Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving the square root of 2 exists(Possible typo in Zorich's Mathematical Analysis)
In the picture above, I don't understand how if $$s^2 \lt 2$$ then $$\left (s + \frac{2-s^2}{3s}\right)^2 \lt 2$$ holds. What insures that $(s + \frac{2-s^2}{3s}) \in X = \{x : x^2 \lt 2\}$ where $x$ is a positive real number ??
And w... | We know $1 < s^2 < 2$ so $0 < 2- s^2 = \triangle < 1$. So $ \frac{\triangle}{3s} < \frac {\triangle}3 < \triangle < 1$ so $(\frac{\triangle}{3s})^2 < \frac{\triangle}{3s}$.
That's the groundwork.
So $(s + \frac {2-s^2}{3s})^2 = (s + \frac {\triangle}{3s})^2$
$= s^2 + 2s*\frac {\triangle}{3s} + (\frac {\triangle}{3s})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Indentify the plane defined by $Re(z\overline{z})=\frac{lm(\overline{z})}{2}$ I tried:
$$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 = \frac{-y}{2} \Leftrightarrow \\
-2x^2 = y$$
But my book... | Let $z=x+iy$ then $z\bar{z} = (x^2 + y^2) + 0i$ .
So $\Re(z\bar{z}) = x^2 + y^2$. Similarly $\bar{z} = x-iy$ so $\Im(\bar{z})=-y$. Hence the required plane is $$x^2 + y^2 = -\frac{y}{2} \iff x^2 + \left(y + \frac{1}{4}\right)^2 = \left(\frac{1}{4}\right)^2$$ which is a circle centered at $(0, -1/4)$ and radius $1/4$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving $\sin x = \sin 2x$ The equation is: $\sin x = \sin 2x$
I recognize that $\sin 2x$ is a double angle and use the double angle identity, so it becomes:
$$\begin{array}{rrcl}
&\sin x &=& 2 \sin x \cos x \\
\implies& \sin x - 2 \sin x \cos x &=& 0
\end{array}$$
Then I am stuck... Not sure how to proceed.
| \begin{align}
\sin x & = \sin 2x \\
&= 2 \sin x \cos x
\end{align}
Then
\begin{align}
\sin x - 2 \sin x \cos x = 0
\end{align}
So
$$\sin x (1 - 2\cos x ) = 0$$
This imples that $\sin x = 0$ or $(1-2\cos x) = 0$.
$\sin x = 0$ implies that $x = n \pi$. Also $(1 - 2\cos x ) = 0$ implies that $\cos x = \frac{1}{2}$ and so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangular series perfect square formula 8n+1 derivation In triangular series
$$1 $$
$$1+2 = 3$$
$$1+2+3 = 6$$
$$1+2+3+4 =10$$
$$\ldots$$
Triangular number in 8n+1 always form perfect square .
Like $8\cdot 1+1 = 9 , 8\cdot 3+1 = 25$ .
How this formula is derived ?
| $$\sum_{i=1}^ki=\sum_{i=1}^k\frac{(i+1)^2-i^2-1}{2}=\frac{(k+1)^2-1^2-k}{2}=\frac{k(k+1)}{2}$$
The $k$th triangular number is $\frac{k(k+1)}{2}$.
$$8\left[\frac{k(k+1)}{2}\right] +1=4k^2+4k+1=(2k+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve integral : $\int\frac{\sqrt{x^2 + 1}}{x^2 + 2}dx$ How to solve integral : $\int\frac{\sqrt{x^2 + 1}}{x^2 + 2}dx$
Integration by parts or substitution?
| Let $x = \tan u$, $dx = \sec^2u ~du$. Substitute in the integral to get
$$\int \frac{\sec^3u}{2+\tan^2u}du.$$
Now
\begin{align}
I = \int \frac{\sec^3u}{2+\tan^2u}du &= \int \frac{\frac{1}{\cos^3u}}{2+\frac{\sin^2u}{\cos^2u}}du\\
&=\int \frac{\frac{1}{\cos^3u}}{\frac{2\cos^2u+\sin^2u}{\cos^2u}}du\\
&= \int \frac{1}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do i find this limit WITHOUT L'hôpitals rule How do I find
$$\lim_{x\to 0}\frac{(x+4)^{3/2}+e^x-9}{x}$$
without l'hôpital rule? I know from l'hôpital the answer is 4.
| Less elegant than zhw'x answer.
For $(x+4)^{3/2}$, use the generalized binomial theorem of Taylor expansion
$$(x+4)^{3/2}=8+3 x+\frac{3 x^2}{16}+O\left(x^3\right)$$ Using the standard Taylor expnasion of $e^x$, we then have $$(x+4)^{3/2}+e^x-9=\left(8+3 x+\frac{3 x^2}{16}+O\left(x^3\right) \right)+\left( 1+x+\frac{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Guess the formula for $\sum\frac 1{(4n-3)(4n+1)}$ and prove by induction
For $n \ge 1$, let
$$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
Guess a simple explicit formula for $a_n$ and prove it by induction.
So I have guessed the formula as : $$\frac{n}{4n+1}$$
But wasn't sur... | Base case:
$$a_1=\frac1{4(1)+1}\color{green}\checkmark$$
Now the inductive step:
$$\begin{align}a_{n+1}&=a_n+\frac1{(4n+1)(4n+5)}\\&=\frac n{4n+1}+\frac1{(4n+1)(4n+5)}\\&=\frac{n(4n+5)}{(4n+1)(4n+5)}+\frac1{(4n+1)(4n+5)}\\&=\frac{4n^2+5n+1}{(4n+1)(4n+5)}\\&=\frac{(4n+1)(n+1)}{(4n+1)(4n+5)}\\&=\frac{n+1}{4(n+1)+1}\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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fine the limits :$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$ fine the limits-without-lhopital rule and Taylor series :
$$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=... | Here is a different way.
You can use standard Taylor series expansions, as $x \to 0$, to get $$\sin 2x-2x\cos x=-\frac{x^3}3+o(x^4) $$ $$\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))=-18x+o(x^2) $$ $$x\sin x \tan x\sin 2x=2x^4+o(x^5)$$ then $$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284923",
"timestamp": "2023-03-29T00:00:00",
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How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
Someone told me that it has $(29+1)/2=15$ solutions. How come? Any other method to solve this?
| First, factorise :
$$x^2-y^2=(x-y)(x+y)=5^{29}.$$
Then by the Fundamental Theorem of Arithmetics, both factors must be powers of $5$, i.e.
$$\left\{\begin{array}{}x+y & = & 5^a \\ x-y & = & 5^b,\end{array}\right.$$with $a+b=29$. Since $x+y>x-y$, we must have $a>b$; hence you will have one system for every $15\leq a\leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relationship between two rotated ellipses centered at the origin Let there be two ellipses centered at the origin, and rotated angles $\alpha$ and $\alpha'$ from the x-coordinate axis. $a, b$ are the semiaxes of the first, $a',b'$ the semiaxes of the second, and the ratio $ab/a'b'$ is called $f$.
The equations for both... | The equation of an ellipse in standard position with semi-axis lengths $a'$ and $b'$ can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\frac1{a'^2}&0\\0&\frac1{b'^2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1.\tag1$$ The trace of the central matrix—the sum of its main diagonal element... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial sum using telescopic series The summation of $(n^2+1)\cdot n!$ using to $N$ using telescopic series.
Question: Find a closed expression in $n$ for:
\begin{gather}
\text{Let } u_n = (n^2+1)\cdot n! \\
\sum_{n=1}^{N}u_n = \sum_{n=1}^{N}(n^2+1)\cdot n!\\
\end{gather}
My Attempt
I tried to split $(n^2+1)\cdot n!$ i... | The key is that $$(n^2+1)\cdot n! = n\cdot(n+1)! - (n-1)\cdot n!.$$
Thus $$\sum_{n=1}^Nu_n=(N\cdot(N+1)!-(N-1)\cdot N!) + ((N-1)\cdot N! - (N-2)\cdot (N-1)!) +\\+ \cdots + (1\cdot2!-0\cdot1!) = N\cdot(N+1)!.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solve equation with absolute value of complex numbers Find all complex numbers $z$ such that $|z| + \overline{z}=9-3i$.
