. This is a isosceles triangle and I know one angle in this triangle : 120 grades. The question is: angle 120 grades must be beside of the vertex of parabola?
Thanks a lot for answer, I am beg... | choose the coordinates so that the equation of the parabola is $y = x^2.$ let the triangle be $OAB$ with $O =(0,0), A = (a, a^2), B= (b, b^2)$ and $\angle AOB = 120^\circ$ by cosine rule, $$AB^2 = OA^2 + OB^2 + OA*OB $$ so that
$$(a-b)^2 + (a^2 - b^2)^2 = a^2 + a^4 + b^2 + b^4 + \sqrt{(a^2 + a^4)(b^2+b^4)} $$
cleaning... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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I need my work checked(properties of operations) Let $x * y = |x + y|.$
$x * y = |x + y| = |y + x| = y * x,$ so $*$ is commutative.
$(x * y) * z = ||x + y| + z| = |x + |x + z|| = x * (y * z),$ so $*$ associative.
$x * e = |x + e| = x,$ so $e = 0.$ Further, $e * x = |0 + x| = x.$ So, $*$ has an identity.
$x * x' = |x +... | For number 5:
it is not associate.
you did not check:
$(x∗y)∗z=x∗(y∗z)$
Also, you said "$xe+1=x$, so $e=0$" well if $e=0$ then we get $1=x$ but we need $x=x$ either way, you are right there is no identity but you did not provide the right reason.
This is important: If there is no identity then there is no inverse. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$ is a real number at a time before the invention of complex numbers I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that ... | I think it's a stretch to claim that this result can be proven without complex numbers, especially because of the presence of $\sqrt{-7}$. Anyway, here's a way to look at it. Let $A = \sqrt[3]{-1 + \sqrt{-7}}$ and $B = \sqrt[3]{-1 - \sqrt{-7}}$. If $t = A + B$, then $$t^3 = A^3 + B^3 + 3AB(A + B) = (-1 + \sqrt{-7}) + (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Does the limit of this double sequence exist? Consider
$$a_{mn}=\frac{m^2n^2}{m^2+n^2}\left(1-\cos\left(\frac{1}{m}\right)\cos\left(\frac{1}{n}\right)\right)$$
Does $\lim_{m,n\to\infty}a_{mn}$ exist?
It can be seen that $$\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{... | We have
\begin{align}a_{mn} &= \frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 + \frac{1}{2m^2} + O\left(\frac{1}{m^4}\right)\right)\left(1 + \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)\right)\\
&=\frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 - \frac{1}{2}\left(\frac{1}{m^2} + \frac{1}{2n^2}\right) + O\left(\frac{1}{m^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find $\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$ Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.
From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boya... | Hint :
by $$u=\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}$$
$$\Rightarrow \sqrt{1-4x}=\frac{1}{2u-1}$$
$$\Rightarrow x=\frac{1}{4}(1-\frac{1}{(2u-1)^2})$$
$$dx=\frac{1}{(2u-1)^3}du$$
you get $$I=\int_1^{\infty}\frac{\ln u}{u(u-1)}du=-\sum_{n=0}^{\infty}\int_1^{\infty}u^{n-1}\ln u du=-\sum_{n=0}^{\infty}-\frac{1}{n^2}=\frac{\pi^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the smallest fraction produced by a sum of fractions with bounded denominator? For $x$ a sum of fractions:
$$
x = \sum_{i=1}^{N}\frac{a_i}{b_i}
$$
for all $a_i, b_i \in \mathbb{Z}$ with $ 0 < b_i \leq D$ and $N$ are non-zero positive integers, I know that the denominator of $x$ will not exceed the least common ... | The LCM of $1, \ldots, 12$ is $27720$. One example with sum $1/27720$ (there are infinitely many others) is
$$ \dfrac{3}{7} + \dfrac{1}{8} + \dfrac{ 2}{9} + \dfrac{3}{10} + \dfrac{1}{11} -\dfrac{14}{12}$$
Hint: if $L(d)$ is the LCM of $1, \ldots, d$, then
$$ \dfrac{1}{L(d)} = \dfrac{\gcd(d,L(d-1))}{d\; L(d-1)}$$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find $z$ that satisfies $z^2=3+4i$ Super basic question but some reason either I'm not doing this right or something is wrong.
The best route usually with these questions is to transform $3+4i$ to $re^{it}$ representation.
