Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ for any $x\in\mathbb{R}\setminus{k\pi}$ where $k\in\mathbb{Z}$.
I wrote $\sin5x$ as $5\cos^4x\sin{x}-10\cos^2 x\sin^3x+\sin^5x$ and now I need to prove that $5\cos^4x-10\cos^2 x\sin^2x... | For $\sin x\ne0,$
\begin{align}
\frac{\sin5x}{\sin x} &= 5\cos^4x-10\cos^2x(1-\cos^2x)+(1-\cos^2x)^2\\
&= 16\cos^4x-12\cos^2x+1\\
&= \left(4\cos^2x-\frac32\right)^2+1-\frac94
\end{align}
Since anything squared is $\ge0$, we have:
\begin{align}
0 &\le \left(4\cos^2x-\frac32\right)^2\\
1-\frac94 &\le \left(4\cos^2x-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the generating function of a series with a binomial coefficient and a exponential coefficient So I am given this series
$$2^8, 2^7 \binom{8}{1}, 2^6 \binom{8}{2}, 2^5 \binom{8}{3}, 2^4 \binom{8}{4}, 2^3 \binom{8}{5}, 2^2 \binom{8}{6}, 2^1 \binom{8}{7}, \binom{8}{8}, 0, 0, 0, 0, ...$$
which I converted to the su... | Hint:
$$ \sum \frac{2^8 \binom{8}{n} x^n}{2^n} = \sum 2^8 \binom{8}{n} \left(\frac x2\right)^n = \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?
I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.
| Note that $N(a+b\omega)=a^2-ab+b^2$ is the sum of squares, because
$$
a^2-ab+b^2=\frac{1}{4}((2a-b)^2+3b^2).
$$
Hence we have to solve the equation $(2a-b)^2+3b^2=4\cdot 7\cdot 13\cdot 19=6916$, which is
straightforward, since we only have to test a few integers $a,b \in \mathbb{N}$. In particular, $3b^2\le 6916$, so t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed ... | Aspects of your question have been covered here about Lebesgue's Identity,
lebesgue's identity
and a general approach to $x_1^2+x_2^2+\dots+x_n^2 = z^2$,
Diophatine equation $x^2+y^2+z^2=t^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there ... | Generating functions to the rescue!
\begin{align*}
\sum_{\substack{n\ge 0 \\ n \text{ even}}} x^n \sum_{i=0}^{n/2} \binom{n-i}{i}2^i &= \sum_{i=0}^\infty 2^i \sum_{\substack{n\ge 2i \\ n \text{ even}}} \binom{n-i}{i} x^n \\
&= \sum_{i=0}^\infty 2^i \sum_{\substack{m\ge 0 \\ m \text{ even}}} \binom{m+i}{i} x^{m+2i} \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to show distributivity in a ring, and what is wrong with my algebra? I am trying to show the following is a commutative ring with unity, however I am encountering a problem.
First, addition and multiplication are defined as:
$$a \oplus b=a+b-1$$$$a \odot b=ab-(a+b)+2$$
I have shown this "addition" is an abelian g... | We want to show that $\odot$ distributes over $\oplus$. However, we can NOT assume that $\odot$ distributes over $+$.
For the first part, we have:
\begin{align*}
a \odot(b \oplus c)
&= a \odot (b + c - 1) \\
&= (a)(b + c - 1) - ((a) + (b + c - 1)) + 2 \\
&= (ab + ac - a) - (a + b + c - 1) + 2 \\
&= ab + ac - 2a - b - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"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
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | using the lemma $$t+\frac{1}{t}\geq 2$$ for $t>0$ we get the inequality $$\tan(x)+\frac{1}{\tan(x)}\geq 2$$
your way is also ok.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
changing $r=1+2r\cos \theta$ to its cartesian equivalent My textbook says the polar equation, $r=1+2r\cos \theta$, its cartesian equivalent is $y^2-3x^2-4x-1=0.$ I understand that I get this if I square $r$; $r^2=x^2+y^2=(1+2x)^2.$ But don't I need to square root this back or something?? I thought the answer is suppose... | You are right that the polar equation $r = (1+2r \cos \theta)$
and the cartesian equation $y^2 - 3x^2 - 4x -1 = 0$ are not equivalent.
The solutions of the first one are a subset of the solutions of the second.
$y^2 - 3x^2 - 4x -1 = 0$ is equivalent to
$r^2=(1+2r \cos \theta)^2$ and that is equivalent to
$$
r = (1+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$ How can I prove this equality?
$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$
| Assuming convergence of both expressions, you have $A^2=x+A$ and $B^2=x-B$, so
$$
(A-B)(A+B)=A^2-B^2=(A+B).
$$
We conclude that $A-B=1$. Then $$A^2+B^2=2x+(A-B)=2x+1,$$ and therefore
$$
AB=\frac{1}{2}(A^2+B^2-(A-B)^2)=\frac{1}{2}(2x+1-1)=x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
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 $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg... | See here for the definining properties of the intersection property. These allow us to compute:
$~~~I_P(x^5+x^4+y^2,x^6-x^5+y^2)\\
=I_P(x^5+x^4+y^2,(x^6-x^5+y^2)-(x^5+x^4+y^2))\\
=I_P(x^5+x^4+y^2,x^4 (x^2-2x-1))\\
=I_P(x^5+x^4+y^2,x^4)+I_P(x^5+x^4+y^2,x^2-2x-1)\\
=I_P(x^5+x^4+y^2,x^4)\\
=I_P(y^2,x^4)\\
=2 \cdot 4 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d... | $$
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x &=\int \ln x d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{\sqrt{x^{2}-1}}{x} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x^{2}} d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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If 8 divides $a^2$ + $3b^2$, prove that both a and b are even. Question: Prove that if $a$ and $b$ are integers and 8 divides $a^2$ + $3b^2$, then both a and b are even.
I'm able to prove that 8 divides $a^2$ + $3b^2$ if both $a$ and $b$ are even using the fundamental theorem of arithmetic and expressing 8 as $2^3$, bu... | $a^2+3b^2=(a+b)^2+2b^2-2ab.$ Clearly $2b^2-2ab$ is even. If one of $a$ or $b$ is odd, then $(a+b)^2$ is odd. But then $$ (a+b)^2+2b^2-2ab$$ is the sum of an odd and even number and hence odd. Thus $8$ cannot divide it.
If they are both odd then we have
\begin{align*}
a^2+3b^2&=(2m+1)^2+3(2n+1)^2 \\
&= 4m^2+4m+1+12n^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Disk/Washer method proof checking This is a homework question, but i am just checking if what i am saying is correct,
The question in the book states that
A sphere of radius $r$ is cut by a plane of $h$ ($h < r$) units above the equator. Find the volume of the solid (spherical segment) above the plane.
Now i know that ... | I found my mistake,
$$
V = 2 \pi \int_a^b (R^2 - r^2)dx
$$
Therefore
$$
V = 4 \pi \int_0^\sqrt{r^2-h^2} (r^2-x^2-h^2) dx
$$
$$
V = 4 \pi [r^2 \sqrt{r^2-h^2} - \frac{(r^2-h^2)^{3/2}}{3}
- h^2 \sqrt{r^2-h^2}]$$
So much easier
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating a product of tangents
(1) Evaluation of $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = $
(2) Evaluation of $\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\... | Like Prove the trigonometric identity $(35)$,
$$\prod_{k=1}^m\tan \left(\frac{k \pi}{2m+1}\right)=\sqrt {2m+1}$$ as all the angles lies in $\left(0,\frac\pi2\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Understanding expanding trig identities I already have the answer, but I am trying to figure out why because it's not making much sense. Or maybe my professor misspoke I am not sure.
