Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{ali... | $$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^2}} &\stackrel{t=\sin x}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\frac{1}{2} \int \frac{(\sin t+\cos t)+(\cos t-\sin t)}{\sin t+\cos t} d t \\
&=\frac{1}{2}\left[\int 1 d t+\int \frac{d(\sin t+\cos t)}{\sin t+\cos t}\right] \\
&=\frac{1}{2}[t+\ln |\sin t+\cos t|]+C \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Spivak, Ch. 4 Graphs, Problem 15: Draw the graph of $f(x)=ax^2+bx+c$? I am asking about this problem because the solution manual seems to have a much more limited solution than mine below, and I wonder if they have been lazy or if I have done something incorrect.
Draw the graph of $f(x)=ax^2+bx+c$.Hint: use methods of... | You're both correct. Spivak's diagram is perfectly fine, as long as you accept the possibility that:
*
*The axes are not necessarily drawn over the lines $x = 0$ and $y = 0$; and
*The axes are not necessarily increasing in the usual directions.
These are both generally considered bad things to do in most cases beca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people ... | If you choose six people and at least two women are selected, then either two women and four men, three women and three men, or four women and two men are selected. The number of ways of selecting $k$ women and $6 - k$ men from four women and seven men is
$$\binom{4}{k}\binom{7}{6 - k}$$
Hence, the number of selection... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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What´s the value of the area of the triangle below? For reference: The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of the triangle (Answer:$9$)
My progress...
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipul... | Use cosine law,
$c^2=a^2+b^2-2ab\cos{\gamma}$
$(\sqrt{26})^2=(3\sqrt2)^2+(\sqrt{20})^2-2\cdot3\sqrt2\sqrt{20}cos\gamma$
$\implies cos(\gamma) =\frac{1}{\sqrt{10}}\Rightarrow sin(\gamma)=\frac{3}{\sqrt{10}}$
$A=\frac12\cdot 3\sqrt2\cdot \sqrt{20}\cdot sin(\gamma)$
$\implies...A=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$.
If there were such an inequality:
$ t_{1} \leq... | For $z\rightarrow0^+$ and $x=y=1$ we obtain: $t\leq\frac{25}{4}$.
We'll prove that $\frac{25}{4}$ is valid.
Indeed, we need to prove that $$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4}.$$
Now, let $x+y+z=2u$ and since our inequality is homogeneous, we can assume $xy+xy+xz=1$
and we need to prove t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding $\lim\limits_{n→∞}\frac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$
Find $\lim\limits_{n→∞}\dfrac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$.
Notice that I don't have $a>b$ or $b>a$, and also no $b≠0$ or $a≠0$, so I don't know if i even can use my solution.
My solution :
\begin{gather*}
\frac {a^n+b^n}{a^n-b^n} = \frac {a^n}{a^n-b^... | Your answer should contain all possible cases, so in general, the answer will not be a single number.
Your solution, by the way, is correct, but it does not cover all cases. If $a>b$, then your proof that the limit is $1$ assumes that $a,b>0$! If $b<a<0$, then the limit $$\lim_{n\to\infty} \frac{a^n}{b^n}$$ is, in fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Verify if $e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)$ is a little-o of $x^{2}$ as $x \to 0^{+}$ using limit I have to verify if $e^{x^{3}} - \sqrt[4]{1-x^{4}} + x \sin(x)$ is a little-o of $x^{2}$ as $x \to 0^{+}$ using limit, i.e. calculating the limit
$$\lim_{x \to 0^{+}} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \... | We can simply using asymptotics relations:
$$\lim_{x \to 0} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)}{x^{2}}\,\,=\,\,\lim_{x \to 0} \frac{\left(e^{x^{3}} - 1\right)+\left(1-\sqrt[4]{1-x^{4}}\right) + x^2 \sin(x)}{x^{2}}\,\,\sim\,\,\lim_{x \to 0^{+}} \frac{x^3+\frac{1}{4}x^4+x^3}{x^2}=\lim_{x\to 0}\frac{o(x^2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding distribution given bivariate normal $f_{xy}$ Let $X$ and $Y$ be distributed as bivariate normal random variables with pdf
$$f_{X, Y}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\biggl(\frac{-1}{2(1-\rho^2)}(x^2-2\rho(xy)+y^2)\biggr)$$
Find the distribution of $aX+bY+c$.
Would this be done by finding the marginal dist... | That is right. There is a more general way to calculate such affine transformations you probably will see later. We have already established that
$$ \begin{pmatrix} X\\ Y \end{pmatrix} \sim N_2\left(
\begin{pmatrix} 0 \\ 0 \end{pmatrix} ,
\begin{pmatrix} 1& \rho\\ \rho&1 \end{pmatrix}\right) $$
then $aX+bY+c = (a,b)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why am I getting the same angle by inputting different values of n in the general solution? Problem:
Solve: $\sqrt{3}\sin x-\cos x=2$, $-2\pi<x<2\pi$
My attempt:
$$\sqrt{3}\sin x-\cos x=2$$
$$\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x=1$$
$$\sin(\frac{\pi}{3}).\sin x-\cos(\frac{\pi}{3})\cos x=1$$
$$\cos(\frac{\pi}{3})\... | $$x=2n\pi\pm\pi-\frac{\pi}{3}\\=-\frac{\pi}{3}+\pi(\color{red}{2n\pm1}).$$
$\color{red}{2n\pm1}:\quad\dots,-7,-7,-5,-5,-3,-3,-1,-1,1,1,3,3,5,5,\ldots.$
When solving trigonometric equations generally and the cosine or secant of the generating solution equals $-1,$ then $\color{red}{2n\pm1}$ occurs, giving rise to repeat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers.
I thought about proving that this number is divisible by $2$, $3$ and $7$ but I don't know how should I do that.
| Put $E=(a^6+b^6-1)(a^6+b^6-2)$; the two factors are consecutive so coprime and $252=2^2\cdot3^2\cdot7$.
$1)$ $E\equiv0\pmod7$. By Fermat's Little Theorem it is quite clear, even if $a$ or $b$ is divisible by $7$.
$2)$ $E\equiv0\pmod{3^2}$. We know that $$(X\pm1)^6=X^6\pm6X^5+15X^4\pm20X^3+15X^2\pm6X+1$$ There are two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the area of region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.
My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{... | $S_{\triangle MBP} = \frac 12 \cdot \frac{BC}{2} \cdot BP \sin{(90^\circ - \frac{\angle B}{2})} = \frac{BC \cdot BP}{4} \cos (\frac{\angle B}{2})$
$S_{\triangle ABP} = \frac1 2 \cdot 3 BC \cdot BP \sin{(90^\circ + \frac{\angle B}{2})} = 6 \cdot S_{\triangle MBP}$
$S_{\triangle ABM} = S_{\triangle ABP} - S_{\triangle M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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A composite function problem The question is:
Suppose $$f(x) = x^2+1,$$ $$g(x) = 3-x.$$
Find the values for $x$ for such that $$(g\circ f)(x) = (f \circ g)(x).$$
I tried banging my head for one hour but my answer doesn't match the one given by the book which is $1/\sqrt{2}$ and $-1/\sqrt{2}$.
I think the answer given i... | $$f(g(x)) = g(x)^2 + 1 = (3-x)^2 + 1$$
$$g(f(x)) = 3 - f(x) = 3 - x^2 - 1$$
Hence we need to find the points $x$ that satisfy
$$(3-x)^2 + 1 = 2 - x^2$$
We get that:
$$9 -6x+ x^2 + 1 -2 + x^2 = 0 \implies$$
$$2x^2 -6x + 8 = 0 \implies$$
$$x^2 - 3x + 4 = 0 \implies$$
$$...$$
you are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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limit, quotient of roots How can I find
$$
\lim_{n \to \infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \sqrt[6]{n - 3}?
$$
If I multiply by $\sqrt{n + 2} + \sqrt{n + 1}$ I could get no divisor, but I cannot get any result on this way. On the other hand I could also transform this expressio... | Put $ m=n+1$.