Thinking about using with the fact that $z=a+bi$ and $|z| = \sqrt{a^2+b^2}$ in some way. Not sure how.
| Write $\;z=x+iy\;$ , so that your equation is
$$\sqrt{x^2+y^2}+x-iy=9-3i\stackrel{\text{compare real and imag. parts}}\implies\begin{cases}\sqrt{x^2+y^2}+x=9\\{}\\y=3\end{cases}\;\;\implies$$
$$\sqrt{x^2+3^2}+x=9\implies x^2+9=81-18x+x^2\implies18x=72\implies x=4$$
and the unique solution is $\;4+3i\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f... | Since the square of each number can only equal $0$, $1$ or $4$, it comes down to selecting either five numbers which equal $1$ (with the four remaining ones being $0$), or select one which equals $2$ and one which equals $1$ (with the seven remaining ones being $0$). As such, the number of valid matrices equals:
$${9 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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completing square for a circle In the following question:
I don't understand how we can get from the original equation to the final equation using completing the square.
Any thoughts as how to get to the final equation?
| One way of doing this is to avoid fractions completely until the last step.
$$\begin{align}
\frac{x+y}{x^2+y^2+1}&=c\\
x+y&=c(x^2+y^2+1)\\
x+y&=cx^2+cy^2+c\\
cx^2-x+cy^2-y+c&=0\\
cx^2-x+cy^2-y&=-c\\
4c(cx^2-x+cy^2-y)&=4c(-c)\\
4c^2x^2-4cx+4c^2y^2-4cy&=-4c^2\\
(2cx)^2-2(2cx)+(2cy)^2-2(2cy)&=-4c^2\\
(2cx)^2-2(2cx)+1+(2cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the solution to the nonlinear PDE: $U - XUx - (1/2)*(Uy)^2 + X^2 = 0$ with $U(X,0) = X^2 - (1/6)*(x^4), 0Step 1: Rewrite the PDE
$P= Ux$ and $Q= Uy$
$$F = U - XP - (1/2)Q^2 + x^2$$
Step 2: Charpits Equations
$dx/dτ = Fp = -x$
$dy/dτ = Fq = -2Q$
$dp/dτ = -Fx-P*Fu = -2x$
$dq/dτ = -Fy-Q*Fu = -Q$
$du/dτ = PFp+QFq = -x... | Hint:
Let $U=X^2-V$ ,
Then $U_X=2X-V_X$
$V_Y=-U_Y$
$\therefore 2X^2-V-X(2X-V_X)-\dfrac{(-V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
$XV_X-V-\dfrac{(V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
Let $V=XW$ ,
Then $V_X=XW_X+W$
$V_Y=XW_Y$
$\therefore X(XW_X+W)-XW-\dfrac{(XW_Y)^2}{2}=0$ with $W(X,0)=\dfrac{X^3}{6}$
$X^2W_X=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $g(x) = \sqrt{1 + x^2}$ is continuous Show that $g(x) = \sqrt{1 + x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon - \delta$ argument.
This is what I was thinking...
Let $p \in \mathbb{R}$. Then $g(x)-g(p) = \sqrt{1 + x^2} - \sqrt{1 + p^2}$
$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \qua... | You can also follow this slightly shorter argument:
Fix an epsilon $\epsilon$.
$$|g(x)-g(p)|=|\sqrt{1+x^2}-\sqrt{1+p^2}|\leq \sqrt{|1+x^2-1-p^2|}\leq \sqrt{|x^2-p^2|}=\sqrt{|x+p||x-p|}$$
You can bound $|x+p|$ by making sure $|x-p|<1$ or $|x+p|<1+2p$. Also, you make sure that $|x-p|<\frac{\epsilon ^2}{1+2p}$ So that yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of all possible values of $\gcd(a-1,a^2+a+1)$
Find the sum of all possible values of
$$\gcd(a-1,a^2+a+1)$$
where $a$ is a positive integer.
| \begin{align}
\gcd(a-1,a^2+a+1)&=\gcd(a-1,a^2+a+1-(a+2)(a-1))\\
&=\gcd(a-1,3)\\
&=\begin{cases} 3 & \textrm{if }a\equiv 1\quad(\textrm{mod }3) \\ 1 & \textrm{if }a\not\equiv 1\quad(\textrm{mod }3) \end{cases}
\end{align}
The sum is $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to write $\frac{x^3+4x}{x^2-4}$ as a serie at $x_0=1$? I did this, but what to do next?
$f(x)=x+\frac{4}{x-2}+\frac{4}{x+2}$
| In this case the $n$-th derivative of $f$ is explicitly calculable:
$$f(x) = x + 4\,\frac{1}{x - 2} + 4\,\frac{1}{x + 2}$$
$$f'(x) = 1 - 4\,\frac{1}{(x - 2)^2} - 4\,\frac{1}{(x + 2)^2}$$
$$f''(x) = 4\,\frac{2}{(x - 2)^3} + 4\,\frac{2}{(x + 2)^3}$$
$$\cdots$$
$$f^{(n)}(x) = (-1)^n4\,\frac{n!}{(x - 2)^{n+1}} + (-1)^n4\... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$
Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$
My idea was to find the minimal polynomial of $\sqrt[3]{5}+\sqrt{2}$ over $\mathbb{Q}$ and to show that $\deg p(x)=6$
Attempt:
Let $u:=\sqrt[3]{5}+\sqrt{2}\\
u-\sqrt[3]{5}=\sqr... | There is no need to compute minimal polynomials.
Let $\alpha = \sqrt[3]{5}+\sqrt{2}$. Then $(\alpha-\sqrt{2})^3=5$ and so $\alpha^3 + 6 \alpha -5 =\sqrt{2} (3\alpha^2 + 2)$, which gives
$(\alpha^3 + 6 \alpha -5)^2 =2 (3\alpha^2 + 2)^2$.
Therefore,
$\alpha$ is a root of a polynomial of degree $6$ and so
$\left[\mathbb{... | {
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"timestamp": "2023-03-29T00:00:00",
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Find this limit of an integral Find $$\lim_{n\to\infty}n\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)$$
it is obvious
$$\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}})dx=B\left(\dfrac{n+1}{2n},\dfrac{2n+1}{4n}\right)$$
| You are missing a crucial factor $2$. You want to compute
$$ \lim_{n\to +\infty}n^2\left[\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)\,\Gamma\left(\frac{1}{2}+\frac{1}{4n}\right)}{2\,\Gamma\left(1+\frac{3}{4n}\right)}-\frac{\Gamma\left(\frac{1}{2}\right)^2}{2}\right] $$
hence it is enough to exploit the Taylor ser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Analyze convergence of the series $\sum\limits_{n=1}^\infty\frac{1}{n(n^\frac14-n^\frac13+n^\frac12)}$ $$\sum_{n=1}^\infty\frac{1}{n(n^\frac{1}{4}-n^\frac{1}{3}+n^\frac{1}{2})}$$
It is possible to use the ratio test or the root test or any of them without give opposite answers ? There is a case in which a method give a... | Just for the fun.