I have $\sin^5x $ and I am trying to express it in terms containing only first powers of sine and cosine.
Expanding into:
$\sin x\cdot \s... | Since $\sin^2 x = \frac{1 -\cos(2x)}{2}$, $$\sin^5 x = \sin x \cdot \left(\frac{1 - \cos(2x)}{2}\right)^2 = \sin x \cdot \frac{1 - 2\cos(2x) +\cos^2(2x)}{4} = \frac{\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2x)}{4}.$$
Since $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$,
\begin{align}\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$ Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$
I am almost done. But I cannot seem to solve the equation for x(t) in terms of t.
This is what I got:
$$
\frac{dx}{\sin x}=2tdt\\
\csc x =2tdt\\
-\ln|\csc x+... | \begin{align*}
\frac{dx}{dt}&=2t \sin x\\
\frac{dx}{\sin x}&=2t \ dt\\
\csc x \ dx&= 2t \ dt\\
-\log(\csc x + \cot x)&=t^2+C_1\\
\log((\frac{1+\cos x}{\sin x})^{-1})&=t^2+C_1\\
\log (\tan\frac{x}{2})&=t^2+C_1\\
\tan{\frac{x}{2}}&=Ce^{t^2}\\
x(t)&=2\arctan(Ce^{t^2})
\end{align*}
For the initial condition:
\begin{align*}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| You're essentially asking to prove $\sqrt{1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k}=2\sqrt{1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k}$
So it's sufficient to prove:
*
*$1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k=4\cdot(1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k)$
*$1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of
$$x^4-2(a+b)x^2+(a-b)^2=0$$
are......
$$x=\sqrt{a}+\sqrt{b}$$
$$x=\sqrt{a}-\sqrt{b}$$
$$x=-\sqrt{a}+\sqrt{b}$$
$$x=-\sqrt{a}-\sqrt{b}$$
When $a$ and $b$ are real numbers(negative or positive)
I prove... | Calculate the determinant:$$4(a+b)^2-4(a-b)^2 = 16ab$$
Then find the roots as following:
$$x^2 = \frac{2(a+b)+\sqrt{16ab}}{2},\frac{2(a+b)-\sqrt{16ab}}{2}$$
$$x^2 = a+b+2\sqrt {ab}, a+b-2\sqrt {ab}$$
$$x = \pm(\sqrt a + \sqrt b), \pm(\sqrt a - \sqrt b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $ I have a task to compute such a sum:
$$ \sum_{j=1}^{n} \cos(j) $$
Of course I know that the answer is $$ \frac{1}{2} (\cos(n)+\cot(\frac{1}{2}) \sin(n)-1) = \frac{\cos(n)}{2}+\frac{1}{2} \cot(\frac{1}{2}) \sin(n)-\frac{1}{2} $$
but I don't have any idea how to ... | You could also use $\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})=2\cos n\sin\frac{1}{2}$; $\;\;$ so if we let $a=2\sin\frac{1}{2}$,
$\cos n =\frac{1}{a}\left[\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})\right]\implies$
$\cos 1+\cos2+\cos 3+\cdots+\cos(n-1)+\cos n=$
$\frac{1}{a}\left[(\sin\frac{3}{2}-\sin\frac{1}{2})+(\sin\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Ways to write "50" A really good friend of mine is an elementary school math teacher. He is turning 50, and we want to put a mathematical expression that equals 50 on his birthday cake but goes beyond the typical "order of operations" problems. Some simple examples are
$$e^{\ln{50}}$$
$$100\sin{\frac{\pi}{6}}$$
$$25... | We also have
\begin{align*}
50 &= 11+12+13+14 \\
&= (8+4)+(8-4)+(8\cdot 4)+(8/4) \\
&= 4^2 + 4^2 + 3^2 + 3^2\\
&= 6^2 + 3^2 + 2^2 + 1^2\\
&= (7+i)(7-i) \\
&= (10-\color{red}{5})(10-\color{red}{0})\\
&= 10(\color{red}{5}+\color{red}{0})\\
&= \sqrt{30^2+40^2}\\
&= \sqrt[3]{170^2+310^2}\\
&= \sqrt[3]{146^2+322^2}\\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 20,
"answer_id": 15
} |
Evaluate $\int\frac{\cot x}{\cos^2 x-\cos x+1}\,\,dx$ $$\int\frac{\cot x}{\cos^2 x-\cos x+1}\,\,dx$$
Please guide me by which term it should be substituted to get the result of this integration. I have tried it by using $\cos x =t$, but it went so long and more problematic.
| Hint :
\begin{align}
\int\frac{\cot x}{\cos^2 x-\cos x+1}\;\mathrm dx&=\int\frac{\frac{\cos x}{\sin x}}{\cos^2 x-\cos x+1}\cdot\frac{\sin^2 x}{\sin^2 x}\;\mathrm dx\\[10pt]
&=\int\frac{\cos x}{\cos^2 x-\cos x+1}\cdot\frac{\sin x}{1-\cos^2 x}\;\mathrm dx\\[10pt]
&=\int\frac{y}{y^2 -y+1}\cdot\frac{\mathrm dy}{y^2 - 1}\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Splitting field of $X^6-7X^4+3X^2+3$ over $\mathbb Q$ and $\mathbb F_{13}$ I want to find the splitting field and the degree of the splitting field over $\mathbb Q$ and $\mathbb F_{13}$ for the polynomial $X^6-7X^4+3X^2+3$.
Over $\mathbb Q$ the polynomial factors as $(X-1)(X+1)(X^4-6X^2-3)$, but how can I find the spli... | Observe that $Y^2-6Y-3$ splits over $\mathbb{F}_{13}$ because its roots are
$$\frac{6\pm\sqrt{36+12}}{2}=3\pm4\sqrt{3}=3\pm4(4)=7\text{ and }12$$
The roots of $X^4-6X^2-3$ are obviously the square roots of the roots of $Y^2-6Y-3$. The element $12$ is already a square in $\mathbb{F}_{13}$ (its square roots are $5$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the limit of function using Taylor series Good evening,
I'm somehow stuck on solving some easy exercises :
$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$
| $$\begin{align}&x\sqrt x\left(\sqrt{x+1}+\sqrt{x-1}-2\sqrt x\right)=x\left(\sqrt{x^2+x}+\sqrt{x^2-x}-2x\right)=\\{}\\
&=x\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{x^2+x}+\sqrt{x^2-x}+2x}=\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2}=\\{}\\
&=\frac{4(x^4-x^2)-4x^4}{\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve the recurrence relation: $2a_n = 7a_{n-1} - 3a_{n-2}; a_0 = a_1 = 1$ $2a_n = 7a_{n-1} - 3a_{n-2}\\
a_0 = a_1 = 1$
My attempt:
$2t^2 - 7t + 3 = 0\\
t = -\frac{1}{2}, -3\\
\\
U_n = b(-\frac{1}{2})^n + d(-3)^n\\
b+d = 1 = -\frac{1}{2}b-3d\\
b = \frac{8}{5}, d = -\frac{3}{5}\\
a_n = \frac{8}{5}(-\frac{1}{2})^n - \fra... | The answer given by Wolfram Alpha does match your textbook. (Try multiplying the $2^{-n}$ through.)