The limit becomes after factorisation and simplification
$$\lim_{m\to\infty}\frac{(1+\frac 1m)^{\frac 13}-1}{(1+\frac 1m)^{\frac 12}-1}(1-\frac 4m)^{\frac 16}$$
Use the result
$$\lim_{N\to \infty}N\Bigl((1+\frac 1N)^\alpha-1\Bigr)=\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\frac{\log_aN-\log_bN}{\log_bN-\log_cN}=\frac{\log_aN}{\log_cN}$ Show that $$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}=\dfrac{\log_aN}{\log_cN}$$ where $a,b$ and $c$ are positive and are consecutive terms of а geometric sequence, $a\ne1,b\ne1,c\ne1,N>0,N\ne1$.
$a,b$ and $c$ are consecutive terms of a geometri... | From where you left -
$ \displaystyle \frac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN} = \frac{\log_aN-2(\log_aN)(\log_{ac}a)}{2(\log_cN)(\log_{ac}c)-\log_cN}$
$ \displaystyle = \frac{\log_aN}{\log_cN} \cdot \frac{1 - 2\log_{ac}a}{2\log_{ac}c-1}$
Now note that by dividing by $\log_{ac}c$,
$ \displaystyle \frac{1 - 2\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?
if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?
Given answer is
$(-13/4,-13/12) \cup [-1]$
My Approach:
for $\log(x^2+2ax)$ to be valid $x^2+2ax>0$
He... | We want to find $a$ such that there is only one $x$ satisfying
$$x^2+2ax=4x-4a-13\tag1$$
$$x^2+2ax\gt 0\tag2$$
$$4x-4a-13\gt 0\tag3$$
The discriminant of $x^2+(2a-4)x+4a+13=0$ has to be non-negative, so it is necessary that $a\in (-\infty,-1]\cup [9,\infty)$.
*
*$a=-1$ is sufficient since $x=3$ is the only solution.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{... | $A(3) = \lim_{m\to\infty} K_m$ where
$$
K_m = \sum_{k=1}^m\frac{1}{6k-5}
+\sum_{k=1}^m\frac{1}{6k-4}
+\sum_{k=1}^m\frac{1}{6k-3}
-\sum_{k=1}^m\frac{1}{6k-2}
-\sum_{k=1}^m\frac{1}{6k-1}
-\sum_{k=1}^m\frac{1}{6k}
$$
Then use estimates like
$$
\sum_{k=1}^m\frac{1}{6k-a} =
\frac{1}{6}\left(\log m-\psi\left(1-\frac{a}{6}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove that: $(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$ Let $a,b,c$ be real numbers such that $ab+bc+ca=3$. Prove that:$$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$$
I have an imperfect solution:
Let $a+b+c=p;ab+bc+ca=q=3;abc=r$
The problem is:
$$r^2-2(q^2-2pr)+4(p^2-2q)+8-3(p^2-2q)\ge18$$
or:
$$r^2-4pr+p^2+2\ge0 $$
By ... | Some Hints:
$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$
$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+2+b^2+2+c^2+2)$
Also,
$(a+b+c)^2$= $a^2+b^2+c^2+6$ (as $ab+bc+ca=3$)
$(a+b+c)^2$= $a^2+2+b^2+2+c^2+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why isn't $]-1, 1]$ the domain of $f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$? I have to find the domain of this function: $$f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$$
My attempt:
*
*because arcsine is the inverse of sine function, then its domain is the image of sine: $[-1, 1]$.
*the absolute value is defined ... | Lying in bed I thought of probably the clearest answer:
The domain of $\arcsin$ is $[-1,1]$ so the range of $\frac {|x|}{x+1}$ must be restricted to $-1 \le \frac {|x|}{x+1} \le 1$.
That requires that $0\le |\frac {|x|}{x+1}|=\frac {|x|}{|x+1|} \le 1$ and that $|x| \le |x+1|$.
If we consider the four possible positive/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Height of the hill The angles of elevation of the top of a distant hill in the forest as seen from three consecutive km stones on a straight horizontal road are 30, 45 and 60 degrees. Find the height of the hill.\
My try: Let km stone C,B, A are at distances $x,x+1,x+2$ from base of hill $OP$
Then $\tan 60=\frac{h}{x}$... | Following up on @Matt's comment, the answer is indeed $\boxed{h = \sqrt{\frac{3}{2}}}$ when you consider a more general scenario.
Setup
(Diagram at the bottom.)
Let $P$ be the peak of the hill that lies $h > 0$ above the origin $O$, which is a horizontal distance $r \geq 0$ from the closest point on the horizontal stra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Is $\sqrt{5-2\sqrt{5}}$ in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$? Is $\sqrt{5-2\sqrt{5}}$ in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$?
I'm told that $\mathbb{Q} \subseteq \mathbb{Q}(\alpha,\sqrt{2})$ is a Galois extension, and so the minimal polynomial of $\alpha$ must split in $\mathbb{Q}(\alpha,\sqrt{2})$.
The m... | Let $k=\mathbb{Q}(\sqrt{2},\sqrt{5})$
Let $\mu, \nu \in k$ and we ask when is
$\sqrt{\mu}\in k(\sqrt{\nu})$ if this happens,
$$\sqrt{\mu}=a+b\sqrt{\nu}$$ where $a, b\in k$ and so
$$\mu=a^2+b^2\nu+2ab\sqrt{\nu}$$ and assuming $\sqrt{\mu}, \sqrt{\nu}\not\in k$ we have $a=0$ and
thus
$$\mu=b^2\nu$$
So we ask is $\sqrt{5-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{... | For part (a), I would suggest starting as $$\sin(y)=\frac{2x}{1+x}$$ Now, you can use this to build a right triangle with the side of $2x$ and the hypotenuse of $1+x$. Now you can find the other side (How?).
Now let's check the desired part: You need to calculate $tan(y/2)$ (you have $y=f(x)= 2 \tan^{-1} (x)$).
From ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\fr... | Let $g(x) = (1-x)/(1+x)$. Then $g$ is self-inverse; i.e., $g^{-1} = g$, but it is not everywhere monotone. Thus, consider $\Pr[g(X) \le y]$ for the case $y \le -1$ versus $y > -1$ separately. In the first case, $g$ is monotone decreasing on $X \in (-\infty, -1)$ and we have $$\Pr[g(X) \le y] = \Pr[g^{-1}(y) \le X < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way... | Welcome to MSE!
What we're doing is moving out from $n^2$ in either direction by $2$s. You should have something analogous to gauss's trick for summing $1 \ldots n$ in mind, where we pair up $1$ and $n$, then we pair up $2$ and $n-1$, etc.
So for instance, let's look at $n = 5$. We get the sequence of numbers
$$5^2 - 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^... |
These are surfaces over {a,b} expected reals. The calculations are straightforward.
The problems are the second and third solutions are complex. The graph shows only the real parts.
Still calculations are straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How many 12-letter words are there with no block $5 \times a$, $4 \times b$ and $3 \times c$ We arrange 12-letter words having at our disposal five letters
$a$, four letters $b$ and three letters $c$. How many words are there without any block $5 \times a$, $4 \times b$ and $3 \times c$.
I need to use inclusion - exclu... | If we let $N$ denote the total number of distinguishable arrangements of $aaaaabbbbccc$, $A$ denote the set of permutations with five consecutive $a$s, $B$ denote the set of permutations with four consecutive $b$s, and $C$ denote the set of permutations with three consecutive $c$s, then, by the Inclusion-Exclusion Prin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Number of car plates having $3$ letters and a $4$-digit number. Car plates have are to be designed by placing $3$ (not necessarily distinct) English letters followed by a $4$-digit number, where the digits of this number can not be simultaneously $0$, that is, the string "$0000$" is excluded. Also, $0$ can be used in t... | Your original answer is correct.