Consider $$A=\frac{1}{x(x^\frac{1}{4}-x^\frac{1}{3}+x^\frac{1}{2})}$$ and make $x=y^{12}$ to get $$A=\frac{1}{y^{15} \left(y^3-y+1\right)}=\frac{1}{y^{18} \left(1-\frac 1{y^2}+\frac 1 {y^3}\right)}$$ Now, make the long division (or use Taylor series) to get $$\frac{1}{ 1-\frac 1{y^2}+\frac 1 {y^3}}=1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integral of $\frac{x^9}{(1+x^3)^{1/3}}$, recurrence relation I need to find $$\int \frac{x^9}{(1+x^3)^{1/3}}$$
I have recurrence relation:
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$$a(m+1+np)J_{m, p} = x^{m+1-n}(ax^n + b)^{p+1} - b(m+1-n) J_{m-n,p}$$
But when I tried to find the integral, I had problems with $$J_{0, p}$... | $$a = 1, n = 3, b = 1, p = -\frac{1}{3}$$
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$m = 9$
$$9 J_{9, p} = x^{7}(x^3 + 1)^{\frac{2}{3}} - 7 J_{6,p}$$
$m = 6$
$$6 J_{6, p} = x^{4}(x^3 + 1)^{\frac{2}{3}} - 4 J_{3,p}$$
$m = 3$
$$3 J_{3, p} = x(x^3 + 1)^{\frac{2}{3}} - J_{0,p}$$
I think we have to stop the reccurrence at thi... | {
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"url": "https://math.stackexchange.com/questions/2308701",
"timestamp": "2023-03-29T00:00:00",
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Is Differentiation of total variation function $V(f, [0,x])$ possible in this condition? I am having difficulty of determining whether differentiation of total variation function $V(f, [0,x])$ is possible on $x = 0$ when $f:[0,1]→R$ is defined as:
$ f(x) = 0$ for $x = 0$ and $f(x) = x^2sin(1/x)$ for $x≠0$
I think it wo... | To compute the derivative of the total variation of $f$ at $x = 0$ as a definitional limit, take a partition with subintervals
$\left[((k+1)\pi )^{-1},(k\pi)^{-1}\right].$ Since $f(x) = x^2 \sin(1/x)$ vanishes at the endpoints, there is an extremum at some point $\xi_k$ where
$$\frac{1}{(k+1)\pi} \leqslant \xi_k \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1... | As you noted: $y =\frac {2^{2k} - 1}3$ is always an integer. So $2y = 2\frac {2^{2k} - 1}3 = 2\frac {2^{2k+1} - 2}{3} = \frac {2^{2k+1} - 1}3 - \frac 13$ is integer so $\frac {2^{2k+1} -1}3$ is not an integer.
So no.
If know modulo arithmetic it's easier.
$y = \frac {2^{2k} -1}3$ being an integer is the same as $3y +1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}... | $$\cos(2x)=\sin\left(\frac{\pi}2-2x\right)\\\frac{\sin(\frac{\pi}2-2x)}{x-\frac{\pi}4}=\frac{2\sin(\frac{\pi}2-2x)}{2x-\frac{\pi}2}$$
Now it should be easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Help with this integral: $\int_{0}^{1}\frac{1}{1+x^{2}}dx$ (Riemann) I'm stuck when I try to solve this integral:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx$$
I try this:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}\frac{1}{n}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=... | We can proceed as follows. First, we expand the term $\frac{n}{k^2+n^2}$ in the geometric series
$$\frac{n}{k^2+n^2}=\sum_{\ell =0}^N (-1)^\ell \frac{k^{2\ell}}{n^{2\ell +1}}+(-1)^{N+1}\frac{k^{2N+2}}{n^{2N+1}(k^2+n^2)}\tag 1$$
Then, summing $(1)$ reveals
$$\begin{align}
\sum_{k=1}^n \frac{n}{k^2+n^2}&=\sum_{k=1}^n\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
I've tried different substitutions like put $Q(x)$ $=$ $k$ for some $k$ and getting the equation $k(x^2-6x+8)$ $=$ $(k-2)(x^2-6x)$ $<=>$ $k$=$(6x-x^2)/4$, but that doesn't gi... | $$\begin{align*}
(x^2-6x+8) Q(x) &\equiv (x^2 -6x)Q(x-2)\\
(x^2 - 6x)[Q(x)-Q(x-2)] &\equiv -8Q(x)
\end{align*}$$
If $Q(x)$ has degree $n > 0$, $Q(x) - Q(x-2)$ has degree $n-1$, and so the left hand side has degree $n+1$. However, the right hand side has degree $n$.
So $n\not>0$, and $Q(x)$ can only be a constant functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Growth series for the lamplighter group This paper gives the growth series of the lamplighter group as
$$ \frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2+x^3)^2(1-x-x^2)} = 1 + 3x + 6x^2 + 8x^3 + 10x^4 + \ldots $$
The first 3 coefficients seem to be correct. However, I get 12 group elements with a word norm of 3. They are the fo... | You are right. On page 6 of that paper, equation $(3.5)$ is given as:
$$f_{G\wr\mathbb{Z}}(x)=\frac{f_G(x)(1-x^2)^2(1+xf_G(x))}{(1-x^2f_G(x))^2(1-xf_G(x))}$$
On substituting $(1+x)$ for $f_G(x)$, I get:
$$f_{{\cal{L}}_2}(x)=\frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2-x^3)^2(1-x-x^2)}$$
And when expanded, the result seems to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Do whole number solutions exist for $3x^2 = y^2$? I am investigating properties of square numbers and would like to find whole number solutions for this equation
$3x^2 = y^2$
or $\sqrt{3x^2} = y$
How do I prove that whole number solutions do not exist or how do I identify them?
| We can rearrange this to get
\begin{align}3x^2&=y^2\\
y&=\pm\sqrt{3x^2}\\
&=\pm\sqrt{3}x\end{align}
If we assume $x$ is a whole number, then $y$ can only be a whole number if $x=y=0$
Equally:
\begin{align}3x^2&=y^2\\
x^2&=\frac {y^2}{3}\\
x&=\pm\sqrt{\frac{y^2}{3}}\\
&=\pm\frac{y}{\sqrt{3}}\end{align}
Again, if we ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Proving of exponential equation If $3^a=21^b$ and $7^c=21^b$
prove that $b=\frac{ac}{(a+c)}$
Can someone please help me prove this ?
Already tried:
$3^a=3^b \times 7^b$ and $7^c=3^b \times 7^b$
Also, $3^a=7^b$.
| from the first, You have
$$3=(21)^{\frac {b}{a}} $$
and from the second,
$$7=(21 )^{\frac {b}{c} }$$
thus by product,
$$21=(21)^{\frac {b}{a}+\frac {b}{c}} $$
hence
$$1=\frac {b}{a}+\frac {b}{c} $$
and
$$\frac {1}{b}=\frac {1}{a}+\frac {1}{c} $$
$$\implies b=\frac {ac}{a+c} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Completing the square but in different situation
Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$
My attempt,
I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$
But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x... | Well, we have:
$$x^2+4\cdot\left(\frac{x}{x-2}\right)^2=45\tag1$$
Bring together using a common denominator:
$$\frac{x^2\cdot\left(x^2-4x+8\right)}{\left(x-2\right)^2}=45\tag2$$
Multiply both sides by $\left(x-2\right)^2$:
$$x^2\cdot\left(x^2-4x+8\right)=45\cdot\left(x-2\right)^2\tag3$$
Expand out terms of the right ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $c+d$ when $a,b,c,d\in\mathbb N$, $ab=2(c+d)$, $cd=2(a+b)$ Question:
\begin{align}
&a,b,c,d\in\mathbb N\\
&ab=2(c+d)\\
&cd=2(a+b)\\
&a+b\ge c+d\\
&\text{(Four numbers don't need to be all different.)}\\\\
&\text{Find all possible values for }c+d
\end{align}
What I tried was
\begin{align}
&x^2-(a+b)x+2(c+d)=0\\
&x^... | Hint
$$0=ab+cd-2(c+d)-2(a+b)\\
8=(a-2)(b-2)+(c-2)(d-2)\\
$$
This leads to only few possibilities to check where all numbers are at least $2$.
Hint 2: If $a=1$ (or any other of the numbers) you have
$$b=2(c+d)\\
cd=2(b+1)$$
hence
$$cd=2(b+1)=2(2(c+d)+1)=4c+4d+2\\
(c-4)(d-4)=18$$
which is easy to solve by factoriations.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Wieferich's criterion for Fermat's Last Theorem I have found the following way to prove some(Wieferich's) criterion for Fermat's Last Theorem and am wondering what would be wrong. My point of doubt is calculation of the Fermat-quotients of $y,z$ being $-1$, since I found these rules on Wikipedia.
Also, should I split ... | A mistake they made is assuming $q_p(s) \equiv q_p(u) \equiv 0 \pmod p$, since $s^{p - 1} \equiv 1/s \pmod {p^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{... | $180°-122.5°=57.5°$, and the angles sum to be $180°$, nothing wrong here.