Your mistake was when you changed the roots of the characteristic to positive; if
$$a_n = A\left(\frac{1}{2}\right)^n + B\left(3\right)^n$$
then the correct method of solving for $A$ and $B$ would give $A=\frac{4}{5}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Mclaurins with $e^{\sin(x)}$ To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:
$e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$
$\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$
$\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$
$\iff e^{\sin(x)... | $$=e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{(x-x^3/6)^3}{6}+\dfrac{(x-x^3/6)^4}{24}+O(x^5)$$
and this term
$$\dfrac{(x-x^3/6)^4}{24}=\dfrac{x^4}{24}+o(x^5)$$
so
$$e^{\sin{x}}=1+x+\dfrac{x^2}{2}+\left(-\dfrac{x^4}{6}+\dfrac{x^4}{24}\right)+o(x^5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To evaluate limit of sequence $\left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$ How do I evaluate the limit of the following sequence
$$a_n = \left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$$
I tried ... | Rewrite $\displaystyle a_n = \prod\limits_{k=1}^{n} \left(1+\frac{k}{n}\right)^{1/n} $
Since, $\displaystyle \lim\limits_{n\to \infty} b_n^{1/n} = \lim\limits_{n\to \infty} \frac{b_{n+1}}{b_n}$ (that is if the limit exists)
We may write:
$\displaystyle \begin{align} \lim\limits_{n\to \infty} a_n =\lim\limits_{n\to \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
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$(1-a)(1-b)(1-c)(1-d)\geq abcd$ for $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that $$(1-a)(1-b)(1-c)(1-d)\geq abcd.$$
I thought about substituting $a=\sqrt{w},b=\sqrt{x}$, etc. (assuming first that $a,b,c,d$ are positive), and then looking at the convexity of the function $f... | Observe that as long as $a,b,c,d \ge 0$: $$\begin{align}(1-a)(1-b) - cd &\ge 1 - a - b + ab - \left(\frac{c^2+d^2}{2}\right) \\ &= 1 - a - b + ab - \left(\frac{1-a^2-b^2}{2}\right) \\ &= \frac{1}{2}(1-a-b)^2 \ge 0\end{align}$$
Similarly, $$(1-c)(1-d) -ab \ge \frac{1}{2}(1-c-d)^2\ge 0$$
Hence, $$(1-a)(1-b)(1-c)(1-d) \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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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
| 1073177
Use Sylvester's dialytic eliminant:
$\begin{vmatrix}
-2&-7x&3x^2&2x^3-20&0\\
0&-2&-7x&3x^2&2x^3-20\\
1&x&x^2-7&0&0\\
0&1&x&x^2-7&0\\
0&0&1&x&x^2-7\\
\end{vmatrix}$
$(=79x^6-1022x^4-80x^3+2891x^2+1120x-972)$
to get rid of the $y$s. Equate it to $0$, solve it for $x$, then substitute each value into one of the or... | {
"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": 4
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If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| A key to this kind of question is a formed by Newton's identities, which relate the power sums $p_k=\sum_{i=1}^nx^i$ for different values of $k$ to the elementary symmetric polynomials $e_k$; one can define $e_k$ as the coefficient of $X^k$ in the expansion of $\prod_{i=1}^n(1+x_iX)$. The relations are most easily stat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+... |
$$$$
Let us start to calculate it.
\begin{eqnarray}
I&=&\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{\frac{a^2+1}{2a}-x}\sqrt{\frac{b^2+1}{2b}-x}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{m-x}\sqrt{n-x}}\\
&=&\frac{1}{\sqrt{ab}}\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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What property of summation is used while solving this problem? Saw this problem on a website.
Can someone explain how the summation is split into summation of summation?
What property of summation was used here?
$$ T(n) = \sum_{k=1}^n \lfloor \sqrt{k} \rfloor. $$
Therefore
$$
\begin{align*}
T(m^2-1) &= \sum_{k=1}^{m^2-... | Both the single sum $\displaystyle \sum_{\ell=1}^{m^2-1}$ and the nested sum $\displaystyle \sum_{r=1}^{m-1} \sum_{\ell=r^2}^{(r+1)^2-1}$ run over the same set of values:
*
*for $r=1$, $\ell$ runs from $1^2 = 1$ to $2^2-1 = 3$;
*for $r=2$, $\ell$ runs from $2^2 = 4$ to $3^2-1 = 8$;
*for $r=3$, $\ell$ runs from $3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$a_n$ is bounded and decreasing my second question from
[An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
Show that $a_{n}$ is decreasing and bounded
note tha... | The reasoning about $a_n$ decreasing is fine; that is, for $n\ge2$, $0\lt\cos\left(\frac\pi{2^n}\right)\lt1$. This also shows that $a_n\gt0$. However, we can do a bit better.
Trigonometric Approach
Since
$$
2\sin\left(\frac\pi{2^k}\right)\cos\left(\frac\pi{2^k}\right)
=\sin\left(\frac\pi{2^{k-1}}\right)
$$
We have
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding solutions to $y^2 = x^3 - 27$ I am trying to find integer solutions to this equation:
$$ y^2 = x^3 - 27 $$
With the other problem I tried I was able to use unique factorization in $\mathbb Z [\sqrt{n}]$. I don't know how to get started on this one. I tried looking online and there's some stuff about elliptic cu... | It seems the following.
This equation is a special case of the Mordell equation, $y^2 =x^3 +A$. A tremendous amount of work has been done on the Mordell equation, and solutions have been tabulated for large ranges of values of $A $, for example, here.
Also it is new for me that you consider a non-natural factorization... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$
For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$
This is IMO 1983 problem 6.
I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except was... | $$ \it{a^2 b(a-b)+b^2c(b-c)+c^2 a(c-a)}$$
$$ \it{={\frac {bc\, \left( a+b-c \right) ^{2} \left( a-b \right) ^{2}+a \left( c+a-b \right) \left\{ b \left( a+b-c \right) \left( a-c \right) ^{2} +a \left( b+c-a \right) \left( b-c \right) ^{2} \right\} }{a \left( c +a-b \right) +b \left( a+b-c \right) }}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How can I evaluate this limit with or without applying derivatives? $$ \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} $$
High school student here! This was a question from our Mathematics exam (prior to learning derivatives). Now there was some sort of a bounty here in our school, but ... | Consider that, since $a-b=\frac{a^3-b^3}{a^2+ab+b^2},$
$$ (n+1)^{\frac{4}{3}}-n^{\frac{4}{3}} = \frac{(n+1)^4-n^4}{3n^{\frac{8}{3}}+O\left(n^{\frac{5}{3}}\right)}=\frac{4}{3}n^{\frac{1}{3}}+O\left(n^{-\frac{2}{3}}\right).\tag{1}$$
By multiplying both sides of the previous equation by $\frac{3}{4}$, then summing over $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| Observing that
$$
f(x)=\frac{1}{x}\sqrt{\frac{1-x}{1+x}}=\frac{1}{x}\sqrt{\frac{1-x}{1+x}\cdot\frac{1-x}{1-x}}=\frac{1}{x}\frac{1-x}{\sqrt{1-x^2}}
$$
so we have
$$
I=\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\mathrm d x=\underbrace{\int \frac{1}{x\sqrt{1-x^2}}\mathrm d x}_J-\underbrace{\int \frac{1}{\sqrt{1-x^2}}\mathrm d ... | {
"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": 2
} |
Evaluating an alternating sum using contour integrals Evaluate:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{3n-1}$$
Using contour integration.
Normally I would use $\pi\csc(\pi z)f(z)$ and evaluate the residue multiply by (-1) and divide by $2$ if the function $f(n)$ were even, in this case it is not even.