There are $26$ choices for each letter, giving $26^3$ ways to fill the three positions allocated for letters. Similarly, if we initially ignore the constraint that the final four positions cannot be filled with $0000$, we would have $10$ choices for each digit, giving $10^4$ ways to fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Number of ways to pick $3$ balls from a box with $8$ balls if $3$ of the balls are identical and the other $5$ are all different
A box contains 8 balls, of which 3 are identical and the remaining 5 are different from each other. 3 balls are to be picked out of the box; the order in which they are picked out does not m... | When the order of selection does not matter, you should be thinking in terms of combinations.
In what follows, suppose that the three identical balls are white and that the remaining balls are blue, green, red, orange, and yellow.
Since we wish to determine the number of distinguishable selections of three balls, your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help on this integral $I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$ $$I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$$
Here is my attempt
$\frac{x}{1-x^2}dx=-\frac{1}{2}d\ln(1-x^2)$, integration by part, we got
$$I=-\frac{1}{2}P-Q$$
Where $P=\int_0^1 \frac{\ln(1-x^2)... | Here is a more general solution. Use the contour setup by Sangchul Lee, it can be shown that
$$\int_0^1 \frac{x \arctan a x}{1-x^2}\ln\frac{1+a^2}{1+a^2x^2}dx=\frac13 \arctan^3 a
$$
Let $a=1$ to obtain
$$\int_0^1 \frac{x \arctan x}{1-x^2}\ln\frac{2}{1+x^2}dx=\frac13 \arctan^3 (1)=\frac{\pi^3}{192}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = ... | If $x>0,$
Let $1-x=\cos2t,dx=2\sin2t\ dt$
so we get $$\int\dfrac{dt}{\sin^4t}dt=\int(\cot^2t+1)\csc^2t\ dt$$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} ... | If $f(x)^3-f(x)^2-x^2f(x)+x^2=0$ for every $x$, you can input the needeed values $x_0=0,-\frac{4}{3},\frac{1}{2}$ and solve the polynomial for $f(x_0)$. In general, this may give you 3 solutions for each $x_0$ (so 27 possible solutions for the problem), but maybe those 27 are the same.
Edit: The conditions $\max(f) = 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Is it possible to find the $n$-derivative of $\csc(m\pi)?$ I am trying to find the $n$-th derivative of $\csc(m\pi)$, so I took few cases:
for simplicity let $x=\cot(m\pi)$ and $y=\csc(m\pi)$,
$$\frac{d^0}{dm^0}\csc(m\pi)=\pi^0(\color{red}{1}x^0y^1)$$
$$\frac{d^1}{dm^1}\csc(m\pi)=-\pi^1 (\color{red}{1}x^1y^1)$$
$$\frac... | Thanks to @Domen for his solution, the answer is
$$\frac{d^n}{dm^n}\csc(m\pi)=(-\pi)^n\csc^{n+1}(m\pi)\sum_{k=0}^{\lfloor{n/2}\rfloor}t(n,k) \cos^{n-2k}(m\pi)$$
where
$$t(n,k)=(2k+1)t(n-1,k)+(n-2k+1)t(n-1,k-1)$$
and
$$t(n,0)=1$$
Different form:
The $n$-th derivative of $\sec(x)$ is given by Wolfram:
$$\frac{d^{ n}}{d{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$A+B+C=2\pi$, prove determinant equals to zero. Given $A$, $B$, $C$ which satisfy $A+B+C=2\pi$, is there an ingenious method to prove that
$$
\det\begin{pmatrix}
1 & 1 & 1 \\
\tan A & \tan B & \tan C \\
\tan 2A & \tan 2B & \tan 2C
\end{pmatrix}=0
$$? By column transformation we have
$$
\det\begin{pmatrix}
... | Well here is my working of the problem. I must have made a sign mistake somewhere, however if I have not, then the equation only holds under special cases. Feedback welcome.
Let $a=\tan A, b=\tan B,c=\tan C$ then
$$\tan 2A=\frac{2a}{1-a^2}$$
etc.
So the matrix is
$$X=\begin{vmatrix}
1&1&1\\
a&b&c\\
\frac{2a}{1-a^2}&\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of the shaded region. (Apparently this is a Chinese primary school math question!)
Rectangle $ABCD$ has an area of $1$. $AE = ED$. $3BF = AB$. What is the shaded area?
| The simplest way, in my opinion, is to write the equations of lines $BE$, $EC$, $FC$, $FD$. Because scaling doesn't change the ratios of areas, we can assume that we have a square of unit side length, where
$ B = (0, 0)$
$ E = (\frac{1}{2}, 1)$
$ C = (1, 0) $
$ F = (0, \frac{1}{3} )$
$ D = (1,1) $
It is straight forwa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Maximize $\prod x_i$ without Lagrange multipliers
Suppose that $\sum_{i=1}^{6} x_i=0$ and $\sum_{i=1}^{6} x_i^2=6$. We wish to maximize $\prod x_i$.
One can show via Lagrange multipliers that at an extremal point, either $x_i=x_j$ or $x_i x_j=-1$, which leads to the maximum of $1/2$ when 4 of the variables are $1/\sq... | Let $p$ denote the maximum of $\prod x_i$. Clearly, $p > 0$.
We only need to consider two cases:
Case 1: $x_1, x_2, x_3, x_4 > 0$ and $x_5, x_6 < 0$
Denote $A = x_1 + x_2 + x_3 + x_4 > 0$.
Then $x_5 + x_6 = -A$.
Clearly, we have $x_5^2 + x_6^2 \ge (x_5 + x_6)^2/2 = A^2/2$.
Also, $x_1^2 + x_2^2 + x_3^2 + x_4^2 \ge (x_1 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1... | A direct splitting as follows is also possible and sometimes quite quick:
\begin{eqnarray*} \color{blue}{\frac{1}{(1 + x)(1 - x)^2}}
& = & \frac{1+x-x}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1-x -1}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1}{(1 + x)(1 - x)} \color{blue}{- \frac{1}{(1 + x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Roll a fair 6 faces die. Stop if the sum of the previous rolls is a multiple of 3. What is the expected time to stop. Given there is a regular fair die, roll the die and stop if the sum of all previous rolls is a multiple of 3. What is the expected number of rolls?
Let $X$ denote the number of rolls until the event {th... | Let $f(x)$ denote the expected number of runs till we get remainder $0$ with $x$ denoting the current remainder mod $3$.
Note that each remainder has $\frac{1}{3}$ chance of occurring.
$$f(1) = \frac{1}{3}(1 + f(2)) + \frac{1}{3}(f(1) + 1) + \frac{1}{3}(1)$$
$$f(2) = \frac{1}{3}(1 + f(1)) + \frac{1}{3}(f(2) + 1) + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Polynomial division modulo $7$
We are given polynomials $f(x)=x^5+2x^4+3x^3+4x^2+5x+6,\ g(x)=3x^3+x$ in the polynomial ring $(\mathbb{Z}/7\mathbb{Z})[x]$. I want to find the polynomials $q,r$ for which $f=gq+r$. Note that $deg(r)<deg(q)$
Attempt:
Either $r=0$, or $r\ne 0$ and $\deg(r)<\deg(q)$. Note that $\deg(q)=\de... | Your coefficient of $24x^2$ in the original problem changed to $4x^2$, but with reduction mod $7$, it should change to $3x^2$. This should also help resolve in your solution that in one place $b=3$ and in another $b=4$.
Edit after revision of problem:
The remainder polynomial is potentially of degree $2$. That is $\de... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\i... | It was shown in this answer that if $n$ is a nonnegative integer, then $$\int_{0}^{\infty} \frac{\sin\left((2n+1)x \right)}{\cosh (x) + \cos (x)} \, \, \frac{\mathrm dx}{x} = \frac{\pi}{4}. $$
Therefore, $$\begin{align} \int_{0}^{\infty}\frac{\sin^{3}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &= \frac{1}{4} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $2\sin\frac{\pi}{4}2\cos\frac{\pi}{4}$ I have a question about $2\sin\frac{\pi}{4}2\cos\frac{\pi}{4}$. I ask because when I try to plug the equation in or substitute the double angle equation in for this problem I can never get the answer to come out. The commutative property says $(a\cdot b)\cdot c = a\cdot(... | We wish to simplify the expression
$$2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right)$$
Let's continue your initial calculation.