Ok, so $r^2=3^2+4^2 = 25$, so $r=5$. And $\frac{4}{3}=\tan(t)$ so that means $t \approx 0.3217$ a... | We can use the double angle formula to find the components of $z$:
$$ \cos \theta = \pm \sqrt{\frac{1 + \cos 2\theta}{2}}$$
$$ \sin \theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}$$
Here $\theta$ is the argument of $z$ and $2\theta$ the argument of $z^2$.
We know $ \cos 2\theta = \frac{3}{5}$ so $\cos \theta = \pm \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simple algebra involving trigonometry, but confusing How do I get from
$$\frac{\sqrt3}{2} + \frac12 \tan x = 2 \tan x \cdot \frac{\sqrt3}{2}$$
to
$$\frac{\sqrt3}{2 \sqrt3 - 1} = \tan x$$
and then to
$$11 \tan x = 6 + \sqrt3$$
| $$\frac{\sqrt{3}}{2\sqrt{3}-1}=\tan(x)$$
$$\frac{\sqrt{3}}{2\sqrt{3}-1}*\frac{2\sqrt{3}+1}{2\sqrt{3}+1}=\tan(x)$$
$$\frac{6+\sqrt{3}}{11}=\tan(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\frac{abc}{1+bc}+\frac{bcd}{1+cd}+\frac{cda}{1+ad}+\frac{dab}{1+ab}\le \frac{1}{17}$ Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$
My attempt:
I figured out that if each of the eleme... | You could approach it as follows:
$$
\sum_{cyc} \frac{abc}{1+bc}=\sum_{cyc} a\left(1-\frac{1}{1+bc}\right)=\sum_{cyc} \left(a-\frac{a}{1+bc}\right)=1-\sum_{cyc} \frac{a}{1+bc}
$$
So the inequality is equivalent to:
$$
1-\sum_{cyc} \frac{a}{1+bc}\le\frac{1}{17}\iff\frac{16}{17}\le\sum_{cyc} \frac{a}{1+bc}
$$
Using CS, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability of $(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$
A fair die is thrown three times. If $a$, $b$, $c$ are the numbers obtained on the die, then what is the probability that $$(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$$
(where $\omega$ is a cube root of unity)
My attempt:
On simplification, $a^... | Hint: Since $a,b,c\in\mathbb R$ and $\omega$ is the complex cube root of unity, we can write
$$a^3+b^3+c^3-3abc=(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)(a+b+c)$$
$$(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)=\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
Now, we only have to find out how many integer solutions are there for $a^3+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition I am having difficulties manipulating the problem so that I can find a $N$ value to choose.
Suppose $x > N$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| \leq \frac{4x^2}{|2x^3 - 5|} + \frac{7}{|2x^3 - 5|} $$
This is basical... | Not quite, you made your denominator larger there, so that step is not obvious. That is to say, $2x^3-5\le 2x^3\iff {1\over 2x^3-5}\ge {1\over 2x^3}$, so you cannot conclude that ${4x^2-7\over 2x^3-5}\le {4x^2\over 2x^3}$. Instead I recommend the following:
For $x>2$ we have $4x^2>4x^2-7$ and $2x^3-5\ge 2x^3-x^3$, i.e.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that the sum is always greater than $1$ Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+ \frac{1}{n^2} \ge 1$, for all natural numbers $n$.
I tried to use mathematical inducion but failed and I tried to figure out a short formula for the sum but I couldnt find any.
| Using the Arithmetic-Harmonic Inequality:
$$\frac{n^2-n+1}{n+(n+1)+\cdots+n^2}\leq\frac{1}{n^2-n+1}\left(\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{n^2}\right)$$
Now, the denominator of the left-hand fraction is:
$$\sum_{k=0}^{n^2-n}(n+k)=\sum_{k=1}^{n^2-n+1}(n+k-1)=\sum_{k=1}^{n^2-n+1}(n-1)+\sum_{k=1}^{n^2-n+1}k=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to solve $y′′′+y'=2-\sin(x)$ I have tried to solve this but with no luck.