I was wondering if ... | $$\begin{align}\int_{0}^{1} \frac{1}{1+x^3}dx&=\frac13\int_0^1\left(\frac1{x+1}-\frac{x-2}{x^2-x+1}\right)dx\\&=\frac13\int_0^1\left(\frac1{x+1}-\frac12\frac{2(x-1/2)}{(x-1/2)^2+3/4}-\frac{3/2}{(x-1/2)^2+3/4}\right)dx\\
&=\frac13\left(\ln|x+1|-\frac12\ln|(x-1/2)^2+3/4|-\frac32\frac1{\sqrt{3/4}}\arctan\frac{x-1/2}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 ... | ${\rm mod}\ 7\!:\,\ \overbrace{55\cdots 55}^{1+3n\rm\,\ fives}\, =\, \dfrac{5(10^{1+3n}\!-1)}9\, \equiv\, \dfrac{-2\,(3^{1+3n}-1)}2 \,\equiv\, -3(\color{#c00}{3^3})^{n}\!+1 \equiv 4\ $ by $\ \color{#c00}{3^3\equiv -1},\ n$ odd
| {
"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": 6
} |
Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\f... | When we exchange $\beta$ and $\gamma$ in the equation, we get the same equation back. To me, that means $\beta = \gamma+4k\pi$, where $k$ is an integer, is a third equation that we can use. Of course, only $k=0$ satisfies the constraints that $\alpha+\beta+\gamma=\pi$ and $\alpha, \gamma, \beta>0$, since they are ang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
| Note that $x$ is positive, so we can write $x=y^2-2$, for some real number $y>\sqrt{2}$ and the equation becomes
$$y^2-2=\sqrt{y^2-1}+y+\sqrt{y^2+1}$$
we obtain thus
$\sqrt{y^2+1}=y^2-y-2-\sqrt{y^2-1}$, hence
$$y^2+1=(y^2-y-2)^2+(y^2-1)-2(y^2-y-2)\sqrt{y^2-1}.$$
This yields
$$2(y^2-y-2)\sqrt{y^2-1}=(y^2-y-2)^2-2$$
so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Slight problem with solving a trigonometric equation. I had to prove the identity $4\cos^3x-3\cos x=\cos 3x$ and then use it to solve the equation $(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)=1$.
After proving the identity I proceeded to simplify the equation, obtaining
\begin{align*}
(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)&... | By multiplying by $\cos(x)$ you created additional solutions precisely at the points where $\cos(x)=0$. This is because at these points you just multiplied your original equation by zero which makes any equation true. Consider for example the equation $x = 1$ which has precisely one solution. Contrast this with the equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove by induction that $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$ Prove by induction $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$
Step 1: Show true for n = 0:
LHS: $\frac{0}{2^{0}}$ = 0
RHS = $2 - \frac{0+2}{2^{0}}$... | $\dfrac{p+1}{2^{p+1}} + 2 - \dfrac{p + 2}{2^{p}} = 2 + \dfrac{p + 1 - 2(p+2)}{2^{p+1}} = 2 + \dfrac{(-p - 3)}{2^{p+1}} = 2 - \dfrac{p+3}{2^{p+1}} = 2 - \dfrac{(p + 1)+2}{2^{p+1}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the volume of the region of a sphere bounded by two planes Calculate the volume of a sphere $x^2+y^2+z^2=R^2$ which is bounded by $z=a$ and $z=b$, where $0\leq a<b<R$ using double integral. I can imagine the picture but I don't know how to set it up.
| I show below how you do this calculation as a difference of two volumes:
The volume is given by $I(a)-I(b)$ where (for $0<s<R$), $I(s)$ is defined to be the volume of the ball $x^2+y^2+z^2\leq R^2$ such that $z>s$. Let us calculate $I(s)$ (you complete the details):
The domain is bounded from below by the plane $z=s$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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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)}$$
Thanks for any help
| Let we sketch an alternative technique. Since:
$$\sum_{n\geq 0} x^{2n} = \frac{1}{1-x^2} = \frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right),\tag{1}$$
we have:
$$\sum_{n\geq 0} \frac{x^{2n+1}}{2n+1} = \frac{1}{2}\left(\log(1+x)-\log(1-x)\right)=\operatorname{arctanh} x,\tag{2} $$
and integrating again:
$$\sum_{n\geq ... | {
"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": 0
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Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
$$n (n-1)( n-2)(n-3)x^4=4$$
Wha... | $$\sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}} = - 5 + 2\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 1} \right)^2 }}{{n!}}} = - 5 + 2e
$$
$$
\begin{array}{l}
s = \sum\limits_{n = 0}^{ + \infty } {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $2^{2x+1} - 2^{x+4} = 2^3 - 2^x$ $$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$
How can I solve an exponential equation that has many terms as the one above. Include more than one method if available.
| The key here is to transform this into a quadratic equation.
$$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$ can be rewritten as $$2 \cdot 2^{2x} - 2^4 \cdot 2^x = 2^3 -2^x.$$
Temporarily we will express $2^x=u$ and we find that the equation turns into $$2u^2 - 16u = 8 - u.$$ Finally writing this as $$2u^2 - 15u - 8 = 0$$ we can fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Prove inequality $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2 $ I cannot prove the following inequality.
Let $a,b,c$ be positive numbers.Prove that:
$\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2. $
I tried to use Cauchy inequality. I think the function... | Using Cauchy inequality:$$\frac{a^3}{b+c}+a(b+c)\geq2a^2(1)$$
$$\frac{b^3}{a+c}+b(a+c)\geq2b^2(2)$$
$$\frac{c^3}{a+b}+c(a+b)\geq2c^2(3)$$
$$(1)+(2)+(3):\frac{a^3}{b+c}+a(b+c)+\frac{b^3}{a+c}+b(a+c)+\frac{c^3}{a+b}+c(a+b)\geq2(a^2+b^2+c^2)$$
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}+2(ab+bc+ac)\geq2(a^2+b^2+c^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Summation to Equation I have a summation and I want to be able to find the sum for given $n$ without having to go through $1,\dots,n$.
$$\sum_{x=1}^{n - 1}x+300\cdot2^{x/7}$$
It's been awhile since I've done summations and I can't figure this one out. Also, this isn't for homework if that makes a difference.
Thanks
| Might as well give an answer. Note that we can split this sum into the following two sums: $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \sum_{x=1}^{n-1} x + \sum_{x=1}^{n-1} 300 \cdot 2^{x/7}.$$
Given that $$\sum_{x=1}^n x = \frac{n(n+1)}{2}$$ we know that $$\sum_{x=1}^{n-1} x = \frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $a^b>b^a$ if $aI would like a proof that if $a<b$ are integers with $2\leq a,b$ we have $a^b>b^a$ unless $a=2,b=3$ or $a=2,b=4$ . I would like to use as little calculus as possible. Here is my current solution:
Case 1: $a>2$
Fix $a$ and start with $b=a$. notice $a^a=a^a$. Now increase $b$ by $1$ to now get $... | Consider $\mathbb{N}^a$ as a Cartesian lattice. Consider the collection of paths of length $b$ that
*
*start at the origin
*grow in legs of length $1$, moving parallel to one of the $a$ axes
*only grow outward from the origin
There are $a^b$ such paths, since for each leg you have $a$ directional options. We can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find and specify the type of all singularities
Find and specify the type of all singularities of the function $$f(z)=\frac{(z-\frac{\pi}{2})z\sin(\frac{1}{z})}{\cos z}$$
I have some difficulties doing this
When it comes to singularities, there are $w_0:=0$ and $z_n:=\frac{\pi}{2}+n\pi$, $n\in\mathbb{Z}$
Do you have a... | First $z\sin(\frac{1}{z})$ has an essential singularity at $z=0$ since $z^n\cdot (z\sin(\frac{1}{z}))$ is not differentiable for any $n>0$ at that point. Regarding the rest of the expression. Notice that for $z=\frac{\pi}{2}+k\pi$ where $k\in\mathbb{Z}$, we have $\cos(z)=0$. On the other hand for $k=0$ we have both the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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An $n \times n$ matrix with rational entries such that $A^{n+1}=I$ I'm working on finding $A \in M_n(\mathbb{Q})$ such that $A^{n+1}=I$. If $n$ is odd, $A=-I$ satisfies the condition. When $n$ is even, clearly it should have eigenvalues $e^{2 \pi ik/(n+1)}(k=1,\cdots ,n)$ as a complex matrix. Now I proved this matrix s... | First step: decompose $x^{n+1}-1$ in irreducible polynomials of $\mathbb{Q}[x]$.