The angles you find has nothing to do with whether there is such a triangle with the three points as vertices. The triangle is already formed by the three points before you measure the angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ and the straight lines $OP,OQ$ make angles $\alpha,\beta$ with the $x$-axis where $O$ is the origin then $\frac{\tan\alpha}{\tan\beta}=$
$(A)-1$
$(B)-2$
$(C)2$
$(... | We have
\begin{align}
\frac{y_2-y_1}{x_2-x_1}&=\frac{3x_1^2}{2y_1}\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3x_1^3\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3y_1^2\\
2y_1y_2+y_1^2&=3x_1^2x_2\\
y_1^3(2y_2+y_1)^3&=27x_1^6x_2^3\\
y_1^3(2y_2+y_1)^3&=27y_1^4y_2^2\\
(2y_2+y_1)^3&=27y_1y_2^2\\
8y_2^3-15y_2^2y_1+6y_2y_1^2+y_1^3&=0\\
(y_2-y_1)^2(8y_2+y_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Witty functional equation
Let
$$f(x) = 2/(4^x + 2)$$
for real numbers $x$. Evaluate
$$f(1/2001) + f(2/2001) + f(3/2001) + \cdots + f(2001/2001)$$
Any idea?
| We have $\displaystyle f(1-x)=\frac{2}{4^{1-x}+2}=\frac{2\cdot 4^x}{4+2\cdot 4^x}=\frac{4^x}{2+ 4^x}$
$\displaystyle f(x)+f(1-x)=\frac{2}{4^x+2}+\frac{4^x}{2+ 4^x}=1$
The required sum is
$$\frac{2000}{2}\times 1+f(1)=1000+\frac{2}{4+2}=\frac{3001}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 1,
"answer_id": 0
} |
Give me some hints in calculation this limit. $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ \tan (x^{2}-y^{2}) - y^{6}}{\sin(x^{6}-y^{6}) - x^{5}y +xy^{5} + \arctan(x^{2}y -xy^{2})}$$
I used a fact that $$\tan \alpha \sim \alpha \\ \arctan \alpha \sim \alpha \\ \sin\alpha \sim \alpha$$
Since now we have $$\lim_{x\to2\ \\ y\to... | $\lim_\limits{x\to2\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}\\
\lim_\limits{x\to2\\\ y\to2}\frac{(x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5) + (x-y)(x+y)}{(x-y)(x^5+x^4y+x^3y2+x^2y^3+xy^5+y^5) - (x-y)(xy)(x^3+x^2y+xy^2+y^3) + xy(x-y)}\\
\lim_\limits{x\to2\\\ y\to2}\frac{(x^5+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $... | for $n \ge k$ let the sum $\sum_{r=k}^n r^3 $ be $S_{k}(n)$ so for $n \ge 0$
$$
S_{-2}(n) = S_0(n) -9
$$
since $S_0(0)=0$ this gives $S_{-2}(0)=e=-9$ and
$$
S_0(n)= an^4+bn^3+cn^2+dn
$$
now $S_0(n)-S_0(n-1) = n^3$
so
$$
\begin{align}
a(4n^3-6n^2+4n-&1) + \\
b(3n^2-3n+&1)+\\
c(2n-&1)+\\
&d= n^3
\end{align}
$$
comparing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we g... | $A^4=I$ implies that $A$ is invertible. Hence $B^2=A^{-1}BA$ . Repeatedly squaring and using the previous step we get $B^{16}=A^{-4}BA^{4}$ which gives $B^{16}=B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 4
} |
Is this correct ${2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}?$ Given this infinite product
$$\lim_{n \to \infty}{2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}\tag1$$
Where $n\ge0$
Experimental on the infinite product ... | In the same spirit as Meet Taraviya's answer, write
$$\prod_{n=0}^{\infty}\frac{4n+2}{4n+2+(-1)^n}=\prod_{k=0}^{\infty}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}$$ Considering the partial product
$$P_m=\prod_{k=0}^{m}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}=\sqrt{\pi }\frac{ \sec \left(\frac{\pi }{8}\right)\, \Gamma \left(2
m+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Surface area of an ellipse I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!
Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid.
$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\... | $\frac {x^2}{9} + \frac {y^2}{4} = 1\\
\frac {d}{dx}(\frac {x^2}{9} + \frac {y^2}{4} = 1)\\
\frac {2x}{9} + \frac {2y}{4}\frac {dy}{dx} = 0\\
\frac {dy}{dx} = - \frac {4 x}{9 y}\\
2\pi \int_{-3}^{3} y\sqrt {1+(\frac {dy}{dx})^2}\ dx\\
2\pi \int_{-3}^{3} \sqrt {y^2+ \frac {16}{81}x^2}\ dx$
Note you have the wrong limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Diagonalizing symmetric real bilinear form I am given the following symmetric matrix:
$$
A=\begin{pmatrix}
1 & 2 & 0 & 1\\
2 & 0 & 3 & 0\\
0 & 3 & -1 & 1\\
1 & 0 & 1 & 4\\
\end{pmatrix}\in M_4(\Bbb R)
$$
Let $f\in Bil(V), f(u,v)=u^tAv.$
I want to find a base $B \subset \B... | The method of repeated completing squares also leads, by nature, to rational entries in this case. However, that is not the end of the story, as the given form is $SL_4 \mathbb Z$ equivalent to a diagonal form. In brief,
$$ (w + 2x+z)^2 + (x-2y+z)^2 - (9x-9y+13z)^2 + 19 (2x-2y+3z)^2 = w^2 - y^2 + 4z^2 + 4wx +6xy +2wz ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | You know that $1-\cos x=2\sin^2\dfrac{x}2$ and $1+\cos x=2\cos^2\dfrac{x}{2}$, so $$\frac{1-\cos x}{1+\cos x}=\frac{2}{2}\left(\frac{\sin\dfrac{x}2}{\cos\dfrac{x}2}\right)^2=\tan^2\frac{x}2.$$
If you couldn't understand, please comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Note that :suppose $a=z+\frac 1z $so
$$z+\frac 1z=a\\z^2+\frac{1}{z^2}=(z+\frac 1z)^2-2z\frac1z=a^2-2\\
z^4+\frac{1}{z^4}=(z^2+\frac{1}{z^2})^2-2=(a^2-2)^2-2\\
z^3+\frac{1}{z^3}=(z+\frac 1z)^3-3z\frac1z(z+\frac 1z)=a^3-3a$$
what you have now is $$(a^2-2)^2-2=47 \to\\(a^2-2)^2=49 \\\begin{cases}a^2-2=7 & a^2=9\\\to a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $... | Let $a=\tan(\frac{\pi}{5})$ and $b= \tan(\frac{2 \pi}{5})$. Also let $s=a^2+b^2$ and $p=(a b)^2$. We have
\begin{eqnarray*}
a^2+b^2 &=&(ab)^2+5 \\
s&=&p+5.
\end{eqnarray*}
Now $a(1-b^2)=-2b$ and $b(1-a^2)=2a$ square these equations and add them together
\begin{eqnarray*}
3(a^2+b^2)+4(ab)^2&=&(ab)^2(a^2+b^2) \\
3s+4p&=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
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find matrix $A$ in other basis Original matrix is:
$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$
original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$
basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$
Where:
$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\... | Call the first and second basis $E$ and $F$ respectively. You are given $$A_E=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$ and the change of coordinates matrix that changes $F$ coordinates to $E$ coordinates is $$Q=\begin{pmatrix}2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$ Then $$A_F=Q^{-1}A_EQ.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continued fraction in 8th root------ any simpler approach? It seems to be a problem from the Putnam Exam. The problem asked to find the exact value of
$$x=\sqrt[8]{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}}$$
And express as $\dfrac{a+b\sqrt{c}}{d}$, in terms of some integers $a,b,c,d$.