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 2 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $n^2+d$ is not square if $d|2n^2$ ? $n,d$ are positive integers. Obverously,if $n\ge d>0$,then
$$n^2<n^2+d<n^2+2n+1=(n+1)^2$$
So,$n^2+1$ is not square.
Naturelly,we consider $d>n$,but I can't get contradiction.And I do not use $d|2n^2$.
If we assume a positive $r$ s.t. $\sqrt{n^2+d}=r$,then we have $n^2+d ... | If $n=0$ then take $d$ any square to get a counterexample. So now let’s assume $n \neq 0$ and let $q\neq 0$ be an integer such that $q d = 2 n^2$. Suppose $n^2 + d = m^2$ for some integer $m$. Multiply both sides by $q^2$ to get $$(q^2 + 2q) n^2 = (q m)^2$$ and rewrite this as $$(q+1)^2 - 1 = \left(\frac{q m}n \righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I fou... | Given matrix is real symmetric and hence diagonalisable. Hence $A=PDP^{-1}$. Thus $A^n$=$PD^nP^{-1}$ where P is the matrix of eigen vectors. D is diagonal and hence its powers are simply the powers of the eigenvalues which are $2,-1,-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and the... | You do not need to consider cases.
Think first that you are looking for the zeros of function
$$f(x)=\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) -k$$ the first derivative
$$f'(x)=\frac{1}{1+(x+1)^2}+\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}=\frac {3 x^4+6 x^2+8 } {\left(x^2+1\right) \left(x^2-2 x+2\right) \left(x^2+2 x+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving integral $\int \frac{\sqrt{x^2 + x}}{x}dx$ (problem 36 in section $6.25$ in Tom Apostol's calculus) Integrals which involve $\sqrt{(cx + d)^2 - a^2}$ could often be simplified if we do a substitution $cx + d = a \sec t$. If we take a concrete example, $\int \frac{\sqrt{x^2 + x}}{x}dx$, then the substitution wou... | In the first step we try to guess the result basing on our personal experience
and check if the derivative gives back the integrand. If we miss, we can correct our guess and see what happens. In case we are unable to make a good guess, we try to simplify the integrand. In our case
$$x^2+x=\left (x+{1\over 2}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to pro... | We use this $fact*$ that:
if m is a natural number and $a>1$ we have:
$$\big(\frac{a^m-1}{a-1}, a-1\big)= (a-1, m)$$
Proof:
Suppose $\big(\frac{a^m-1}{a-1}, a-1\big)=d$
Using following identity:
$\frac{a^m-1}{a-1}=(x^{m-1}-1)+(x^{m-2}-1)+(x^{m-3}-1)+\cdot\cdot\cdot+(a-1)+m$ $\space\space\space\space\space (1)$
and the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Let $\left,\left$ be real sequences satisfied as following conditions： $\left<a_i\right>, \left<b_i\right>$ be real sequences with
$$a_1^2+a_2^2+\cdots+a_n^2=1，\\
b_1^2+b_2^2+\cdots+b_n^2=1，\\
a_1b_1+a_2b_2+\cdots+a_nb_n=0.$$
Prove that $(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2\leq n$.
My attempt：
I try to prove i... | Setup/ rephrasing the question:
Consider the vectors $ a = ( a_1, \ldots, a_n), b = (b_1, \ldots , b_n)$. The conditions state that these are orthogonal unit vectors, $ \angle (a, b) = 90^\circ$.
Consider unit vector $c = ( \frac{1}{\sqrt{n} } , \ldots \frac{1}{\sqrt{n}} )$ and $-c$.
Since $ \angle (a, c) + \angle (a, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing the AM-GM inequality, $\textit{Problem From The Book}$, 19.17, but using integrals The chapter of this problem is Solving Elementary Inequality Using Integrals. After I typed the problem: I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.
For part $a$,I can... | $\newcommand{\d}{\,\mathrm{d}}$The first step is in evaluating $I$. Let's first note that each $x_i$ and $A$ itself are strictly positive so we only care about evaluating $I$ in the case $x,a\gt0$. A partial fraction decomposition is in order - I leave the derivation at the bottom of the post for readability:
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An interesting system of simultaneous equations with two unknowns. $ \text{We are going to solve the system (E):}$
$\left\{\begin{aligned} \displaystyle \quad x^{n}+y^{n}&=1 \cdots(1) \\x^{n+1}+y^{n+1}&=1 \cdots(2)\end{aligned}\right., \quad $ where $m,n\in N.$
$\displaystyle \begin{aligned}x&=\frac{x^{n+1}}{x^{n}}\... | I believe, there is more simple solution than mine.
My solution is based on evident consequence of system: $(1-x^m)^n=(1-x^n)^m$.
Lemma. If $0\leq x \leq 1$ and $m > n$ then $(1-x^m)^n=(1-x^n)^m$ has only two solutions $x=0$ and $x=1$.
Proof. The fact that $x=0$ and $x=1$ are solutions is easy to check. Let's consider ... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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why does the golden ratio, Φ=√(81/80)+√(1/80)+1/2? I have been trying to understand the mathematics of the golden ratio starting with the following 345 relationship.
Can anybody help me understand why $ Φ = \sqrt{81/80} + \sqrt{1/80} + 1/2 $ ?
I think it is something to do with these angles but its quite complicated... | As suggested by @Blue:
$$\sqrt{\frac{81}{80}} + \sqrt{\frac{1}{80}} + \frac12 \ \ \ = \ \ \ \frac{\sqrt{81}}{\sqrt{80}} + \frac{\sqrt{1}}{\sqrt{80}} + \frac12$$
$$=\frac{9}{\sqrt{80}} + \frac{1}{\sqrt{80}} + \frac12 \ \ \ = \ \ \ \frac{10}{\sqrt{80}} + \frac12$$
$$=\frac{10}{\sqrt{16 \times 5}} + \frac12 \ \ \ = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k... | Complement to @ECL's answer:
*
*For $k = 1$:
Let
$$g(x) := f(x) - \frac{f(3\pi/2) - f(\pi/2)}{\pi}x - \frac{\frac{3\pi}{2}f(\pi/2) - \frac{\pi}{2}f(3\pi/2)}{\pi}.$$
We have $g(\pi/2) = g(3\pi/2) = 0$.
Also, $g(x)$ is convex on $[0, 2\pi]$.
For $x \in [\pi/2, 3\pi/2]$, letting $t = \frac{3}{2} - \frac{x}{\pi} \in [0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
How to evaluate $\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$, where $B(x, y)$ is the Beta function? I am trying to evaluate this sum:
$$\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$$ where $B(x, y)$ is the Beta function.
Checking with WolframAlpha gives beautiful result: $4-4G$... | Let's denote
$$I=\int_{0}^{1}\frac{1}{\sqrt{1-x}}\sum_{k=1}^{\infty}\frac{x^{k-1}}{(2k+1)^2}\,dx$$
Using $\frac{1}{(2k+1)^2}=-\int_0^1t^{2k}\ln t dt$
$$I=-\int_{0}^{1}\frac{dx}{\sqrt{1-x}}\sum_{k=1}^{\infty}x^{k-1}\int_0^1t^{2k}\ln t dt$$
Changing the order of summation and integration and performing summation first ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Simplifying $\sqrt{1+\sqrt{5}(6-2\sqrt{5})^{1/4}}$ I have the radical
$$\sqrt{1+\sqrt{5}(6-2\sqrt{5})^{1/4}}$$ for exam preparation (middle school):
I need to simplify it in natural numbers.