So far I just get,
$$y_p(x) = A \sin x + B \cos x \\
y'_p(x) = A \cos x - B \sin x \\
y''_p(x) =-A \sin x - B \cos x \\
y'''_p(x) =-A \cos x + B \sin x $$
$$y'''+ y' = -A \cos x + B \sin x + A \cos x - B \sin x \\ \\
= \cos... | Rewrite the equation as $$(D^3+D)y=2-\sin(x)$$
Differentiate twice to obtain
$$\begin{align}D^2(D^3+D)y&=\sin(x)\\
\implies (D^2+1)(D^3+D)y&=2\\
\implies D(D^2+1)(D^3+D)y=D^2(D^2+1)^2y&=0\end{align}$$
Which I leave to you to show has the solution
$$y=C_1+C_2x+(C_3+C_4x)\sin(x)+(C_5+C_6x)\cos(x)$$
Plug this back into th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Monkeys Dividing Pile of Bananas Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally b... | Let $A,B,C$ be the banana's removed by monkeys $1,2,3$ respectively and $x$ be the number of bananas monkey $3$ has at the end.
Then:
$\frac{1}{8}A+\frac{3}{8}B+\frac{1}{12}C=x$
$\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=2x$
$\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=3x$
Solving the system we get:
$A=\frac{22}{17}x$
$B=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What are all the twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime? It seems that $p = 3$ and $q = 5$ ($pq - 2 = 13$) are the only solutions. However I'm having a difficult time proving this.
I have that all primes can be represented as $3k + 1$ or $3k + 2$ so if $p = 3k + 2$ then $q = 3k + 4$ and
\beg... | We notice that for $p=3$ and $q=5$ we have a solution.
Let $p \geq 5$
Then p has form $p=6k \pm 1$ where $k \in N$
If $p=6k+1$ then $q=6k+3=3(2k+1)$ so $q$ is not a prime then. That leaves us with $p=6k-1$. Then $q=6k+1$ so $pq-2=(6k-1)(6k+1)-2=36k^2-3=3(12k^2-1)$ which is not a prime.
We conclude that there is no so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How to prove this floor function equation? How can I prove the following equation?
$$
\lfloor nx \rfloor =
\lfloor x \rfloor +
\Big\lfloor x + \frac{1}{n} \Big\rfloor +
\Big\lfloor x + \frac{2}{n} \Big\rfloor +
\Big\lfloor x + \frac{3}{n} \Big\rfloor +
\Big\lfloor x + \frac{4}{n} \Big\rfloor +
\Big\l... | Try to rewrite $x$ as
$$x=\lfloor x \rfloor + \frac kn + \alpha$$
where $0\le k<n$ and $0\le\alpha<\frac1n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integration by parts or using u-sub? Integrate: $4(x^3+1)^{\frac{1}{3}}(x^5)$
I tried using u-sub but it is getting me absolutely nowhere. I also tried using integration by parts and eventually was able to evaluate the integral but it does not seem to simplify to the answer. Thank you for your help.
| Maybe this will help: let $u = x^{3} + 1$. Then $u - 1 = x^{3}$, and $du = 3x^{2}\, dx$ (i.e., $\frac{1}{3} \, du = x^{2} \, dx$), so we get:
$\int 4(x^{3} + 1)^{\frac{1}{3}}(x^{3}x^{2}) \,dx = \int \frac{4}{3}(u)^{\frac{1}{3}}(u - 1) \,du = \frac{4}{3} \int u^{\frac{4}{3}} - u^{\frac{1}{3}} \,du = \frac{4}{3}(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Area of triangle bounded by line and degenerate "crossed lines" conic The question is
Show that the two lines given by
$$(A^2 - 3B^2)x^2 + 8ABxy +(B^2 - 3A^2)y^2=0$$
and the line given by $$Ax+By+C=0$$ determine an equilateral triangle of area $$\frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
I tried factorizing the pair of... | The crossed lines defined by
$$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 \qquad(\star)$$
pass through the origin, so one of our vertices is $O(0,0)$.
Convenient manipulation allows us to re-write $(\star)$ thusly:
$$x^2 + y^2 = \frac{4( Ax + B y )^2}{3(A^2+B^2)} \qquad(\star\star)$$
If $P(x,y)$ satisfi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proving that the sum $\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$ Proving that the sum
$$\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$$
When $m$=integer even number
I know that the $\frac{\pi^2}{8}$ comes from $\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2}$ and I know how to prove it ... | $\sin(n\frac{\pi}{2}) \in \{-1, 0, 1\}$
For even $m$,
$\sin^m(n\frac{\pi}{2}) = \begin{cases}
0 : n \text{ is even} \\
1 : n \text{ is odd}
\end{cases}$
So, we can rewrite this summation as:
$$
\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2} \\
= \sum_{n=1}^{\infty }\frac{1}{n^2} - \sum_{n=1}^{\infty }\frac{1}{(2n)^2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation
$$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$
has a (real) root, prove the inequality
$$ a^2 + b^2 \geq 8. $$
I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway,... | Let $x$ be a root.