My a... | Let $x=a- \frac{1}{x}$ ; this satisfies $x^2-ax+1=0$ ... & $x^2$ satisfies
\begin{eqnarray*}
(x^2+1)^2=(ax)^2 \\
(\color{blue}{x^2})^2-(a^2-2)\color{blue}{x^2}+1=0
\end{eqnarray*}
So to "square root a continued fraction" we need to solve $a^2-2=b$ ... eighth root so we need to do this three times
\begin{eqnarray*}
a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer an... | A non power/taylor series method using Leibniz' generalize product rule:
$$ (fg)^{n}(x) = \sum_{r = 0}^{n} \binom{n}{r} f^{n-r}(x) g^{r}(x)$$
Let $f(x) = \dfrac{1}{x^2 + x + 1}$ and $g(x) = x^2 + x +1 $
Then $f(x) g(x) = 1$. We apply the rule to $fg$.
If $a_k = f^{k}(0)$ the we get the recurrence relation
$$a_k + ka_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
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"answer_id": 5
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Conjectured primality test II This question is closely related to my previous question .
Can you provide a proof or a counterexample for the following claim :
Let $n$ be a natural number , $n>1$ and $n \not\in \{4,8,9\}$ . Then $n$ is prime if and only if
$\displaystyle\sum_{k=1}^{n}\left(2^k+1\right)^{n-1} \equiv n \... | This is a partial answer.
This answer proves that if $n$ is a prime and $b\ge 2$ is an integer, then
$$\displaystyle\sum_{k=1}^{n}\left(b^k+1\right)^{n-1} \equiv n \pmod{\frac{b^n-1}{b-1}}$$
Proof : (This proof is similar to the one in this answer to your previous question.)
For $n=2$, we have
$$\displaystyle\sum_{k=1... | {
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"url": "https://math.stackexchange.com/questions/2345805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| $(a+b)^2 + (b+c)^2 + (a+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=2(a^2+b^2+c^2)+2(ab+bc+ac)=2(a^2+b^2+c^2)+2 \cdot 1=2((a+b+c)^2-2(ab+bc+ac))=2 \cdot (2)+2=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 6
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Differential equation via Laplace transform $$xy''+(2x+3)y'+(x+3)y=3e^{-x}$$
$$L[xy'']+L[(2x+3)y']+L[(x+3)y]=L[3e^{-x}]$$
$$-\frac{d}{dp}(p^2Y)-\frac{d}{dp}(2pY)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2)-\frac{dY}{dp}(2p)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2+2p+1)=\frac{3}{p+1}$$
$$-dY=-\frac{3d... | Consider the equation
$$t \, y'' + (2 \, t + 3) \, y' + (t+3) \, y = a \, e^{-t}$$
then by the standard Laplace transform this becomes
\begin{align}
- \partial_{s}(s^2 \, \overline{y}) + y(0) - 2 \, \partial_{s}(s \, \overline{y}) + 3 \, s \, \overline{y} - 3 \, y(0) - \overline{y}' + 3 \overline{y} &= \frac{a}{s+1}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure
| Let $R>3$ be the circumradius of $ABC$ and $a,b,c$ its side lengths. We have
$$ a=2\sqrt{R^2-(R-2)^2},\qquad b=2\sqrt{R^2-(R-3)^2},\qquad c=2\sqrt{R^2-(R-1)^2} $$
hence
$$ a^2=16R-16,\qquad b^2=24R-36,\qquad c^2 = 8R-4 $$
and we also have
$$ R^2 = \frac{a^2 b^2 c^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} $$
so $R$ is the root... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac... | This would be one way of going about it but using a different fourier series:
$$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$
Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just use odd nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an o... | There is nothing strange in having different expressions for an antiderivative: an antiderivative of $1/x$ over $(0,\infty)$ is $\log x$ as well as $\log(3x)$, because antiderivatives are determined (over an interval) up to an additive constant. However, your second computation is wrong.
An antiderivative of $1/(1-x^2)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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let $f$ be a differentiable function. Compute $\frac{d}{dx}g(2)$, where $g(x) = \frac{f(2x)}{x}$. let $f$ be a differentiable function and $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$
Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative
$$\dfrac{d}{dx}g(2)=?$$
I know that :
$$\dfrac{d}{dx}g(2)=\dfrac{... | You have an extra $4$ in the numerator here:
i know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$
If $g(x) = \dfrac{f(2x)}x$, then
\begin{align*}
\frac d{dx} g(x) &= \frac d{dx} \left(\frac{f(2x)}x\right)\\[0.3cm]
&= \frac{x \frac d{dx} f(2x) - f(2x) \frac d{dx}x}{x^2}\\[0.3cm]
&= \frac{2x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of this indefinite integral. $$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
$$
where both $a$ and $l$ are constants.
I've tried simplifying :
$$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(l^2+a^2+x^2-a^2-x^2)dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(\sqr... | Take $l^2+a^2 = r^2$ and $x=r\tan (\theta)$ so that the indefinite integral becomes $$\int \frac{r \sec^2(\theta) d\theta}{(r \sec(\theta))(r^2 \tan^2(\theta) + a^2)} = \int \frac{\sec(\theta) d\theta}{r^2 \sec^2(\theta) - l^2} = \int \frac{\cos(\theta) d\theta}{r^2-l^2 \cos^2(\theta)} =$$ $$\int \frac{\cos (\theta) d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get the $\phi$ from $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$? $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,
where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.
I want to know how to get to this result.
I'm able to derive $c$ by taking the derivative of the e... | For fun, here's a trigonograph, which leads to a counterpart cosine identity:
$$\begin{align}
a \sin x + b \sin(\theta+x) &= c\sin(\phi+x) \\
a \cos x + b \cos(\theta+x) &= c\cos(\phi+x)
\end{align}$$
where
$$c^2 = a^2 + b^2 - 2 a b \cos(180^\circ-\theta) = a^2 + b^2 + 2 a b \cos \theta \qquad\text{and}\qquad \tan\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume of the region above the cone $ z=\sqrt{x^2+y^2} $ and below the paraboloid $ z=2-x^2-y^2 $ . Find the volume of the region above the Cone $ z=\sqrt{x^2+y^2} $ and below the Paraboloid $ z=2-x^2-y^2 $ .
The options are (i) $\frac{13 \pi}{6}$, (ii) $\frac{7 \pi}{3}$, (iii) $\frac{13 \pi}{3}$, (iv) $4 \... | Converting to cylindrical coordinates, the paraboloid is given by the equation $z = 2-r^2$ and the cone is given by the equation $z=r$. These curves intersect when $$2-r^2 = r \implies r=1\, \text{ or } r=-2$$ Since $r\geq0$, $r=1$. The volume is computed to be
\begin{align}
V &= \int_0^{2\pi}\int_0^1\int_r^{2-r^2}r\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is ev... | $$x^2+x+1=0$$
Let's just consider$$x=\exp\left(\frac{i\pi}{3}\right).$$
$$1990 \equiv 1989+1 \equiv 1 \mod 3$$
$$200 \equiv 198+2 \equiv 2 \mod 3$$
Hence $$x^{1990}+x^{200}+1 = x+x^2+1=0$$
$\exp\left(-\frac{i\pi}{3} \right)$ being the conjugate must be another solution as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational We have to show that roots of $$abc^2x^2 + (3a^2 c + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational.
This can be possible if the discriminant is a perfect square. SO I tried converting it into perfect square but failed:
$$\text{Discriminant}=(3a^2c+b^2c... | The product of the roots is
$$
\frac{-6a^2 -ab +2b^2}{abc^2}
= \frac{(b - 2 a) (3 a + 2 b)}{a b c^2}
$$
which immediately suggests what the roots are.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F... | $$a^{ 2 }+{ a }^{ 2 }+b={ b }^{ 2 }+ab+b\\ { a }^{ 2 }-{ b }^{ 2 }=ab-{ a }^{ 2 }\\ \left( a-b \right) \left( a+b \right) =a\left( b-a \right) \\ \left( a-b \right) \left( a+b \right) +a\left( a-b \right) =0\\ \left( a-b \right) \left( 2a+b \right) =0\\ a\neq b,b=-2a\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $n^a+mn^b+1$ divides $n^{a+b}-1$, prove $m=1$ and $b=2a$ Let $n,m,a,b $ be positive integers such that: $n\geq 2$, $a\leq b $ and $n^a+mn^b+1$ divides $n^{a+b}-1$. Prove that $m=1$ and $b=2a $.
