Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $CF = r(\frac{1-\cos x}{\sin x})$ in a geometry problem without using $\tan\frac{x}{2}$
This problem is from IMO $1985$ Problem $1$.
A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.
Let $\angle DAO$ be $\theta... | Solution without trigonometry.
Let $P\in AB$ such that $AP=AD$.
We can assume also that $P$ is placed between $A$ and $O$ as on your picture.
Thus, $$\measuredangle APD=\frac{1}{2}(180^{\circ}-\measuredangle A)=x,$$ which says that $DPOC$ is cyclic, which gives:
$$\measuredangle CPO=\measuredangle CDO=y$$ and since $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to show $\frac{\cos (3(x - \frac{\pi}{4}))}{\cos(x - \frac{\pi}{4})} = \frac{\sin 3x - \cos 3x}{\sin x + \cos x}$
$$\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\cos(x - \frac{\pi}{4})} = \frac{\sin 3x - \cos 3x}{\sin x + \cos x}$$
My attempt:
\begin{align}
LHS &=\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\co... | Using the angle addition formula $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$ we have
$$\frac{\cos(3(x-\frac{\pi}{4}))}{\cos(x-\frac{\pi}{4})}=\frac{\cos(3x)\cos(\frac{3\pi}{4})+\sin(3x)\sin(\frac{3\pi}{4})}{\cos(x)\cos(\frac{\pi}{4})+\sin(x)\sin(\frac{\pi}{4})}$$
$$=\frac{-\frac{\sqrt{2}}{2}\cos(3x)+\frac{\sqrt{2}}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given $f(x,y) = x\sqrt{3 - x^2 - y^2}$ find and sketch the domain? I have been trying to do this but I am having a hard time. All I have is that $\sqrt{3-x^2-y^2}\geq0$
The next step I want to do is the following:
\begin{aligned}(\sqrt{3-x^2-y^2})^2\geq0^2\end{aligned}
\begin{aligned}3 -x^2-y^2\geq0\end{aligned}
\begin... | The domain of the function is the set of points $(x,y)$ on which the function is well defined. Now, you know that you can only take the square root of a non-negative number
Hence,
$$ 3-x^2-y^2 \geq 0$$
$$\implies x^2+y^2 \leq 3$$
Only if $(x,y)$ satisfies this can you get a real value for $f(x,y)$
As for sketching, thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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showing all gcd $(n^3-n, 2n^2-1)$ I've been stuck on this problem for a long time.
I must show all gcd of $(n^3-n, 2n^2-1)$. I know that $n^3-n=n(n+1)(n-1)$ and so I must find all gcd of $(n(n+1)(n-1), 2n^2-1)$. But I don't know how. Can somebody help me ?
| I'd do it the way I did below without factoring. But factor was a good thought and it will work:
$n^3-n = n(n+1)(n-1)$ so lets see what factors $n,n+1, n-1$ have with $2n^2 -1$.
First of all any (nontrivial) factor of $n$ will be a factor of $2n^2$ and will not be a factor of $2n^2 -1$. So $n$ and $2n^2 -1$ have no f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | You have nice answers by expansion, so I present a solution without it.
$$
\sin^6x + \cos^6x
$$
$$
= \sin^6x + (1-\sin^2 x)^3
$$
$$
= \sin^6x + 1 - 3\sin^2 x+3\sin^4 x-\sin^6 x
$$
$$
= 1 - 3\sin^2 x + 3\sin^4 x
$$
$$
= 1 - 3\sin^2(1 - \sin^2 x)
$$
$$
= 1 - 3\sin^2 x\cos^2 x
$$
$$
= 1 - \frac{3}{4}\sin^22x
$$Hence Pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculate ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $ Let ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $, for n=1,2,3,.....Then
(A) ${S_n} ... | We have $$S_n \le I \le T_n$$ because $f(x)$ here is a decreasing function.