*
*About your example, we can generalize as follows: let $p$ be a prime number and let $A\in M_{p-1}(\mathbb{Q})$ s.t. $A^p=I_{p-1}$. One has $x^p-1=(x-1)cyclo(p,x)$. Thus the minimal polynomial of $A$ is $x-1$ and $A=I_4$ or $r(x)=cycl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Simplify $\cos 1^\circ + \cos 3^\circ + \cdots+ \cos 43^\circ$? I am currently working on a problem and reduced part of the equations down to
$$\cos(1^\circ)+\cos(3^\circ)+\cdots+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ)$$
How can I calculate this easily using the product-to-sum formula for $\cos(x)+\cos(y)$?
| The trick is to multiply the whole sum by $2\sin 1^\circ$. Since:
$$ 2\sin 1^\circ \cos 1^{\circ} = \sin 2^\circ - \sin 0^\circ,$$
$$ 2\sin 1^\circ \cos 3^{\circ} = \sin 4^\circ - \sin 2^\circ,$$
$$\ldots $$
$$ 2\sin 1^\circ \cos 43^{\circ} = \sin 44^\circ - \sin 42^\circ,$$
by adding these identites we get that the or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
| Here are the steps
$$
\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right] \left[{\frac{\sqrt{1 + x + x^2} + 1}{\sqrt{1 + x + x^2} + 1}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{1 + x + x^2 - 1}{ x\left(\sqrt{1 + x + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate rational function with multiple complex roots I want to integrate
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx.
$$
By WolframAlpha I know the solution is
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx = \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}.
$$
Do derive it on my own, I did a partial fraction decomposition and... | To avoid multiple complex conjugate roots, you can use the formula in Application to symbolic integration
\begin{align}
\frac{x^2}{(1+n^2x^2)^2}
&=\frac{d}{dx}\left[\frac{Ax+B}{1+n^2x^2}\right]+\frac{Cx+D}{1+n^2x^2}\\
&=\frac{Cn^2x^3+(D-A)n^2x^2+(C-2Bn^2)x+(D+A)}{(1+n^2x^2)^2}
\end{align}
so that
$$\left\{
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume... | $$\begin{array}{l}
\left| x \right| < 1:\frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty {x^k } = 1 + \sum\limits_{k = 1}^\infty {x^k } \\
\Rightarrow \sum\limits_{k = 1}^\infty {kx^k } + \sum\limits_{k = 1}^\infty {x^k } = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{x}{{1 - x}} \\
\Rightarrow \sum\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$ Let $a,b,c\in{\mathbb{R^+}}$. Prove that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$
I tried to expand both, but I did not g... | As Macavity says, this is the same as
$$
\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3
$$
which is the same after rearrangement as
$$
\underbrace{\left(\frac ab-1\right)^2+\left(\frac bc-1\right)^2+\left(\frac ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $ Calculate $S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!
| Since $$\frac{1}{k(k+2)} = \frac{1}{2} \frac{2}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$$
for all $k$, we have
$$\sum_{k = 1}^n \frac{1}{k(k+1)(k+2)} = \frac{1}{2}\sum_{k = 1}^n \left(\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right),$$
which telescopes to $$\frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n... | {
"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": 0
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Sum $\sum_{n=1}^{\infty}\ln(1+2^{-2^n})$ How can I solve this: Find the sum of:
$$\sum_{n=0}^{\infty}\ln(1+2^{-2^n})=?$$ and $$\sum_{n=1}^{\infty}{(-1)^{n+1}}\cdot{1\over n}=$$
Can you please give me not the solution of the problem, but just how I can begin. Thanks.
| $$I=\left(1+\dfrac{1}{2^{2^0}}\right)\left(1+\dfrac{1}{2^{2^1}}\right)\left((1+\dfrac{1}{2^{2^2}}\right)\cdots\left(1+\dfrac{1}{2^{2^n}}\right)$$
use this
$$(x+y)(x-y)=x^2-y^2$$
then
$$\dfrac{(1-\dfrac{1}{2^{2^0}})I}{1-\dfrac{1}{2^{2^0}}}=\dfrac{1-\dfrac{1}{2^{2^{n+1}}}}{1-\dfrac{1}{2}}\to 2,n\to \infty$$
second 2
Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Street Fighter: is the game balanced? Suppose that $A$ is a matrix that describes the matchup information of any pair of Street Fighter characters e.g., considering $3$ characters, assume that the first row/collumn is associated with a character like Ryu, the second row/collumn is associated with Chun-li, the third row... | Let's start by subtracting $5$ from each entry to turn it into a payoff matrix for a zero-sum game: $$A'=\begin{pmatrix} 0 & -2 & 4 \\ 2 & 0 & -1 \\ -4 & 1 & 0 \end{pmatrix}$$
Now we can look for a mixed strategy (a probability with which we select each character) which is a Nash equilibrium. That boils down to
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
| Taking the contrapositive, we suppose $x+y+z<\sqrt 3$ and want to prove
$$xy+yz+zx<\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)<1$$
Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z<3$ and we want to prove
$$(xy+yz+zx)^2(x^2+y^2+z^2)<27$$
And that's a simple AM-GM:
$$(xy+yz+zx)^2(x^2+y^2+z... | {
"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": 0
} |
Trigonometric functions - finding solutions Question:
Find the general solution for the equation:
$$\sin x + 2\sin2x - \sin3x = 3$$
Approach:
Well using the identity of $\sin a - \sin b $, I merged together $\sin x - \sin3x$
And as $\sin 2x = 2\sin x\cos x$, I got up till here:
$$2\sin x(2\cos x - \cos 2x) = 3$$
Ed... | There are no real solutions. $s = \sin(x)$ must satisfy
$16s^6-24s^3-12s^2+12s+9=0$, which has no real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Why's the derivative of $f(x) = x^3-5x-2 $ not $3x^2-7$? I wanted to resolve this problem :
$$
f(x) = 3 x^2 - 5 x - 2
$$
to a derivative, and I did it like this :
$
\begin{align}
f(x) &= x^3-5x-2 \\
f'(x) &= 3x^2-5-2 \\
&= 3x^2-7
\end{align}
$
but once I checked the correction, I found this:
$$
f'(x) = 3 x^2 ... | The derivative of a constant is zero, so if $$f(x) = x^3 - 5x - 2 = x^3 - 5 x^1 - 2,$$ the correct derivative is $$f'(x) = 3x^{3-1} - 5 \cdot 1 x^{1-1} - \color{red}{0} = 3x^2 - 5x^0 - 0 = 3x^2 - 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Show that $x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$ For $x, y \ge 0$ prove that:
$$x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$$
What I think would apply is the AM-GM Inequality, so first,
$$(x^2 + y^2 + 1)^2 \le (x^3 + y + 1)(y^3 + x + 1)$$
$$\implies (x^3 + y + 1)(y^3 + x + 1) - (x^2 + y^2 + 1)... | Solution 1:
Only use Cauchy-Schwarz inequality :
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$
let $$a_{1}=x^{\frac{3}{2}},a_{2}=y^{\frac{1}{2}},a_{3}=1$$
$$b_{1}=x^{\frac{1}{2}},b_{2}=y^{\frac{3}{2}},b_{3}=1$$
so we have
$$(x^3+y+1)(x+y^3+1)\ge (x^2+y^2+1)^2$$
so
$$\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Let $p=a+b$, and suppose that $p$ divides at least one of $a^a+b^b$ or $a^b+b^a$.