We could factor the divisor to proceed by $n^{a+b}-1=(n-1)(1+n+n^2+...+n^{a+b-1})$ but I cannot see another way we co... | Suppose $k(n^a + mn^b + 1) = n^{a+b} - 1$. Since $k$ is the ratio of two positive integers (using $n\ge 2$), it is certainly positive, and also clearly $k < n^{a+b}/(mn^b) = n^a/m$. In particular $0 < k < n^a$.
On the other hand, looking at this equation modulo $n^a$ (using the fact that $a \le b$ and hence $n^a \mid... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all the natural numbers $n$ such that $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime How many natural numbers $n$ exist for which $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime?
I thought a lot but I don't how to calculate the correct part.
| So it's pretty clear that $n = 3k$ only works for $k=1$ i.e. $n=3$.
For $n = 3k+1$ we have that $n^2 = 9k^2 + 6k + 1$, so $\lfloor \frac{n^2}{3}\rfloor = 3k^2 + 2k$. If $k$ is even then this number is divisible by $2$. If $k$ is odd, then this number is divisible by $k$, which means that we still have a solution when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies... | Let $x \neq 0$
$$f\left(\frac1x\right)=\frac{1/x}{1+1/x^2}=\frac{x}{1+x^2}=f(x)$$
Thus, $f$ is not injective. Your mistake was $x+xy^2=y+yx^2 \implies x=y$. This is certainly not true.
$$g(-2)=\frac{(-2)^2}{1+(-2)^2}=\frac{4}{1+4}=g(2)$$
Hence, $g$ is not injective.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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arctan series multisection by roots of unity I'm trying to multisect the series for $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots$ using the method of roots of unity, as described in the paper linked in one of the comments in How to express a power series in closed form. I wan... | We have
\begin{eqnarray*}
\tan^{-1}(x)= x- \frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
and also
\begin{eqnarray*}
\tanh^{-1}(x)= x+ \frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
Now just add these to obtain
\begin{eqnarray*}
\frac{\tan^{-1}(x)+\tanh^{... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \thet... | $|\frac {z}{4} + \frac {6}{z}| \ge 0$
I say that there exist a $z$ such that $1\le|z|\le 7$ and $\frac {z}{4} + \frac {6}{z} = 0$
and any such $z$ must minimize the objective.
$z = i 2\sqrt {6} $
To maximize the objective
$|\frac {z}{4} + \frac {6}{z}| \le |\frac {z}{4}| + |\frac {6}{z}|$
if $z$ is real then:
$|\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute $\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$
Compute $\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$
As I see it there are two possibilities:
*
*$g(z)=\frac{\log(z+4)}{(z+i)^2}$ is holomorphic on $D_1(i)$, so we have
$\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}=\oint_{|z-i|=1}\frac{g(... | Both forms are indeed equivalent. Simply note that
$$\begin{align}
\frac{d}{dz}\left((z-i)^2f(z)\right)&=2(z-i)f(z)+(z-i)^2f'(z)\\\\
&=2(z-i)\frac{\log(z+4)}{(z^2+1)^2}+(z-i)^2\left(\frac{1}{(z+4)(z^2+1)^2}-\frac{4z\log(z+4)}{(z^2+1)^3}\right)\\\\
&=\frac{1}{(z+4)(z+i)^2}+\frac{2\log(z+4)}{(z+i)^3}\left(\frac{z+i}{z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364982",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | Yes Possible.
Evaluate
$$\lim_{ x\to \infty} \left( tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
} |
Proof by induction of $\sum_{k=2}^n (k-1)(k)\binom{n}{k} = n(n-1)2^{n-2}$ I've been struggling with this sum for an while, plugging $n+1$ instead of $n$, knowing that $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$ and after some manipulation I've found this sum.
$$2\sum_{k=1}^{n} k^2\binom{n}{k}$$
I couldn't see an... | Applying Pascal's identity gives you
$$ \sum_k k(k - 1) \binom{n + 1}{k} = \sum_k k(k - 1) \binom{n}{k} + \sum_k k(k - 1) \binom{n}{k - 1}. $$
If we call the sum $f(n)$ then this says that
$$ f(n + 1) = f(n) + \text{ a sum that looks very much like } f(n). \tag{1}$$
We can deal with the second sum by changing variables... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Wrong solution using Cramer's rule? I have tried to solve the following systems of equations:
$$
\left\{
\begin{array}{rrrcl}
x & + y & + z &=& 1 \\
2x & + 5y & + 2z &=& 4 \\
4x & + 2y & + 3z &=& 5
\end{array}
\right.
$$
The determinant of the matrix of the coefficients is $3$ and the other determin... | At least$$\Delta=1\cdot5\cdot3+1\cdot2\cdot2+4\cdot1\cdot2-4\cdot5\cdot1-1\cdot2\cdot2-3\cdot2\cdot1=-3$$
$$\Delta_x=1\cdot5\cdot3+1\cdot2\cdot5+4\cdot1\cdot2-5\cdot5\cdot1-1\cdot2\cdot2-3\cdot4\cdot1=-8$$
$$\Delta_y=1\cdot4\cdot3+1\cdot4\cdot2+5\cdot1\cdot2-4\cdot4\cdot1-1\cdot2\cdot5-3\cdot2\cdot1=-2$$ and
$$\Delta_z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Primality test using Chebyshev and Legendre polynomials Inspired by Theorem 5 in this paper I have formulated the following claim :
Let $n$ be an odd number and $n>1$ . Let $T_n(x)$ be Chebyshev polynomial of the first kind and let $P_n(x)$ be Legendre polynomial , then $n$ is a prime number if and only if the followi... | This is a partial answer.
This answer proves that if $n$ is an odd prime, then $P_n(3)\equiv 3\pmod n$.
Using that $\binom nk\equiv 0\pmod n$ for $1\le k\le n-1$, we have
$$\begin{align}P_n(3)&=\frac{1}{2^n}\sum_{k=0}^{n}\binom nk^2(3-1)^{n-k}(3+1)^k\\\\&=\frac{1}{2^n}\sum_{k=0}^{n}\binom nk^2\cdot 2^{n-k}\cdot 2^{2k}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Mu... | help from the graph :
$$x^2+y^2=10 \to \text {Circle}$$
$$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{3}\to y=\dfrac{3x}{4x-3}\to \text{Hemographic
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of the bounded decreasing sequence
Find the limit of the bounded decreasing sequence defined by
$a_1=3 \\a_{n+1}= \frac{1}{4-a_n}+1$
Can anyone teach me how to do this question? Thanks.
The answer is $\frac{5-\sqrt5}{2}$.
| We'll prove that $a_n\leq3$.
Indeed, $a_1\leq3$ and $a_{n+1}=\frac{1}{4-a_n}+1\leq\frac{1}{4-3}+1=2<3$.
In another hand $a_1>\frac{5-\sqrt5}{2}$ and
$$a_{n+1}-\frac{5-\sqrt{5}}{2}=\frac{1}{4-a_n}-\frac{3-\sqrt5}{2}=\frac{1}{4-a_n}-\frac{2}{3+\sqrt5}=$$
$$= \frac{1}{4-a_n}-\frac{1}{4-\frac{5-\sqrt5}{2}}=\frac{a_n-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest value of $n$ such that $19 \,|\, 10^n+1$ A riddle I'm working on reduces to this question. I don't have a number theory background and don't really know how to approach this kind of problem.
In fact I'm not sure such an $n$ exists. I feel like it does, but my computer has been chugging on an R script ... | The pattern
$\begin{array}{l}
10^3 + 1 = 1001 = 999+2 \\
10^4 + 1 = 10001 = 9999+2\\
10^5 + 1 = 100001 = 99999+2\\
\dots \\
10^9 + 1 = 999999999+2
\end{array}$
Multiples of $19$:
\begin{array}{r|ccccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
19n & 19 & 38 & 57 & 76 & 95 & 114 & 133 & 152 & 171... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| Let $x=at$ and by Partial-Fraction Decomposition we get
\begin{align*}\int_0^a \frac{x^4}{(x^2+a^2)^4}dx&=\frac{1}{a^3}\int_0^1 \frac{t^4}{(t^2+1)^4}dt\\
&=\frac{1}{a^3}\int_0^1\left(\frac{1}{(t^2+1)^2}-\frac{2}{(t^2+1)^3}+\frac{1}{(t^2+1)^4}\right)\,dt
\\&=\frac{1}{16a^3}\left[\arctan(t)
+\frac{t}{t^2+1}-\frac{(14/3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\s... | Use $\sin{x}=\frac{2t}{1+t^2}$ and $\cos{x}=\frac{1-t^2}{1+t^2}$, where $t=\tan\frac{x}{2}$.