By squeez law, we have
$$I=\lim_{n \to \infty} T_n=\lim_{n \to \infty} S_n=\int_{0}^{1} f(x) dx $$
$$f(x)=\frac{1}{x^2+x+1}$$
So $$I=\int_{0}^{1} \frac{dx}{(x+1/2)^2+3/4}=\frac{2}{\sqrt{3}}\tan^{=1}\frac{(2x+1)}{\sqrt{3}}=\frac{2}{\sqrt{3}}[\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$
So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1... | Answer :
((x / (1+x² ))+((x³/ (1+x))=((x+x² +x³ +x⁵) / ((1+x)(1+x²) ))=((x+x² +x³ +x⁵) / (x+x² +x³ +1))=1
Because x^5=1
$((x² / (1+x⁴))+((x⁴/ (1+x³ )) = ((x² +x⁵+x⁴+x⁸) / (1+x³ +x⁴+x^7)) =((x² +1+x⁴+x³) / (1+x³+x⁴+x²)) =1$
Because $x^5 =1$
$\frac{x}{1+x^2 }+\frac{x^3 }{1+x} +\frac{x^2 }{1 +x^4}+\frac{x^4}{1+x^3 }=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across... | Hint:
$$x+y+z=xyz\iff x=\dfrac{y+z}{1-yz}$$
If $\dfrac{2x}{1-x^2}=a$ etc. assuming $x^2,y^2,z^2\ne1$, we need $$a+b+c=abc\iff\dfrac1{bc}+\dfrac1{ca}+\dfrac1{ab}=1$$
$$\implies\dfrac1a\left(\dfrac1b+\dfrac1c\right)$$
$$=\dfrac{(1-x^2)}{2x}\left(\dfrac{1-y^2}{2y}+\dfrac{1-z^2}{2z}\right)$$
$$=\dfrac{1-\left(\dfrac{y+z}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 2
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$A^{n}$ matrix problem without using $A=PDP^{-1}$ Eigendecomposition i have the following problem
Find $A^n$ for the following matrix
\begin{equation}
A=\begin{pmatrix}
0 & 0 & 1\\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix}
\end{equation}
I have tried the following, calculating for $n=1,2,3,4,5,6$
\begin{equation}
A^1=\begin... | The following solution is using the eigenvalues in an implicit way.
$$A^3=2A^2+A-I_3$$
Which can be found by inspection or by calculating the characteristic polynomial.
Then, by long division
$$X^n=Q(X)(X^3-2X^2-X+1)+aX^2+bX+c$$
You can find $a,b,c$ by setting $X=\lambda_{1,2,3}$ the solutions to $X^3-2X^2-X+1$ in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Stuck and confused with $\int \frac{u^{1/2}}{(u+1)^2} du$ I'm trying to find the expectation value $\langle x^2 \rangle_{\psi}$ for
$$\psi(x,0) = \sqrt{\frac{2}{\pi}}a \cdot \frac{1}{x^2+a^2}$$
and I know the result is $a^2$. I'm terribly stuck integrating this. Currently:
$$\langle x^2 \rangle_{\psi} = \int\limits_{-\... | The term $\langle x^2\rangle_\psi$ as defined by the OP can be obtained through standard method of integration of rational functions.
Here is one method:
$$
\int \frac{x^2}{(x^2+a^2)^2}\,dx =\int \frac{dx}{x^2+a^2} -a^2\int\frac{dx}{(x^2+a^2)^2}
$$
Setting $I_n=\int\frac{dx}{(x^2+a^2)^n}$, we get by integration by part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$ How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2
I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it accidentally.
Currently I have figured out ... | Let $2\cos(\frac{\pi}{11})$ represented with the help of half angle cosine formula as follows
$2\cos(\frac{\pi}{11}) = \sqrt{2 + 2\cos(\frac{2\pi}{11})}=\sqrt{2 + \sqrt{2+ 2\cos(\frac{4\pi}{11})}}$
$\sqrt{2 + \sqrt{2+\sqrt{2+ 2\cos(\frac{8\pi}{11})}}}$
Now $\frac{8\pi}{11}$ is more than $\frac{\pi}{2}$ which makes $2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determinant by cofactor expansion Vs Row reduction For the following matrix
$$ \begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & -4 \\
-4 & 1 & 1 \\
\end{pmatrix} $$
The determinant by cofactor expansion is 25 . But by EROs reduction I'm always getting -25 ..here is one trial :
--> R3+4R2 & R1-R2
$$ \begin{pmatrix}
1 ... | You can only replace the row $R_i$ with $R_i + k R_j$ (not row $R_j$).
If you replaced row $R_j$ instead, the determinant is multiplied by a factor of $k$.