Then $p$ divides the product
$$(a^a+b^b)(a^b+b^a)=a^p+b^p+(ab)^a+(ab)^b $$
By Fermat's little theorem
*
*$a^p= a\mod p$
*$b^p= b\mod p$
*$a^p+b^p= a+b = 0 \mod p$
So (remember we suppose that $p$ divides the product) we have :
$$ (ab)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$? I've struggled to solve this exercise
$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .
$$$$
I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $... | since
$$b_{n+1}=(5a_{n}-1)b_{n}\Longrightarrow a_{n}=\dfrac{1}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)$$
take in
$$a_{n+1}=\dfrac{2a_{n}-1}{5a_{n}-1}\Longrightarrow \dfrac{1}{5}\left(\dfrac{b_{n+2}}{b_{n+1}}+1\right)=\dfrac{\frac{2}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)-1}{\frac{b_{n+1}}{b_{n}}}$$
$$\Longrightarrow \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Where does the sum of $\sin(n)$ formula come from? I have seen Lagrange's formula for the sum of $\sin(n)$ from $1$ to $n$ during one of my classes last week, but I never saw how it came to be. I tried googling it to find a proof but couldn't seem to find any as it kept bringing up his other work instead and just state... | A comment on a follow-up question suggests using the following identity:
$$
2 \sin \left(\tfrac12\right) \sin (k)
= \cos\left(k - \tfrac12\right) - \cos\left(k + \tfrac12\right).
$$
Then
\begin{align}
2 \sin \left(\tfrac12\right)&\left(\sin(1) + \sin(2) + \sin(3) + \cdots + \sin(n-1) + \sin(n)\right) \\
&= \left(\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Trying to understand a step in Ramanujan's Proof of Bertrand's Postulate regarding the gamma function My question relates to this step in the proof here:
But it is easy to see that
$$\log \Gamma(x)-2\log\Gamma(\frac12x+\frac12) \le \log\left\lfloor x\right\rfloor!-2\log\left\lfloor\frac12x\right\rfloor!$$
if I use $\... | I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$
$$\log\Gamma(x) \le \log\Gamma(x + 1 - \left\{ x\right\}) = \log\Gamma(\left\lfloor x\right\rfloor + 1) = \log\left\lfloor x\right\rfloor!$$
$$2\log\Gamma(\dfrac{x}{2} + \dfrac{1}{2}) \ge 2\log\Gamma(\dfrac{x}{2}+1 - \left\{\dfrac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is ... A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.
| Manuel Lafond’s suggestion that you try some small values of $n$ is always worth pursuing if you don’t see a cleverer approach. By actual calculation you can fairly readily discover that $$\begin{align*}
S_0&=\frac12=1-\frac12\;,\\
S_1&=\frac9{10}=1-\frac1{10}\;,\\
S_2&=\frac{409}{410}=1-\frac1{410}\;,\text{ and}\\
S_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Probability question please guide There are $5$ red and $3$ blue chips in a bowl. The red ones are numbered $1,2,3,4,5$ and the blue ones as $1,2,3$ respectively. if $2$ chips are drawn without repacement, find the probability that these chips have either the same number or the same colour.
What i think ::
For same col... | The total number of ways to choose $2$ out of $8$ chips is $\binom82=28$.
The number of ways to choose $2$ chips with the same color is $\binom52+\binom32=13$.
The number of ways to choose $2$ chips with the same number is $\min(5,3)=3$.
So the probability to choose $2$ chips with the same color or number is $\frac{3+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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? (... | $$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \\
\log \frac{1}{1-x} = x + \frac{x^2}{2} + \frac{x^3}{3} +\frac{x^4}{4} + \cdots $$
and so
$$\frac{1}{2}(\, \log(1+x) + \log \frac{1}{1-x}\,) = x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots$$
or
$$ x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots= \log \sqrt{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Need help understanding Fibonacci Fast Doubling Proof From this website, http://www.nayuki.io/page/fast-fibonacci-algorithms (fast doubling proof close to the bottom of the page).
I have understood the proof for the most part but I am struggling to see how this part of the proof works especially when the the F(n) funct... | Start with:
$$F(n+1) = F(n) + F(n-1)$$
Rewrite as:
$$\begin{align} F(n + 1) &= F(n) + F(n-1) \\
F(n) &= F(n) \end{align}$$
which is:
$$\begin{align}
\begin{bmatrix} F(n+1) \\ F(n)\end{bmatrix}
&=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} F(n) \\ F(n-1)\end{bmatrix} \\ &=
\begin{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Cylindrical limits of integration for a particular triple integral In cylindrical coordinates, what would be the limits of integration for the triple integral serving to find the volume of the region in $\mathbb R^3$ bounded by:
$x^2 + y^2 = y$ and the sphere of center $O$ and radius $1$, in the first octant?
$\theta$ ... | $x^2+y^2 = y$ describes an infinite cylinder $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$
The interior of intersection of two regions is thus given by
$$
\mathcal{R} = \left\{ x^2 + y^2 + z^2 \leqslant 1, x^2 + \left(y - \frac{1}{2}\right)^2 \leqslant \left(\frac{1}{2}\right)^2, x \geqslant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving an underdetermined system of equations I've had a little break in solving systems of equations and so I wanted to verify here that my own answer to this problem is 100% correct and done :) So I have the following problem:
Solve the system of equations:
$$3x_1+x_2-4x_3+5x_4=2$$ $$2x_1-3x_2-2x_3+3x_4=5$$
My a... | Note that
$$
\DeclareMathOperator{rref}{rref}
\rref
\begin{bmatrix}
3 & 1 & -4 & 5 & 2 \\
2 & -3 & -2 & 3 & 5
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & -\frac{14}{11} & \frac{18}{11} & 1 \\
0 & 1 & -\frac{2}{11} & \frac{1}{11} & -1
\end{bmatrix}
$$
This implies
\begin{align*}
x_1 &= \frac{14}{11}\,x_3-\frac{18}{11}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
| Note that our expression is equal to $2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)$. Find an angle $\phi$ such that $\sin\phi=\frac{1}{2}$ and $\cos \phi=\frac{\sqrt{3}}{2}$. Then our expression is equal to $2\sin(\phi+x)$.
Remark: The start you mentioned in a comment will work, though it is a bit messier. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Given any $a,b$, find $A,B$ such that $a\sin(x) + b\cos(x) = A\sin(x+B)$ I understand that if $f''+f = 0, f(0) = a$, $f'(0) = b$ we can have $f = b\sin +a\cos$. so if $f(x) = A\sin(x+B)$, we choose $a = A\cos(B)$ and $b = A\sin(B)$
How do we find $a$ and $b$ conversely?
| Let $\theta$ be such that $\cos \theta = a/\sqrt{a^2 + b^2}$ and $\sin \theta = b/\sqrt{a^2 + b^2}$. Then $$a\sin(x) + b\cos(x) = \sqrt{a^2 + b^2}(\cos(\theta)\sin(x) + \sin(\theta)\cos(x)) = \sqrt{a^2 + b^2}\sin(x + \theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\frac{a-bc}{a+bc}+\frac{b-ac}{b+ac}+\frac{c-ab}{c+ab}\le \frac32$ Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\le \dfrac32$.
My attempt:
First I thought if each term could be less than $\dfrac12$. That did not ... | Using Cauchy-Schwarz Inequality:$$\left(\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\right)\left(\sum\limits_{cyc} bc(a+bc)\right)\ge \left(\sum\limits_{cyc} bc\right)^2$$
So it suffices to prove: $\displaystyle 4\left(\sum\limits_{cyc} bc\right)^2 \ge 3\left(\sum\limits_{cyc} bc(a+bc)\right) = 9abc + 3\sum\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$ The limit is:
$$
\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}
$$
When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:
$$
\frac{1}{5x^{\frac45}}
$$
My question is:
*
*How to do it without calculator.