One of roots is $\sqrt2-1$ and the second is very ugly.
For $\tan\frac{x}{2}=\sqrt2-1$ we obtain $\frac{x}{2}=22.5^{\circ}+180^{\circ}k,$ where $k\in\mathbb Z$,
which gives $x=\frac{\pi}{4}$.
The second root does not give solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$
My a... | $\sin{y}=3\sin{x}$ and $\cos{y}=2\cos{x}$.
Thus,
$$1=\sin^2y+\cos^2y=9\sin^2x+4\cos^2x=4+5\sin^2x,$$
which is impossible.
If $\sin{y}=\frac{1}{3}\sin{x}$ we obtain:
$$1=\sin^2y+\cos^2y=\frac{1}{9}\sin^2x+4\cos^2x=\frac{1}{9}+\frac{35}{9}\sin^2x.$$
Now, you can get $\sin^2x$, $\sin^2y$ and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Maximize linear function over disk of radius $2$
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.
$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt... | $$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 \leq 4\end{array}$$
Since the objective function is linear and nonzero, its gradient never vanishes. Thus, the maximum is attained at the boundary of the feasible region, i.e., on the circle of radius $2$ centered at the origin
$$\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Sequence defined recursively: $\sum_{k=1}^n x_k = \frac{1}{\sqrt{x_{n+1}}}$ Let $(x_n)_{n \ge 1}$ defined as follows:
$$x_1 \gt 0, x_1+x_2+\dots+x_n=\frac {1}{\sqrt {x_{n+1}}}.$$ Compute the limit $\lim _ {n \to \infty} n^2x_n^3.$
MY TRY: I thought about using Stolz-Cesaro lemma, but I couldn't get to an appropriate fo... | This answer will use the Stolz-Cesaro lemma.
$$L=\lim_{n \to \infty} n^2x_n^3$$
$$S_n=\sum_{k=1}^n{x_k} \quad\quad x_{n+1}=S_n^{-2}$$
We will try to find a recurrence relation that is easier to manipulate:
$$\frac{x_{n+1}}{x_n}=\frac{S_n^{-2}}{S_{n-1}^{-2}}=\frac{S_{n-1}^{2}}{S_n^{2}}=\left (\frac{S_{n-1}}{S_n}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving System of quadratic equations If $b²-4ac=0$ ($a \neq 0$, and $a, b, c \in \mathbb {R}$) and $x, y $ satisfy the system $$ax²+(b+3)x+c=3y$$ and $$ay²+(b+3)y+c=3x$$ then the value of $x/y$ is...?
| Assuming $x,y \in \mathbb{R}$, then $x/y = 1$.
Your equations give
$$
ax^2+bx+c = 3(y-x) = -3(x-y) = -(ay^2+by+c),
$$
so
$$ ax^2+bx+c = -(ay^2+by+c). $$
The assumption $b^2-4ac=0$, $a\neq 0$, means that one of the following must be true:
*
*$ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2$,
*$ax^2+bx+c = (\sqrt{a}x-\sqrt{c})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Inequality : $(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$ Find the largest integer $k$ such that
$$(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$$
for all positive real numbers $a, b, c$.
My attempt :
By Holder inequality, $(a^3+1+2)(2+b^3+4)(4+8+c^3)\geq(\sqrt[3]{8}a+\sqrt[3]{8}b+\sqrt[3]{8}c)^3$
$\Leftrightarrow(a^3+3)(b^3+6)(c^3+12)\ge... | The equality should be for $(a,1,\sqrt[3]2)||(\sqrt[3]2,b,\sqrt[4]4)||(\sqrt[3]4,2,c),$ which is impossible.
Let $a_1+a_2=3$, $b_1+b_2=6$, $c_1+c_2=12$, $b_1c_1=a_1c_2=a_2b_2$, where $a_i\geq0$, $b_i\geq0$ and $c_i\geq0$.
Thus, for the equality occurring after using Holder we need
$$\left(a,\sqrt[3]{a_1},\sqrt[3]{a_2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rationalize $\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$ I am having trouble rationalizing the denominator of $$\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$$
I tried grouping the denominator as $(2 + 2\sqrt[6]{2}) + 2\sqrt[3]{2}$ and multiplying top and bottom by $(2 + 2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\s... | Let $\sqrt[6]2=x\implies\sqrt[3]2=x^2$
We have $$\dfrac1{2+2x+2x^2}=\dfrac{1-x}{2(1-x^3)}=\dfrac{(1-x)(1+x^3)}{2(1-x^6)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Transforming $-1+ \cos x +\sin x$ into $-2\sin^2 (x/2) +2\sin(x/2)\cos(x/2)$ This trigonometrical part was included in a complex variable problem. I tried to find how
$$ -1+ \cos (x) +\sin (x)$$
becomes
$$-2\sin^2 (x/2) +2\sin(x/2)\cos(x/2)$$
Can I let $1 = \sin^2 x +\cos^2x$ ?
Any help'll be appreciated.
| The identity $ \sin(2 \theta) = 2 \sin( \theta ) \cos( \theta ) $ give us that
$$
\sin(x) = \sin(2 \cdot x/2) = 2 \sin(x/2) \cos(x/2)
$$
Similary, we use that $ cos(2 \theta) = \cos^2( \theta ) - \sin^2( \theta ) $ to obtain
$$
-1 + \cos(x) = -1 + \cos(2 \cdot x/2) = -1 + ( \cos^2(x/2) - \sin^2(x/2) )
$$
Finally from t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\lim_{x\to 0} \frac{e^x-1-x}{x^2}$ without L'Hôpital's Rule. I want to solve this limit without the use of L'Hôpital's rule:
$$\lim_{x\to 0} \frac{e^x-1-x}{x^2}.$$
| I would Taylor Expand
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
and plug in to find
$$\lim_{x\to 0}\frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x}{x^2}=\lim_{x\to 0}\frac{\sum_{n=2}^\infty \frac{x^n}{n!}}{x^2}=\lim_{x\to 0}\frac{\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots}{x^2}=\lim_{x\to 0} \frac{1}{2}+\frac{1}{6}x+\frac{1}{24}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\be... | Write $t=\tan\frac{\theta}{2}$ so $\sin\theta=\frac{2t}{1+t^2},\,\cos\theta=\frac{1-t^2}{1+t^2}$. Hence $$\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1+t^2+2t-1+t^2}{1+t^2+2t+1-t^2}=\frac{2t+2t^2}{2+2t}=t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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An inequality with $\cos$ and triangle sides Here is the problem:
Let $ABC$ be a triangle with sides $a, b, c$. Show that $\dfrac{\cos A}{a^3}+\dfrac{\cos B}{b^3}+\dfrac{\cos C}{c^3}\geq\dfrac{3}{2abc}.$
Here's my attempt:
By the cosine formula, we have $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ etc, which the left hand sid... | $$2abc\sum_{cyc}\frac {a^2+b^2-c^2}{2abc^3}=$$ $$=\sum_{cyc}(\frac {a^2}{c^2}+\frac {b^2}{c^2}-1)=$$ $$=-3+\sum_{cyc}(\frac {a^2}{b^2}+\frac {b^2}{a^2})=$$ $$=-3+\sum_{cyc}(2+(\frac {a}{b}-\frac {b}{a})^2\;)=$$ $$=+3+\sum_{cyc}(\frac {a}{b}-\frac {b}{a})^2\geq 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given $m$, find three numbers $a,b,c$ such that $c-b=m(b-a)$ and $a+b,a+c,b+c$ are all squares I've worked with the next problem
*
*Given an integer $m$, find three integers $a,b,c$, with $b\neq a$, such that $c-b=m(b-a)$ and
$$a+b=x^2, \ \ a+c=y^2, \ \ b+c=z^2,$$
where $x,y, z \in \mathbb{Z}$.