My attempt is:
We know the rule:
$(a-b)^2=a^2-2ab+b^2$
Let's $a^2+b^2=6$, then $2ab=2\sqrt5$ and $ab=\sqrt5$, suppose that $b=1$... | As noted by Gerry Myerson, the expression is likely supposed to be
$$\sqrt{(1+\sqrt5)\sqrt{6-2\sqrt5}}$$
$$= \sqrt{(\sqrt{5} + 1)(\sqrt{5} - 1)}$$
$$= \sqrt{5 - 1}$$
$$= 2$$
Remark: My intention to answer this question is less to answer the question (which I believe the OP is completely capable of doing themselves) and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Parallelogram with an angle and the midpoint of a side $M$ is the midpoint of the side $AB$ of a parallelogram $ABCD$. Find the perimeter of the parallelogram if $MD=4,MC=6$ and $\measuredangle BAD=60^\circ$.
Let $M$ be the midpoint of $AB\Rightarrow AM=BM=x$ and $AD=BC=y$. We are supposed to find $P_{ABCD}=2(AB+AD)=2... | There are two possible values of perimeter.
$x^2+y^2-xy=16$, $x^2+y^2+xy=36$ $\Rightarrow x^2+y^2=26$, $xy=10$ $\Rightarrow (x+y)^2=46$, $(x-y)^2=6$ $\Rightarrow$
$x+y=\sqrt{46}$, $x-y=\pm\sqrt{6}$ $\Rightarrow 4x+2y=3(x+y)+(x-y)=3\sqrt{46}\pm\sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all integer solutions of $3^a +7 = 2^b$ I want to find all integer solutions of $3^a + 7 = 2^b$
I have found (by brute force) the two solutions
$3^0 + 7 = 2^3$ and
$3^2 + 7 = 2^4$
but I want to see if there are more solutions. I have found that if (b mod 3) = b' then (a mod 6) must be 2b', and that (a mod 4) c... | A negative $b$ or a negative $a$ cannot lead to a solution. So $a,b\ge 0$. For $a=0$ we obtain OP's first solution $3^0+7=2^3$. Else $a\ge 1$. Then taken modulo $3$ the powers of two ($2\equiv -1\ [3]$) for $b=0,1,2,3,\dots$ are $1,-1,1,-1,\dots$, so the involved two-power $b$ is even, $b=2B$ for some natural $B$.
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Need pure geometric solution for proof on 10-20-40-50 angle problem $D$ is a point in $\triangle{ABC}$ so that $\angle{ABD}=10^{\circ}$, $\angle{DBC}=20^{\circ}$, $\angle{BCD}=40^{\circ}$, $\angle{DAC}=50^{\circ}$.
Find $\angle{BAD}$.
This problem is easily done with trigonometric Ceva theorem as:
$$
\begin{aligned}
&\... | Let's make following construction. $E$ is point of $BC$ such that $BE=BD$, $F$ is point symmetric to $E$ around $CD$. $G$ is point symmetric to $D$ around $AB$. Angular measures which can be restored from problem statement are given in picture.
$\angle DFC=100$°$=2\angle DAC$, then $F$ is center of circumcircle of tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Showing $\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$ for $x,y,z>0$ and $xyz=1$
Let $x,y,z>0$ with $ xyz=1$ then prove that,
$$\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$$
Let $$5≤\sqrt{4(x^2+y^2+z^2)+6}\implies x^2+y^2+z^2≥\frac {25-6}{4}=\frac {19}{4}$$ then the inequality is always true.
This also shows... | Take $x=y=z=1$ and you will easily see that the right hand side is larger than $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Closed-form solution of recurrence relation I have the first-order non-homogeneous recurrence (defined for $n \geq 1$):
$$f(n) = f(n-1) \; \frac{n-1}{n} + 1$$
with base case $f(1) = 1$.
Looking at the values of the sequence ‒ $1, 1.5, 2, 2.5$ ‒ one can easily see the closed form is $f(n) = \frac{1}{2}(n+1)$.
But how wh... | You just have to use this formula
$$f(n) = f(n-1)\frac{n-1}{n} + 1$$ recursively that is,
\begin{aligned}
f(n) &= f(n-1)\frac{n-1}{n} + 1 \\
&= (f(n-2)\frac{n-2}{n-1} +1)*(\frac{n-1}{n}) + 1 \\
&=
f(n-2)\frac{n-2}{n} + \frac{n-1}{n} + 1
\end{aligned}
If you use it n-1 times and the condition that f(1) = 1 you get
$$f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Probability of Full House in 5 Cards Given a Pair of Aces What is the probability of a full house in a 5 card hand, given the first 2 cards are aces?
My thought process is that there are 4 other events that could occur: | Method 1: If the first two cards selected from the deck are aces, then of the remaining $50$ cards, two are aces. There are $$\binom{50}{3}$$ ways to select three cards from the $50$ that remain. To obtain a full house, either one of the two remaining aces and two cards from one of the remaining $12$ ranks must be s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$
I tried by first using AM-HM inequality on $a, b, c$ to get the following result.
$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$
$\implies (a+b+c)(\... | Before I start my solution, I want to say that the proof you have just showed was nice try but it's hard to prove the claim. If you want to show that $A>B$, it's meaningless to show that $A>C$ and $B>C$.
Alright, so I'll show my solution.
\begin{align}
& \text{Claim. } 9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca). \\
& \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Reference Request: Integral of Gaussian over Unit Sphere I am looking for a reference for integrals of the form
\begin{equation}
\tag{1} \int_{S^{n-1}} \mathcal{N}_{\omega} ( \mu , \Sigma ) d \omega
\end{equation}
where $S^{n-1}$ is the sphere in $\mathbb{R}^n$ and
\begin{equation}
\mathcal{N}_x (\mu , \Sigma) = \frac{... | For the $n = 2$ case, (2) in the OP can be evaluated as follows. Denote $\tilde{\mu} = \mu$. The integral in question may be reduced to
\begin{equation}
\tag{1} e^{- \frac{1}{2} \left( \frac{\mu_1^2}{\sigma_1^2} + \frac{\mu_2^2}{\sigma_2^2} \right) } \int_0^{2 \pi} e^{ - \frac{1}{2} \left( \frac{\cos^2{\theta}}{\sigma_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solve $\sin(3x)\cos(6x)-\cos(3x)\sin(6x)=-.9$ I made a graph of an equation and also solved the equation algebraically. Even though I can find one answer, I am having a hard time finding all the rest. All of the answers have to fall in the interval $[0, 2\pi)$. Here is my math and graph
$$\sin(3x)\cos(6x)-\cos(3x)\si... | We wish to solve the equation $\sin(3x)\cos(6x) - \sin(6x)\cos(3x) = -0.9$ in the interval $[0, 2\pi)$.
\begin{align*}
\sin(3x)\cos(6x) - \sin(6x)\cos(3x) & = -0.9\\
\sin(3x - 6x) & = -0.9 && \text{since $\sin\alpha\cos\beta - \cos\alpha\sin\beta = \sin(\alpha - \beta)$}\\
\sin(-3x) & = -0.9 && \text{subtract}\\
-\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4433287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solution of the recurrence relation $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$ $y_0 = 0$ and $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of this reccurent equation is $y_n = 1 - \frac{1}{2^n}$, accordingly with the software. But I do not understand the minus sign since it would be $y_n = \frac{1}{2} + \frac{1}{2}\... | $$\begin{split}
y_n &= \frac 1 2 + \frac {y_{n-1}} 2\\
&= \frac 1 2 +\frac 1 2\left( \frac 1 2 +\frac {y_{n-2}} 2\right)\\
&= \frac 1 2 + \frac 1 2 \left(\frac 1 2 +\frac 1 2 \left (\frac 1 2 +\frac {y_{n-2}} 2 \right) \right)\\
&= \dots\\
&= \frac 1 2 + \frac 1 4 + \frac 1 8 +\dots + \frac 1 {2^{n}} + \frac{y_0}{2^n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can you please help me whether $N-R$ method or other methods converges? Consider the system of equations
\begin{align}
&f(x,y)=x+\frac{y^4}{2}+\frac{x^{32}}{4}+\frac{y^{128}}{8}=0 \\
&g(x,y)=y+\frac{x^8}{2}+\frac{y^{32}}{4}+\frac{x^{256}}{8}=0.