*Sho... | $$\lim_{h \to 0} \frac{(x(1+\frac{h}{x}))^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \frac{x^{1/5}(1+\frac{h}{x})^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \bigl( x^{1/5} \cdot \frac{(1+\frac{h}{x})^{1/5} -1}{h} \bigr) = x^{1/5}\lim_{h \to 0} \frac{(1+\frac{h}{x})^{1/5} -1}{h} =x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Is the orthogonal decomposition of a symmetric matrix unique? Suppose I have a real symmetric matrix $M$, and it can be decomposed in two ways: $M = AUA^T$ and $M = BVB^T$, where
*
*$U$ and $V$ are diagonal matrices
*The columns of $A$ are orthonormal to each other
*The columns of $B$ are also orthonormal to each ... | No. Unless all of the eigenvalues are the same, $U$ is not necessarily equal to $V$ because we can permute the diagonal entries to get a different diagonal matrix.
Also, $A$ is not necessarily equal to $B.$ Following is an example in which even if we fix $U,$ we can find an infinite number of orthogonal matrices $A$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$? When integrating $\int \frac{1-x}{x^2-1} dx$ Maple rewrote it as $-\int\frac{1}{x+1}dx$
How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$?
| $\dfrac{1-x}{x^2-1} = \dfrac{-(x-1)}{(x+1)(x-1)}$ and if $ x \ne 1 $, then $\dfrac{-1}{x+1}$.
With this, $\int\dfrac{1-x}{x^2-1}dx = -\int\dfrac{1}{x+1}=- \ln|x+1|+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\mathbb Z[\sqrt{-5}]$ is not a UFD Prove that the ring of integers of $\mathbb Q (\sqrt{-5})$ does not have unique factorisation.
Since $-5\equiv 3\pmod 4$, I know that the ring of integers of $\mathbb Q (\sqrt{-5})$ is $\mathbb Z [\sqrt{-5}]$.
I assume the way to prove that this does not have a unique factorisation ... | In a UFD, an element is irreducible if and only if it is prime.
Observe that $2$ is irreducible in $\mathbb Z + \mathbb Z\sqrt{-5}$: Suppose $$2 = (a + b \sqrt{-5})(c + b \sqrt{-5}),$$ taking the norm of both sides gives us $$4 = (a^2 + 5b^2)(c^2 + 5d^2)$$ which means $a^2 + 5b^2 = 1, 2$ or $4$. If $a^2 + 5b^2 = 1$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$
If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$
I have tried the following,in decreasing order of success:
1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$
2)Substituting $1=abc$ yields nothing
3)Substituti... | Hint:
Indeed, as @Macavity pointed out, the sum $s(a,b,c)$ only has circular symmetry. In fact it is not hard to show ( the same method as below) that if $a\le b\le c$ then
$s(a,b,c)\le s(c,b,a)$.
Here is a method amenable to a computer algebra system. As suggested, substitute $a=u/v$, $b=v/w$, $c=w/u$. Get an equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
First order partial differential equation - how to finish the solution? I have problems solving this simple pde. I don't know what I'm doing wrong:
$$yu_x - u_y = y, \ \ \ u(x,0 ) = \frac{1}{x}$$
Here is how I do it. The equation is equivalent to this one:
$$\frac{dx}{y} = \frac{dy}{-1} = \frac{du}{y}$$
So $\frac{dx}{y... | Use method of characteristics.
First, reparameterise your curve letting
$$x \to x(s) \\
y \to y(s)$$
Hence you have
$$u = u(x(s), y(s))$$
Taking the total derivative wrt to $s$
$$\begin{align}
\implies \frac{d}{ds} u &= \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds} \\
&= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the method to finding the scale factor of enlargement and rotation of a 2D matrix? The matrix M is defined by:
\begin{bmatrix}
-1 & -1 \\
1 & -1 \\
\end{bmatrix}
Assuming the matrix represents an enlargement followed by a rotation
My idea here was to make an equation so you're left with simultaneous equation... | If you suspect that this matrix is a scaling followed by a rotation, you can apply it to some basis vectors to get a clue.
For instance multiplying your matrix on $[1,0]^T$ yields $[-1, 1]$. And applying it to $[0,1]^T$ yields $[-1, -1]$.
The scale factor for $[1,0]^T$ is $\sqrt{2}$. The rotation can be figured out fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of natural solutions to the equation $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000$ I'm trying to tackle some question and I think I solved it, but I'm not sure and would like someone to check my work.
The question goes like this:
Find the number of non-negative solutions of the equation $x_1\cdot{x_2}\cdot{x_3}\c... | Let's try it for $100=x_1\cdot x_2$. Then the numbers of solutions to $a_1+a_2=2$ is $3$ and the number of solutions to $b_1+b_2=2$ is $3$. Your answer would give us $3+3=6$ answers.
Let's now enumerate:
$$100\cdot 1\\50\cdot 2\\25\cdot 4\\20\cdot 5\\10\cdot 10\\5\cdot 20\\4\cdot 25\\2\cdot 50\\1\cdot 100$$
That's nine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding number of integer solutions using Generating Functions This is a problem for a practice test my professor gave me.
$$\text{How many integer solutions are there to } x_1+x_2+x_3+x_4 \leq 50 \\ \text{with } x_i \geq 2 \text{ for all } i = 1,2,3,4 \text{ and } x_1,x_2 \leq 7 \text{?}$$
This is how I approached th... | The actual generating function should be $$\frac{(1-x^6)^2}{(1-x)^5}$$ When you said, "To further simplify...," you forgot that $$1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x},$$ not simply $1-x^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Rank of a matrix summation Assume $M\in\{0,1\}^{n\times n}$ of rank $r$. Is there an example of $M$ such that $rk(M+M')/rk(M)> 2$? Is there an example of $M$ such that $rk(M)/rk(M+M')> 2$?
| Due to the rank inequalities, the answer to the first question is no:
$$
rank(M+M') \le rank(M) + rank(M') = 2 rank(M).
$$
Here is an example of an $8\times 8$ matrix $M$ such that $rank(M)=8$, $rank(M+M^T)=2$:
$$
M:=
\left(
\begin{array}{cc|cc|cc|cc}
0& 0& 1& 0& 0& 1& 1& 0\\
1& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ My problem is to find the sum of the series
$$
S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots
$$
where the terms are the reciprocals of the positive integers whose on... | Given what you have observed, note the following:
\begin{align}
S &= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}{3^n}\\[1em]
&= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}\frac{1}{3^n}\\[1em]
&= \sum_{m=0}^\infty\frac{1}{2^m}\sum_{n=0}^\infty\frac{1}{3^n}\\[1em]
&= \frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove this floor function equation? How can I prove the following equation?
$$
\lfloor nx \rfloor =
\lfloor x \rfloor +
\Big\lfloor x + \frac{1}{n} \Big\rfloor +
\Big\lfloor x + \frac{2}{n} \Big\rfloor +
\Big\lfloor x + \frac{3}{n} \Big\rfloor +
\Big\lfloor x + \frac{4}{n} \Big\rfloor +
\Big\l... | Replacing $x$ by $x':=x+{1\over n}$ adds $1$ on the left hand side. On the right hand side the first term $\lfloor x\rfloor$ disappears, and on the far right the term $\lfloor x+1\rfloor$ appears. It follows that the right hand side increases by $1$ as well.
Therefore it suffices to prove the formula for $0\leq x<{1\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to solve this quadratic congruence modulo a non-prime number
$(1)$ Find all $x$, that solve $7x^2 + x + 22 \equiv 0 \pmod{60}$.