This is related to the elementary matrix multiplications that underlie the row reduction methods.
Hence for example $1$, under row operations $R_3 + 4R_2 \to R_3$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving an inequality is true (precalculus) How do I prove that
$(|x|+2)(|x^2+9|)-9|x^2-2| \ge 0$?
I tried using properties of absolute values such as triangle inequalities but so far I've got no luck.
The actual question was to prove that
$|\frac{x^2-2}{x^2+9}| \le \frac{|x|+2}{9}$. I tried using triangle inequality p... | Using $f(x) = (|x|+2)(x^2+9)-9|x^2-2|$ as defined in the other answer and the observation that $f(x) = f(-x)$, we prove it when $x \ge 0$ using a slightly different way: completing squares.
For $0 \le x \le \sqrt 2$, we have:
$$f(x) = (x+2)(x^2+9)-9(2-x^2) = x^3+11x^2+9x\ge0 \text{ since }x \ge 0$$
For $x \ge \sqrt 2$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Bias and MSE of $\hat{\theta} = \min(X_1, \ldots, X_n)$ Let $X_1, ... X_n$ iid with pdf given by $$p_{\theta, r} = r \theta^r x^{- (r+1)} \mathbb{1}\{x \geq \theta\}$$ for $\theta > 0$, and some $r > 2$ that is known. Then $\hat{\theta} = \min(X_1, \ldots, X_n) = X_{(1)}$.
I want to determine the bias and MSE of $\hat{... | $$
p_{\theta, r}(x) \, dx = r \left( \frac x \theta\right)^{- (r+1)}\left( \frac{dx} \theta \right) \mathbb{1}\{x \geq \theta\}.
$$
$$
\text{So for } x\ge\theta, \qquad \Pr(X>x) = \left( \frac x \theta \right)^{-r}
$$
$\min\{X_1,\ldots,X_n\} > x$ if and only if all of $X_1,\ldots, X_n$ are${}> x.$
That has probability ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using integrating factors to solve difficult differential equations Consider the differential equatioN:
$$(2x^2 y -2y^4)dx+ ( 2x^3 + 3xy^3) dy = 0$$
This equation is of the form:
$$ Q dx + P dy=0$$
Now, it's easy to see that this differential is not exact by using the commutativity of second order partial condition. ... | I get ("+" in last term).
$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$
Let $\eta=x^ay^b$. Then
$$\left( -2 b-3 a-11\right) \, {{x}^{a}}\, {{y}^{b+3}}+\left( 2 b-2 a-4\right) \, {{x}^{a+2}}\, {{y}^{b}}=0$$
Solve system $-2 b-3 a-11=0,\quad 2 b-2 a-4=0$.
Then $a=-3,\quad b=-1$. Integrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | As written, no simplification is apparent, and this is due to the presence of the radicals. Now considering the identity
$$a-b=\frac{a^2-b^2}{a+b}$$ there is a hope of getting rid of them. In the case of your numerator,
$$\sqrt{6-x}-2=\frac{(\sqrt{6-x})^2-2^2}{\sqrt{6-x}+2}=\frac{2-x}{\sqrt{6-x}+2}.$$
Now the radical i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 1
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How to increase speed of convergence of series? I'm trying to speed up convergence of series:
$$
a_{n} = \frac{n(n-0.3)}{(n^{2} + 0.3)^{2}}
$$
Without series speed up:
$$
R_{method} = \left|A - S_{N_{0}}\right| = \sum_{n = N_{0} + 1}^{\infty}\frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} \leq \sum_{n = N_{0} + 1}^{\infty}\frac{n^2... | Your error seems to lie in this computation:
$$\int_{N_{1}}^{\infty}\frac{30n^3+60n^2+9}{10n^6}dn= \frac{25(3N_{1}+4)N_{1}^2+ 9}{25N_{1}^5}$$
Indeed a primitive function of the integrand is:
$$-\frac32 \frac{1}{n^2}-\frac12 \frac{1}{n^3}-\frac{9}{10} \frac{1}{n^5}=- \frac{75n^3+25n^2+9}{50 n^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the equation of the circle I've been experiencing a difficulties in answering this. I hope someone will help me in solving this
Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.
| HINT
Let start from the general equation
$$(x-x_C)^2+(y-y_C)^2=R^2$$
and plug in the values $(x,y)$ for the given points to obtain three equations in the three uknowns $x_C$, $y_C$ and $R$.