\end{align}
I want to solve it using Newton-Raphson process or any other me... | By performing the contour plot for some truncated approximations, we can infer for which the NR algorithm will work.
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9}=0 \\
}
$$
Here NR works with one solution.
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9} + \frac{y^{2187}}{27}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$\frac{1}{|x-y|}=\frac{1}{\pi^3} \int_{R^3}\frac{1}{|x-z|^2}\frac{1}{|y-z|^2}dz$ Prove the following identity for $x,y \in R^3:$ $$\dfrac{1}{|x-y|}=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|x-z|^2}\dfrac{1}{|y-z|^2}dz$$
I tried multiple ways for example use green function,
put $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3),z=(z_1,z_2,z... | One of the options, probably, not the most rational one, is to evaluate the integral directly. If we choose this option, we have to define the system of coordinate - to choose the center point and axis direction. We can choose $y$ as a center point, and direct the axis $Z$ along the vector $\vec a=\vec x-\vec y$. Denot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The solutions of the equation $ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $ are? I tried this:
$$ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $$
$$ 2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ 4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ \sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}} $$
Here, I tried to set $\sin x = t$
... | As you obtained,
$ \displaystyle t-t^3 = \frac{2}{3\sqrt{3}}~, \text {where } t = \sin x$
$ \displaystyle t - t^3 = \frac{1}{\sqrt3} - \frac{1}{3 \sqrt3} \implies \left(\frac{1}{\sqrt3}\right)^3 - t^3 = \left(\frac{1}{\sqrt3} - t\right)$
Now using the fact that $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$
One of the obvious sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Orthogonal projection of an ellipsoïd from N to 2 dimensional space Suppose we have a $N\times N$ symmetric-positive-definite matrix $A$, representing an ellipsoïd in $N$ dimensional space. How to find the matrix $A_{xy}$ corresponding to orthogonal projection of ellispoid on $xy$-plan ?
I have already consulted this p... | Suppose M is a symmetric $N \times N$ matrix representing an ellipsoid : $$X^T M X = 1$$ with $X \in \mathbb{R}^N$.
1st method of projection : using a projection matrix P
$$P_{xy} = \begin{pmatrix} 1 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 0\end{pmatrix} = \begin{pmatrix} \overrightarrow{i} \\ \overrightarrow{j} \end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did... | Another standard approach is to use the Binomial expansion. For large positive $k,$
$$k-\sqrt{k^{2}+1}= k-\sqrt{k^{2}\left(1+k^{-2}\right)} = k-k\left(1+k^{-2}\right)^{1/2}$$
$$ = k - k\left[1+ \left( \frac{1}{2}\right) \frac{1}{k^2} + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)}{2!} \left(\frac{1}{k^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have ... | Hint: $2+9-11=0$, also $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding minimum value of $x^2+y^2+xy+x-4y+9$
What is the minimum value of $f(x,y)=x^2+y^2+xy+x-4y+9$ ?
I tried completing squares,
$$x^2+y^2+xy+x-4y+9=\frac12(x^2+2xy+y^2+x^2+2x+1+y^2-8y+16+1)=\frac12[(x+y)^2+(x+1)^2+(y-4)^2+1]$$But not sure how to continue.
| Hint:\begin{align}f(x+y,x-y)&=3x^2-3x+y^2+5y+9\\&=3\left(x-\frac12\right)^2+\left(y+\frac52\right)^2+2\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that the following equation does not have integer solutions I want to prove that the following equation has no integer solutions $a,b,c$: $$-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc = 0$$ apart from the naive solution $a=b=c=0$.
The context, in case it helps, is the following: I am dealing with the f... | For a contradiction assume $a, b, c$ is a solution.
First of all, if $a, b, c$ have any common factor, we can factor it out so without loss generality assume $a, b, c$ have no common factor.
Now, reduce the equation $\mod 2$:
\begin{align}
-a^3 - b^3 -c^3 + ab^2 - ac^2 + bc^2 -2a^2c + 3abc &\equiv\\
a + b + c + ab + a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the expected value of $\dfrac{X}{Y}$ Below is a problem I did. I believe I did it correctly and I am hoping that somebody here can either confirm that I did it right or tell me where I went wrong.
Problem:
Let $X$ be a random variable that is uniformly distributed on the interval $[1,2]$. Let $Y$ be a random va... | Your answer looks perfect to me! I did a quick numerical simulation and it agrees with you.
>> b = 1 + rand(1000000,1);
>> a = 1 + rand(1000000,1);
>> mean(a./b)
ans =
1.0396
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x)}{x^2\sin(x)}$ without L'hôpital's rule I need help finding this limit:
$$\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x))}{x^2\sin(x)}$$
I've used L'Hôpital's rule and the solution is $\pi^3/2$. However I'm asked to solved it without using it an... | We use the result that $$\lim_{x \to 0}\frac{\sin x}{x}=1$$
from this result we also have, $$\lim_{x \to 0}\frac{\sin \pi x}{\pi x}=1\text{ and } \lim_{x \to 0} \frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=1$$
Now,$$\lim_{x \to 0}\frac{\sin \pi x \cdot(1-\cos\pi x)}{x^2 \cdot \sin x}=\lim_{x \to 0}\frac{\sin \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
main axis transformation with the conic $Γ:−6x^2−6yx+4x+2y^2−4y+1=0$ The purpose of this exercise is to reduce the conic $Γ:−6x^2−6yx+4x+2y^2−4y+1=0$ to the canonical expression.
What I already have is the Eigenvectors $-7,3$ and the eigenvectors $\begin{pmatrix} 3 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 3 \end{pm... | The given conic equation is of the form
$ r^T Q r + r^T b + c = 0 $
with
$Q = \begin{bmatrix} -6 && - 3 \\ - 3 && 2 \end{bmatrix} $
$ b = [ 4, -4]^T $
$ c = 1 $
From $Q$ we determine the rotation angle $\theta $ using the formula
$ \theta = \dfrac{1}{2} \tan^{-1} \left( \dfrac{ 2 Q_{12} }{Q_{11} - Q_{22}} \right) = \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternate proof that $\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin(\frac{3A}2)\sin(\frac{3B}2)\sin(\frac{3C}2)$ for $A$, $B$, $C$ the angles of a triangle
$A$, $B$, $C$ being the angles of a triangle, we need to prove that:
$$\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin\left(\frac{3A}2\right)\sin\left(\frac{3B}2\right)\sin\left(\frac{3C}... | It's easier to work backwards. Still very much "a lot of algebra", though.
$$1 - 4 \sin(\frac{3A}{2}) \sin(\frac{3B}{2}) \sin(\frac{3C}{2})$$
$$= 1 - 4(\frac{e^{i(3/2)A} - e^{-i(3/2)A}}{2i})(\frac{e^{i(3/2)B} - e^{-i(3/2)B}}{2i})(\frac{e^{i(3/2)C} - e^{-i(3/2)C}}{2i})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)A} - e^{-i(3/2)A})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Green's theorem and translated angular form on the path $\gamma:[0,3\pi/2]\to\Bbb R^2,\gamma(t)=(t,\pi\cos t)$-strange result
I would like to compute the following integral $$\int_\gamma-\frac{y}{(x-\pi)^2+y^2}dx+\frac{x-\pi}{(x-\pi)^2+y^2}dy,$$ where $\gamma:\left[0,\frac{3\pi}2\right]\to\Bbb R^2,\gamma(t)=(t,\pi\cos... | Your $\gamma_2$ ends at $\left(\frac{3\pi}{2},1\right)$, but presumably you want it to end at $(\frac{3\pi}{2},\pi)$ to connect to the start of $\gamma_3$. Correction:
$$ \gamma_2:[0,\pi]\to\mathbb{R}^2,\ \gamma_2(t)=\left(\frac{3\pi}2, t\right) $$
$$ \int_{\gamma_2} \omega = \int_0^\pi \frac{\frac{3\pi}2-\pi}{\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving ${d^2 x \over dt^2}=-\omega^2x +\alpha x^2,$ On solving a Lagrangian, I obtained the Lagrangian equation of motion as
$${d^2 x \over dt^2}=-\omega^2x +\alpha x^2,$$ Where $\omega$ and $\alpha$ are constants and t is the time.