Thus, by C-S $$(x^2+1)^2=-x(ax^2+b^2)\leq\sqrt{x^2(ax^2+b)^2}\leq\sqrt{x^2(x^4+1)(a^2+b^2)}.$$
Id est, it's enough to prove that:
$$\frac{(x^2+1)^4}{x^2(x^4+1)}\geq8$$ or
$$(x^2-1)^4\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Deriving an expression for $\cos^4 x + \sin^4 x$
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$
I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + ... | $\cos^2 x \sin^2 x = (1 - \sin^2 x) \sin^ 2 x = \sin^2 x - \sin^ 4 x$ and by the same argument $\cos^2 x \sin^2 x = \cos^2 x - \cos^ 4 x$. Take the average of these two identities to obtain
$$
\cos^2 x \sin^2 x = \frac{1}{2} - \frac{\cos^4x + \sin^4 x}{2} \, .
$$
Now substitute this and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Real part of Arcsine of a Complex variable How do we show: $\operatorname{Re}({\arcsin{z}}) = \frac{1}{2}(\sqrt{x^2+y^2+2x+1} -\sqrt{x^2+y^2-2x+1})$ ?
None of the things I tried including using the formula of inverse sine seems to work... Please Help
| Solving the equation $(\cosh^2 v)^2-(x^2+y^2+1)\cosh^2 v +x^2=0\,\,$ we have
$$
\cosh ^2v=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right)$$
since $\cosh^2 v=1+\sinh^2 v\ge 1$.
Noting $$
\left(\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right)\right)^2=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation, hat consists on an arithmetical progression. $$1+x+x^2+x^3+\cdots+x^{99}=0.$$
I said to prove with $0+1+2+3+\cdots+99=0$. How should I proceed?
| another observation $$1+x+x^2+x^3+...+x^{99}=\\(1+x^2+x^4+...+x^{98})+(x+x^3+x^5+...+x^{99})=0\\(1+x^2+x^4+...+x^{98})+x(1+x^2+x^4+...+x^{98})=\\(1+x^2+x^4+...+x^{98})(1+x)=0\\\rightarrow (1+x)=0 \rightarrow x=-1$$note that ! $$1+x^2+x^4+...+x^{98} \neq 0\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Unable to calculate pseudo-inverse $A^TA$ I'm trying to calculate the pseudo-inverse of $A^TA$ as described in this paper:
The SVD is particularly simple to calculate when
the matrix is of the form $A^TA$ because $U=V$ and the rows of $U$
are the eigenvectors of $A^TA$ and the singular values in $D$ are the
eige... | Pick up the thread at the eigenvectors or the product matrix. The ordering was mixed. It should be:
$$
v_{1} =
\left[
\begin{array}{c}
1.76322 \\
1 \\
\end{array}
\right],
\qquad
v_{2} =
\left[
\begin{array}{c}
-0.567144 \\
1 \\
\end{array}
\right].
$$
Normalized, these are the column vectors of the domain m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prime numbers and divisibility Prove that for all prime numbers $p,q>3$ holds
$$48|p^4-q^4$$
My solution:
$$48=2^43$$
$$p^4-q^4=(p-q)(p+q)(p^2+q^2)$$
Since adding or substracting odd numbers, we get a even number, $p^4-p^4$ is divisible by $2^3$. How about another $2$ and $3$?
| For $3$, note that $2^4 \equiv 1^4 \equiv 1 \pmod 3$, so $p^4 - q^4 \equiv 1 - 1 \equiv 0 \pmod 3$.
For $2^4$, you have $(p + q) \equiv (p - q) \equiv 0 \pmod 2$. Since $p, q$ are odd, they are equivalent to $1$ or $3 \pmod 4$. $3^2 \equiv 1^2 \equiv 1 \pmod 4$ so $p^2 - q^2 \equiv 1 - 1 \equiv 0 \pmod 4$.
Putting thes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Fractions in Questions and Answers
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