We obtain the system
*
*$2x_1x_C+2y_1y_C-x_1^2-y_1^2=x_C^2+y_C^2-R^2$
*$2x_2x_C+2y_2y_C-x_2^2-y_2^2=x_C^2+y_C^2-R^2$
*$2x_3x_C... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from F... | "Further mathematics" sounds like IB. So, they may suggest a solution using linear recursions.
The given polynomial
$$x^4-\color{green}{1}\cdot x^3 +\color{blue}{0}\cdot x^2 +\color{blue}{0}\cdot x - 1$$ belongs to the linear recursion
$$x_{n+4} = x_{n+3}+x_n$$
with (using Vieta)
*
*$x_0 = \alpha^0 + \beta^0 + \gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relation between prime roots and residues mod 7 in the prime root chapter from my book it was mentioned that the powers of the primitive root mod 7 equals the residues if you are calculating $\frac{1}{7}$.
It is easy to see that $3$ is a primitive root mod $7$ with
$$ 3^1 \equiv 3, 3^2 \equiv 2, 3^3 \equiv 6, 3^4\equiv... | It does work for the other primitive root $5$, provided you work in base $12$ instead of $10$. The sequence of powers is
$$5, 4, 6, 2, 3, 1$$.
Now calculate $1/7$ in base $12$:
$$12 = 1\cdot7 + 5,\quad 5\cdot12 = 60 = 8\cdot 7 + 4,\quad 4\cdot12 = 48 = 6\cdot 7 + 6, \ldots$$
The sequence $5, 4, 6, \ldots$ turns up agai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$
I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked with WA
Is there an ... | Since $|z|=1$ we substitute $z=e^{i\theta}$, to get:
$$
f(\theta)=|1+e^{i\theta}|+|1-e^{i\theta}+e^{2i\theta}|\\
= \sqrt{ (1 + \cos \theta)^2 + \sin^2\theta} + \sqrt{(1 - \cos\theta+ \cos2\theta)^2 + (-\sin\theta + \sin2\theta )^2)}\\
=\sqrt{|2+2\cos\theta |}+|(1-2\cos\theta)|
$$
We can denote $u=\cos\theta$ and invest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Prove $(x^2 + xy + y^2)^2$ divides $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$. As stated in the title, I want to show the (bivariate) polynomial $g(x, y) = (x^2 + xy + y^2)^2$ divides the polynomial $f(x, y) = (x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$, where $n \geq 0$.
Naturally, I resorted induction. For $n = 1$,... | Let $ y = k $ = constant and let $f(x)= (x+k)^{6n+1} - x^{6n+1}-k^{6n+1}$. We know that
$x^2 + xk + k^2= (x-kw)(x-kw^2)$ where $w$, $w^2$ are cube roots of unity. If we put $x=kw $ and $kw^2$ in $f(x)$, it gives the value $0$. Therefore $(x-kw)(x-kw^2)$ divides $f(x)$ . $\frac{df(x)}{dx}$ = $(6n+1)((x+k)^{6n} -x^{6n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3... | Let's look at 3 cases:
1) n is divisible by 3: (n+3) is divisible by 3 and one of (n+2) or (n+3) must be even and, therefore, divisible by 2. Being divisible by both 2 and 3, the expression is divisible by 6.
2) n is not divisible by 3 but (n-1) is: (n+2) is divisible by 3 and one of (n+2) or (n+3) must be even.