I've not been able... | Let
\begin{eqnarray*}
a&=&2((mp^2+2mpq-q^2)^2+(mp^2+q^2)^2-(-mp^2+2pq+q^2)^2) \\
b&=&2((mp^2+2mpq-q^2)^2-(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
c&=&2(-(mp^2+2mpq-q^2)^2+(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
\end{eqnarray*}
One can easily verify that
\begin{eqnarray*}
a+b&=& 4(mp^2+2mpq-q^2)^2 \\
b+c&=&4(-mp^2+2pq+q^2)^2 \\
c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqr... | Denote $u = \frac{x^3+y^3+z^3}{3xyz}$ and $v = \frac{xy+yz+zx}{x^2+y^2+z^2}$.
Then $u\ge 1$ and $0 \le v \le 1$.
We need to prove that $\sqrt[3]{3u} + \sqrt{v} \ge 1 + \sqrt[3]{3}$ or
$\sqrt[3]{3u} - \sqrt[3]{3} \ge 1 - \sqrt{v}$ or
$$\frac{3u - 3}{3^{2/3}(u^{2/3} + u^{1/3} + 1)} \ge \frac{1 - v}{1 + \sqrt{v}}.$$
By us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Maclaurin series with zero denominator when evaluating derivative I have following function:
$\displaystyle f(x) = \frac{\ln(1+x^2) - x^2}{\sqrt{1+x^4} - 1}$
As you can see, when doing the quotient rule for the denominator in your head, the derivative of this function results to $0$ in the denominator when evaluating f... | $$\eqalign{\ln(1+x^2) &= - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4}+ \ldots\cr
\sqrt{1+x^4} - 1 &= \frac{x^4}{2} - \frac{x^8}{8} + \ldots\cr
\frac{\ln(1+x^2)}{\sqrt{1+x^4}-1} &= \frac{x^4 \left(-\dfrac{1}{2} + \dfrac{x^2}{3} + \ldots\right)}{x^4 \left(\dfrac{1}{2} - \dfrac{x^4}{8} + \ldots\right)}\cr
&= -1 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
First four nonzero terms of the McLaurin expansion of $\frac{xe^x}{\sin x}$ at $x_0=0$
Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series.
$$ \frac{xe^x}{\sin x}$$
Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x... | I think that your expansion is partially incorrect. After setting $f(0)=1$, note that for $x\not=0$,
\begin{align*}
f(x)=\frac{x e^x}{\sin(x)}&=x \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(x-\frac{x^3}{6}+o(x^4)\right)^{-1}\\
&=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1-\frac{x^2}{6}+o(x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$
Prove that:
$$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$$
I computed the indefinite integral:
$$\int\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}(\ln 2)^2}+C$$
How can I continue from here?
| $$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}>\int\limits_{1}^{+\infty}\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}\ln^2 2}|_1^{+\infty}=\frac{1}{4\ln^22}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| Partial answer, complete for $n$ even.
Suppose $n\geq2$. Clearly, $k>n$.
Then $k!$ is even, so there is an even number of odd numbers on the left. So $n\equiv0,3\pmod4$.
Suppose $n\geq4$ is even. Odd squares are $1$ mod $8$, and $8\mid k!$, so $n/2$ is divisible by $8$. Odd $16$th powers are $1$ mod $32$ and $32\mid k!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | $$a=\sqrt{6-\sqrt{20}}$$
$$\implies a^2=6-\sqrt{20}$$
$$\implies(a^2-6)^2=20$$
$$\implies a^4-12a^2+16=0$$
$$\implies (a^2-2a-4)(a^2+2a-4)=0$$
so it will be one of the four possibilities $\pm\sqrt{5}\pm1$, and since $\sqrt{6-\sqrt{25}} \lt \sqrt{6-\sqrt{20}} \lt \sqrt{6-\sqrt{4}}$, you want the one which is in $[1,2]$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
A quicker approach to the integral $\int{dx\over{(x^3+1)^3}}$ Source: A question bank on challenging integral problems for high school students.
Problem:
Evaluate the indefinite integral
$$\int{dx\over{(x^3+1)^3}}$$
Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fra... | We have:
$$\int\frac{dx}{(x^3 + 1)^3} = \int \frac1{x^2}\frac{x^2}{(x^3 + 1)^3}dx$$
with
$$u = \frac1{x^2}, dv = \frac{x^2}{(x^3 + 1)^3}dx$$
this becomes:
$$\text{something } - \frac13\int\frac{1}{x^3(x^3 + 1)^2}dx$$
then:
$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \int\frac{x^3 + 1 - x^3}{x^3(x^3 + 1)^2}dx = \int\frac{dx}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Positive definiteness of difference of inverse matrices Let $A$ and $B$ be two $n \times n$ symmetric and positive definite matrices.
If $A \prec B$, then is it true that $B^{-1} \prec A^{-1}$?
Here, $A \prec B$ means that $B-A$ is positive definite.
| For ease of notation I'll use $\ge$ instead of $\succeq$ and so on.
Lemma 1: $A \le B \Rightarrow C^T A C \le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x \ge 0$ for any conformable vector $x$ so that $C^T(B-A)C \ge 0.$
Lemma 2: $I \preceq B \Rightarrow$ $B$ is invertible and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Infinite trigonometry Summation: $ \sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right) $ I would like to evaluate
$$
\sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)
$$
Summation image please view before solving
I saw a pattern and realized the answer will converge a... | This may be seen as a telescoping series, one may write for $N\ge1$,
$$
\small{\sum_{k=1}^N\! \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)\!=\sum_{k=1}^N \!\left(\! \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+1)}\! \right)\!+\!\sum_{k=1}^N\! \left(\! \cos \frac{\pi}{2(k+1)}-\cos \frac{\pi}{2(k+2)} \!\right)}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $z\in\mathbb{C}$ such that $|z|=2i(\bar{z}+1)$ I was trying to solve the following problem:
Find $z\in\mathbb{C}$ such that $$|z|=2i(\bar{z}+1).$$
I tried to solve the problem by letting $z=x+iy$ and found out that the solutions are $z=-1+i\frac{1}{\sqrt{3}},-1-i\frac{1}{\sqrt{3}}$. But as tried to justify the s... | The equation you should get is
$$\sqrt{x^2+y^2} = 2i(x-iy + 1)\\
\sqrt{x^2 + y^2} = y + 2(x+1)i$$
which means that
*
*$2y=\sqrt{x^2+y^2}$
*$2(x+1)=0$
From the second equation, you get $x=-1$, and the first equation becomes
$$2y=\sqrt{y^2+1}$$
Now, you SQUARE the equation and get $$4y^2=y^2+1$$
and get $$y^2=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\... | Use $\lfloor x \rfloor + \lfloor x + 1/2 \rfloor = \lfloor 2x \rfloor$ identity . So $$\lfloor \frac{4a-10}{15} \rfloor = a-4 $$ Try $a = 0 , 1,2,3, \dots$ The only solution is $a = 4$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that there are exactly two solutions to the equation $x^3 = x^2$. This is Velleman's exercise 3.6.12.c:
Prove that there are exactly two solutions to the equation $x^3 = x^2$.
Here's my proof of it:
Proof.
Existence. Let $x = 0$ then clearly $x^3 = x^2$ and let $x = 1$ then clearly $x^3 = x^2$.
Uniqueness. We ch... | Yes, your proof is correct.
But here's a more standard proof . . .
\begin{align*}
&x^3=x^2\\[4pt]
\iff\;&x^3-x^2=0\\[4pt]
\iff\;&x^2(x-1)=0\\[4pt]
\iff\;&x^2=0\;\;\text{or}\;\;x-1=0\\[4pt]
\iff\;&x=0\;\;\text{or}\;\;x=1\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| The problem is that
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 \ne 3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = (397)_9. $$
What we are trying to do is solve
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 = a \cdot 9^2 + b \cdot 9^1 + c \cdot 9^0 = (abc)_9 $$
To find $a$ we take as many copies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
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.