Could anyone please help me to find an analytic solution to this differential equatio... | Writing the ode as
\begin{align*}
x^{\prime \prime}&=\left(\alpha x-\omega^{2}\right) x
\end{align*}
Multiplying both sides by $x^{\prime}$ gives
\begin{align*}
x^{\prime} x^{\prime \prime}&=\left(\alpha x-\omega^{2}\right) x x^{\prime}
\end{align*}
Integrating both sides w.r.t. $t$ gives
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n $. I am to determin $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n.$
I was checking the sequence for the very first elements:
$n=1$: $\;\frac{2^1+(-3)^1}{(-2)^1+3^1}=-1:$
$n=2: \;\frac{2^2+(-3)^2}{(-2)^2+3... | Note that if $n$ is even, then $(-2)^n = (-1)^n 2^n = 2^n$; similarly, $(-3)^n = (-1)^n 3^n = 3^n$. Hence $$a_n = \frac{2^n + (-3)^n}{(-2)^n + 3^n} = 1, \quad n \text{ even}.$$
And if $n$ is odd, then $(-2)^n = -2^n$, and $(-3)^n = -3^n$, hence $$a_n = \frac{2^n - 3^n}{-2^n + 3^n} = -1, \quad n \text{ odd}.$$
Conseque... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425)
$
Or, in two row format, get:
$$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & ... | Here
$$\begin{align}
\sigma^2&=\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\cdot\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\\
&=\begin{pmatrix}1&2&3&4&5\\1&4&3&5&2\end{pmatrix}\\
&=(245).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x... | A bit tricky
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$ Integrate $$g(x)=\int f(x)\,dx=\frac{1}{4} \left(\frac{2}{x+1}-\log (1-x)+\log (x+1)\right)$$
$$g(x)=\frac 12 +\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4 n} x^n$$ Differentitate
$$f(x)=g'(x)=\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4} x^{n-1}$$
Now, to make $x^{n-1}=x^{202... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | From Vieta’s rules, $a+b=7, ab=2$. $$a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)= 7(49-2\cdot 3)=7\cdot 43=301 $$ Again, $$(a^3+b^3)^2=a^6+b^6+2a^3b^3$$$$301^2=a^6+b^6+2(2)^3$$ so $$a^6+b^6=301^2-16=90585. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 4
} |
Is a diagonal matrix with "$12$"s along the diagonal an identity matrix? Is C following an identity matrix?
$$A = \begin{bmatrix}
2& -1 & 12 \\
3 & 6 & -9\\
1& 1& 3
\end{bmatrix}$$
$$B = \begin{bmatrix}
9 & 5 & -21 \\
-6 & -2 & 18\\
-1& -1& 5
\end{bmatrix}$$
$$C = \begin{bmatrix}
12& 0 & 0 \\
0 & 12 & ... | In this situation, $C$ is a scalar matrix that is, as you say, $12$ times the identity matrix, i.e. $C = 12I$. This is not the same thing as the identity, but it has some special properties in common with the identity matrix. For example, it commutes with any other $3 \times 3$-matrix, i.e. $CX = XC$ for any $X$.
Also,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Showing $\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral
Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds
$... | Use AM_GM
$F=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}\ge 4 \sqrt[4]{\frac{s-a}{s-c}\frac{s-b}{s-d}\frac{s-c}{s-a}\frac{s-d}{s-b}}=4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int { \arctan{t} \over \sqrt{t} } $ Integrate: $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt$$
My attempt:
I first substituted $y = \sqrt{t}$
then $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt = 2\int { \arctan{y^2}}\,\mathrm dy$$
then I integrated by parts:
$u = \arctan{y^2}$, $v' = 1$
$$ = 2\left(... | Another simplification can be done for the term
$$ \int \frac{y^2}{1+y^4} dy$$
As
Divde by $y^2$
$$\int \frac{dy}{y^2 + \frac{1}{y^2}}$$
Following by
$$ \frac{1}{2}\int \frac{2+\frac{1}{y^2} - \frac{1}{y^2}}{y^2+\frac{1}{y^2}+2 -2}$$
Separating
$$\frac{1}{2}\left [\int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
prove that $\arctan\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\pi}{4}-x$, where $0I tried to solve this
$$\begin{align} \arctan\frac{\cos x-\sin x}{\cos x+\sin x}&=\arctan\frac{1-\tan x}{1+\tan x}\\&=\arctan\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\\&=\arctan\tan\left(\frac{\pi}{4}-x\right)\\&=\frac{\pi... | Another approach. Let $f(x)=\arctan(\frac{\cos x-\sin x}{\cos x+\sin x})$, with $x\in (-\pi/4,3\pi/4)$.
$$f'(x)=\frac{1}{1+\frac{\cos x-\sin x}{\cos x+\sin x}^2}\frac{(\sin x-\cos x)(\cos x +\sin x)+(\sin x-\cos x)(\cos x-\sin x)}{(\cos x+\sin x)^2}$$
It's easy to simplifie that expression and get that $f'(x)=-1$, it i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of lattice points with the same parity inside a triangle Given a point $(a,b)$ with positive coordinates, I'd like to count the number of lattice points $(x,y)$ with the same parity (i.e., $x \equiv y \ (mod \ 2)$) inside the triangle $(0,0)(a,0)(a,b)$. How to compute it fast?
Note that without the parity requir... | After some trials, I come up with a solution similar to the Euclidean algorithm.
We want to calculate $\sum_{0 \leq x \leq a}\lfloor \frac{\frac{b}{a}x+(x\%2)}{2} \rfloor$ where $\%$ means the standard modulo operator. We separate into two parts: $x$ is odd and $x$ is even. For the first case, we write as $\sum_{0 \leq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Closed form of $\sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}$, an alternating harmonic series with the signs determined by a binomial coeffcient In a comment to Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ I proposed to study this alter... | For statement 1:
Write $k = k_N2^N + k_{N-1}2^{N-1} + \cdots + k_12 + k_0$ for the binary expansion of $k$ (and similarly for $p$). We may appeal to Lucas's theorem to show that $\binom{k}{p} \equiv 1 \pmod{2}$ if and only if for every $i$, if $p_i = 1$ then $k_i = 1$.
Now, let $1 < p < 2^q$ which is not a pure power ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The... | The LHS is always non-negative, but the RHS is negative for real $\,x\,$, so no real solutions exist. Let $\,2x-15=z \in \mathbb C \setminus \mathbb R\,$, then substituting $\,x = (z+15)/2\,$ in the original equation:
$$
z^2 + 40 z + 407 = -4 |z| \tag{1}
$$
Taking complex conjugates and subtracting $\,(1) - \overline{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find another method of solving the shaded area. The diagram shows a square of side length $10\;\rm cm$. A quarter circle, of radius $10\;\rm cm$, is drawn from each vertex of the square. Find the exact area of the shaded region.
And This is my answer.
The answer is right, but I am searching for other ways.