I tried to solve this by first considering the prime factorization $60 = 2^2\cdot 3\cdot 5$ and then using the Chinese Remainder Theorem, that is for $n_1, ..., n_k$ coprime and $a_1, ..... | By CRT $7x^2+x\equiv -22\equiv 38 \bmod 60$ if and only if:
$x(7x+1)\equiv x(3x+1)\equiv 2 \bmod 4$
$x(7x+1)\equiv x(x+1) \equiv 2 \bmod 3$
$x(7x+1)\equiv x(2x+1) \equiv 3 \bmod 5$
Finding the solutions to each of these can be done by hand.
You want numbers such that:
$x\equiv 2$ or $3\bmod 4$
$x\equiv 1 \bmod 3$
$x\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$dx$ being a desginator (with respect to $x$) or being a term? I am confused as to what $dx$ truly is. I am doing some u-substitution problems and this is what I came across:
$$\int 2x(x-1)^{1/2}\,dx$$
$u=x-1$ and therefore $du=1$
when we substitute we get:
$$2 \int (u^{3/2}+u^{1/2}) \, du$$
(here the du simply replace... | Consider
$$\begin{array}{lll}
\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx&=&\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int (x^2)^2(2x)(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\\
\end{array}$$
Next, let's make the substitution
$u=x^2+1$, implying that $du/d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Partial Integration $ \int \frac{x\cos x}{\sin^3x}dx $ The problem:
$$ \int \frac{x\cos x}{\sin^3x}dx $$
Can someone give me a hint on how to solve this without using cosecant?
The solution provided is:
$$ -\frac{1}{2}\left(\frac{x}{\sin^2x}+\cot x\right)\:+\:C $$
| $$\begin{gathered}
I = \int {\frac{{x\cos x}}
{{{{\sin }^3}x}}dx} = \int {\frac{{x\cos x}}
{{\sin x}}.\frac{1}
{{{{\sin }^2}x}}dx} = - \int {\frac{{x\cos x}}
{{\sin x}}.d\left( {\cot x} \right)} \hfill \\
= - \frac{{x\cos x}}
{{\sin x}}.\cot x + \int {\cot xd\left( {\frac{{x\cos x}}
{{\sin x}}} \right)} \hfi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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solve for x in the following Quadratic equations I am not able to solve this problem for my son. Is ther any error in the question itself.
$\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$.
Thanks in advance..
| $\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$
$\implies \dfrac{1}{(x+5)}-\dfrac{1}{(x+4)}=\dfrac{1}{(x+2)}-\dfrac{1}{(x+1)} $
$\implies \dfrac{x+4-x-5}{(x+4)(x+5)}=\dfrac{x+1-x-2}{(x+1)(x+2)}$
$\implies (x+4)(x+5)=(x+1)(x+2)$
$\implies x^2+9x+20=x^2+3x+2$
$\implies 9x-3x+20-2=0$
Or,$$x=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Problem with understanding theorem on Riccati Equation. `The matrices $A,B,C,D,X$ are real, square, $n \times n$.
I have trouble understanding theorem 7.1.2 from Lancaster & Rodman "Algebraic Riccati Equations".
The part that I understand is as follows.
A matrix $X$ solves equation $XBX + XA -DX - C = 0$ iff we can fin... | $J$ is not the Jordan block in this context. If it was, there would be dimension mismatch. Since $T$ is $2n \times 2n$, so is $V$. Therefore, $Y$ or $Z$ could not be square.
I believe the authors tried to prove this: Suppose the Riccati equation has a solution. We need to show that there exists matrices $Y$, $Z$ and $J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show by induction on $n$ that: $$1^4 + 2^4 +\cdots+n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
I proved true for case $n=1$ , assumed true for $n=k$ , but cannot get things to work out. I tried putting the right side over a common denominator and going from there but I'm sure how much it helps.
| First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^4=\frac{6\cdot1^5+15\cdot1^4+10\cdot1^3-1}{30}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^4=$
$\color{red}{\sum\limits_{k=1}^{n}k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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 + ... | $$\cos^4x+\sin^4x\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-\frac12\sin^22x=1-\frac12\left(\frac12(1-\cos4x)\right)=\frac34+\frac14\cos4x$$
As:
$$\sin^2x+\cos^2x=1;\\(a+b)^2=a^2+b^2+2ab;\\\sin2x=2\sin x\cos x;\\\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$$
Also:
$$\cos^4x+\sin^4x=\left(\frac{e^{ix}+e^{-ix}}2\right)^4+\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Euler phi Function and Floor Function Let $\phi(x)$ be the Euler phi function and $\lfloor x\rfloor $ the floor function
Count the sum of $$ \phi(1)\cdot\lfloor \frac{2015}{1} \rfloor + \phi(2)\cdot\lfloor \frac{2015}{2} \rfloor + \phi(3)\cdot\lfloor \frac{2015}{3} \rfloor + \phi(4)\cdot\lfloor \frac{2015}{4} \rfloor ... | Let
$$f(n)=\sum_{k=1}^n\varphi(k)\left\lfloor\frac{n}k\right\rfloor\;,$$
so that you’re interested in $f(2015)$. Calculate the first few values:
$$\begin{align*}
f(1)&=1\\
f(2)&=2+1=3\\
f(3)&=3+1+2=6\\
f(4)&=4+2+2+2=10\\
f(5)&=5+2+2+2+4=15
\end{align*}$$
There is a very clear pattern here: at least for $n=1,2,3,4,5$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the equation of the plane knowing that it passes through 3 points I have to find the equation of the plane that passes through $(0, 0, 0), (4, 0, -2), (0, 8, -6)$.
I have done the following:
The equation of the plane is of the form $$ax+by+cz+d=0$$
Since the points $(0, 0, 0), (4, 0, -2), (0, 8, -6)$ are points... | Hint: first you find the normal to the plane: $N = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 8 \\-6 \end{pmatrix} = (a,b,c)$, then your plane is:
$a(x-0)+b(y-0)+c(z-0) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
| $$\begin{gathered}
\int {\frac{1}
{{3{{\sin }^2}x + 5{{\cos }^2}x}}dx} = \int {\frac{1}
{{\left( {3\frac{{{{\sin }^2}x}}
{{{{\cos }^2}x}} + 5} \right){{\cos }^2}x}}dx} = \int {\frac{1}
{{3{{\tan }^2}x + 5}}d\left( {\tan x} \right)} \hfill \\
= \frac{1}
{3}\int {\frac{1}
{{{{\tan }^2}x + \frac{5}
{3}}}d\left( {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Implicit differentiation of $(x^2 + y^2)^2 = (x-y)^2$ $$
(x^2 + y^2)^2 = (x-y)^2$$
Wolfram alpha yields this answer:
$$
y'(x) = \frac{(-2 x^3-2 x y^2+x-y)}{((2 x^2-1) y+x+2 y^3)}$$
But it's impossible to get $-y$ in the denominator
Actually, my answer is pretty much the same, except it's $+ y$ in the denominator
When d... | $$(x^2+y^2)^2=(x-y)^2$$
$$x^4+2x^2y^2+y^4=x^2-2xy+y^2$$
$$4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}=2x-2y-2x\frac{dy}{dx}+2y\frac{dy}{dx}$$
$$y'(4x^2y+4y^3)+4x^3+4xy^2=y'(-2x+2y)+2x-2y$$
$$y'(4x^2y+4y^3+2x-2y)=-4x^3-4xy^2+2x-2y$$
$$y'=\frac{-4x^3-4xy^2+2x-2y}{4x^2y+4y^3+2x-2y}$$
Take out factor of $2$
$$y'=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.
| With question as it is now, it holds because $\frac{b^2+c^2}{2}\geq(\frac{b+c}{2})^2$. Can you show why the last inequality holds?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.