Thanks.... |
Consider circle center on D it intesect circl center on C at point G. The area under first circle and axis can be fd as follows:
Equation of corcle :
$x^2+(y-10))^2=10^2$
Or:
$y=10-\sqrt{100-x^2}$
If $y=\sqrt{a^2-x^2}$ then:
$S=\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac {a^2}2 \sin^{-1}\frac xa$
$$S=\int^{5}_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Radicals in a Cyclotomic Field Extension I was looking for the theorems which describe about all the kinds of radicals contained in a cyclotomic extension. With radical I mean the number, say $x$, is not in $\mathbb{Q}$ but some power, say $x^n$, is in $\mathbb{Q}$. For example, $\sqrt{2}$ is contained in $\mathbb{Q}(\... | For $a \in \mathbf Z$, $\sqrt{a}$ is contained in some cyclotomic field (specifically, $\mathbf Q(\zeta_{4|a|})$. For nonzero $a \in \mathbf Z$, if $x^n - a$ is irreducible over $\mathbf Q$ and $n > 1$ is not a power of $2$ then $\sqrt[n]{a}$ (that means an $n$th root of $a$, i.e., a root of $x^n - a$) is not contained... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sum\limits_{cyc}{} \frac{a^3}{b+c} \geq \sum_\limits{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}$ If $ a,b,c>0 $, prove that :$$\sum_{cyc}{} \frac{a^3}{b+c} \geq \sum_{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}.$$
my attempt:
After uniting the denominator and dividing by$(a+b)(a+c)(b+c)$
$\Leftright... | Your proof is OK. But it is simpler to manipulate cyclic sum.
It suffices to prove that
$$\sum_{\mathrm{cyc}} [a^3(a + b)(a + c) - (a^4 + b^3c - b^2c^2 + bc^3)(b + c)] \ge 0$$
or
$$\sum_{\mathrm{cyc}} (a^5 + a^3bc - b^4c - bc^4) \ge 0$$
or (because $\sum_{\mathrm{cyc}} b^4c = \sum_{\mathrm{cyc}} a^4 b$
and $\sum_{\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(... | $$ \lim_{n->\infty}(\sqrt{n^{2}+3n-3} - n) = \lim_{n->\infty}\frac{3n-3}{\sqrt{n^{2}+3n-3} + n} = \lim_{n->\infty}\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n}-\frac{3}{n^2}} + 1} = \frac{3}{2}. $$
Notice that in the last limit as $n$ goes to infinity, the numerator goes to $3$ and the denominator goes to $2$. The second li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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What is the number of different sums of the items of a set of consecutive natural numbers? I have been playing around with sets of consecutive natural numbers like say $S=\{1,2,3,...,10\}$ and I have come up with this problem for which however I have not yet found an answer and I don't know if there is one. My problem ... | Assuming you mean sums of $2$ or more elements, you can get any value between $3$ and $n(n+1)/2.$
If $U\subsetneq \{1,2,3,\ dots,n\},$ we can show either:
*
*$1\not\in U,$ or
*$\exists k<n$ such that $k\in U,$ $k+1\notin U$
In the first case, we can create $U’=U\cup\{1\},$ and the second case we can set $U’=U\setmin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$ I am trying to find solutions of the following system of diophantine equations:
$$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$
where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\the... | Take,
$2p=2n^2+1-w$
$2q=2n^2+1+w$
$2r=16m^2$
Where,
$w^2=4n^4+4n^2-8m^2+1$ ---$(1)$
Eqn (1), is satisfied at, $(m,n,w)=(3,2,3)$
Hence,
$p+q+r=8m^2+2n^2+1$
For, $(m,n,w)=(3,2,3)$
$p+q+r=(3)^4$
$8pqr=(2p)(2q)(2r)$
=$(2n^2+1-w)(2n^2+1+w)(16m^2)$
For, $(m,n,w)=(3,2,3)$, we get:
$pqr=(6)^4$
$(p,q,r)=(3,6,72)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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A unspotted mistake involving simple Harmonic numbers There is a mistake in the following calculations. However, I can't find it so I'd like to ask for help
\begin{align}
S&=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2n-1}-\frac{1}{2n}\\
&=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n-1}-\left(\frac... | $1 \over 1-x$ should be $1 \over 1-x^2$ on the fourth line.
$$
\sum_{k=0}^{n-1} x^{2k} = {1-x^{2n}\over 1-x^2}
$$
Disclaimer: I didn't follow through with all the calculations to check that fixing this arrives at the correct answer.
| {
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"url": "https://math.stackexchange.com/questions/4497056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}} = 2\cos\left(\frac {\pi }{2^{n+1}}\right)$ without induction I wanted to know the proof without induction / substitution method for the equation
$$\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{n-times}}= 2\cos\left(\frac {\pi }{2^{n+1}}\right)$$
... | Observe that $2+\sqrt{2} = 2\left(1+\dfrac{1}{\sqrt{2}}\right)=2\left(1+\cos(\frac{\pi}{4})\right)= 2(2\cos^2\left(\frac{\pi}{8}\right))$. Thus when you take the innermost square root you get $2\cos\left(\frac{\pi}{8}\right)$. And repeat this $n$ times you will get to the formula you want to have.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| I will make a guess that the proof you were looking at has the following equation in one of its steps:
$$ \left(x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}. $$
This tells us that $x+\dfrac{b}{2a}$ is one of the square roots
of $\dfrac{b^2-4ac}{4a^2}.$
But it could be either the positive or negative square root.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{... | $\int x^2 \sqrt{x^3 +1}~dx = \frac{2}{9}(x^3+1)^{\frac{3}{2}} +c$ by direct integration
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt 2 + \sqrt[3]2$ is not rational How do I prove that the following is not rational?
$$x=\sqrt 2 + \sqrt[3]2$$
To prove a simpler case like $\sqrt{2}=a/b$, I can raise both sides to the power of 2 and get $a^2=2b^2$, therefore both $a$ and $b$ must be even numbers which can't be true.
| Let
$$x=\sqrt 2 + \sqrt[3]2$$
then
$$x-\sqrt 2 = \sqrt[3]2$$
$$\left(x-\sqrt 2 \right)^3=2$$
$$x^3-3x^2\left( \sqrt 2 \right)+3x(2)-2\left( \sqrt 2 \right)=2$$
$$\sqrt 2= \frac {x^3+6x-2}{3x^2+2}$$
$x$ cannot be rational because $\frac {x^3+6x-2}{3x^2+2}$ will then be rational and yet $\sqrt 2$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $a^3+5a$ is an integer I've been given the following task. Let
$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$
Show that $a^3+5a$ is an integer.
I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I ... | Let $x=\sqrt[3]{1+\sqrt{\frac{152}{27}}}$ and $y=\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$ .
Then
$$\begin{aligned}a^3+5a&=(x-y)^3+5(x-y)\\
&=x^3-3xy(x-y)-y^3+5(x-y)\\
&=x^3-y^3+(5-3xy)(x-y)
\end{aligned}$$
Now,
$\begin{aligned}5-3xy&=5-3\sqrt[3]{\left(1+\sqrt{\frac{152}{27}}\right)\left(-1+\sqrt{\frac{152}{27}}\right)}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !}... | The obvious alternative is the residue theorem. With $y=2x$ your integral is $$\int_0^\pi\frac{2dy}{(3-\cos y)^2}=\int_0^{2\pi}\frac{dy}{(3-\cos y)^2}\stackrel{z=e^{iy}}{=}\oint_{|z|=1}\frac{-4izdz}{(z^2-6z+1)^2}.$$There is one enclosed pole, the second-order $3-2\sqrt{2}$, so the integral is$$8\pi\lim_{z\to3-2\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.
My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1... | You've got the right idea, but as Kenta S's comment indicates, it's $d \mid 3$ (since $x \equiv -1 \pmod{x+1}$ means $x^4 + x^2 + 1 \equiv (-1)^4 + (-1)^2 + 1 \equiv 3 \pmod{x + 1}$), not $d \mid 2$, so the other option is $d = 3$. For that case, for some integers $m$ and $n$, we have
$$x^4 + x^2 + 1 = 3m^2, \; \; x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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