3) n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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The sum of the solutions of $\sin\left(2x\right)=\frac{\sqrt{3}}{2}$ over the interval$ [–π, d]$ is$–π.$ find the value of $d$ The sum of the solutions of $\sin\left(2x\right)=\frac{\sqrt{3}}{2}$ over the interval$ [–π, d]$ is$–π.$ find the value of $d$
I think the first thing I must know is the number of solutions. I... | The solutions in
$$[-\pi,\infty)$$ are
$$-\frac{5\pi}6,-\frac{2\pi}3,\frac\pi6,\frac\pi3,\frac{7\pi}6,\frac{4\pi}3,\cdots$$
and if you accumulate them you obtain
$$-\frac{5\pi}6,-\frac{3\pi}2,-\frac{4\pi}3,-\pi,\frac\pi6,\frac{3\pi}2\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I do not know how to solve for Induction Conclusion A sequence has $x_1=8$, $x_2=32$, $x_{n}= 2x_{n-1}+3x_{n-2}$ for $n \ge 3.$ Prove, for all $i$ of Naturals, $X_i = 2 (-1)^i + 10 \cdot 3^{i-1}$
I got bases covered, and I got the inductive step as $X_{i} = 2 * (-1)^k + 10 * 3^{k-1}$. I do not know how to follow up wit... | Let $A_n = x_n-x_{n-2}$ and $B_n = x_n+x_{n-1}$
We're given that $$x_n=2x_{n-1}+3x_{n-2}$$
$$\implies x_n-x_{n-2}=2(x_{n-1}+x_{n-2})$$
$$\implies A_n=2B_{n-1}$$
Also, using the definitions of $A_n$ and $B_n$:
$$B_n - A_n = B_{n-1}$$
Thus,
$$B_n = 3B_{n-1}$$
So, for general $B_n$:
$$B_n = 3^{n-2}\cdot B_{2}$$
$$\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And... | If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi ik}{n}}$
So the roots are
$$x_1=(2)^\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rflo... | $$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$
Let $f(k)=\lfloor\log_2k\rfloor$. Since $\log$ is increasing we know that
*
*$f(2^0)=f(1)=0$
*$f(2^1)=f(2)=f(3)=1$
*$f(2^2)=f(4)=f(5)=f(6)=f(7)=2$
*$f(2^3)=f(8)=f(9)=\cdots=f(15)=3$
*$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Why substitution in irrational equation doesn't give equivalent equation? I have two examples of irrational equations:
The first example: $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$
In solution, they take cube of both sides and do following:
\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(... | These problems don't happen if you'll use the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Since, $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we see that $\sum\limits_{cyc}(a-b)^2=0$ for $a=b=c$ only and it can give a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * ... | Here is a variation using the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.
We start with $f(x)$ and obtain for $n\geq 3$:
\begin{align*}
\color{blue}{[x^n]f(x)}&=[x^n]x^3\sum_{k=0}^\infty\binom{k+4}{4}(-1)^kx^k\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{1}\\
&=[x^{n-3}]\sum_{k=0}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it valid to apply L'hopital rule to evaluate the limit? $$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\... | Short answer:
$$\tan x\sim x$$
and
$$\sqrt{x^2+x+1}-1\sim\frac x2.$$
Hence the factor $1-\cos x$ makes the limit tend to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Infinite product of $\sqrt{2}$ I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to e... | It follows from the Euler/Weierstrass factorisation for sine: for $x \in \mathbb{R}$
$$\frac{\sin{x}}{x} = \prod_{j \ge 1} \left(1-\frac{x^2}{\pi^2 j^2}\right)$$
Separating the odd even parts we can rewrite your product as:
$$\begin{aligned} \mathcal{P} & = \prod_{k \ge 1 } \left(1 - \frac{(-1)^k}{2k-1}\right) \\& = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $ Solve the ODE with the initial conditions:
$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$
I did the substitution:
$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$
Putting in the ODE:
$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = ... | $$y’ = -\frac{1}{y} + 2$$
$$y’ = -\frac{1}{y} + \dfrac {2y}y$$
$$y’ = \frac{2y-1}{y}$$
$$\dfrac {y}{2y-1}dy = dx$$
Then integrate both sides.
$$\int \dfrac {y \, dy}{2y-1} = \int dx$$
Substitute $u=2y-1 \implies du=2dy$
and $y=\dfrac {u+1}2$
$$\dfrac 14 \int \dfrac {u+1}{u}du = \int dx$$
$$\dfrac 14 \int \left (1+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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I need to compute the integral $\int _0^s\:x^2$ using Riemann sums. I need to compute the integral $\int _0^s\:x^2$ using Riemann sums.
a) Consider the function $y=x^2$ on the interval [0, s] and divide it into n equal intervals. Each of these subintervals has length $\frac{s}{n}$.
b) Find the sum $U_n$ of all rectan... | First a comment. If you assume that $x \mapsto x^2$ is Riemann integrable, you don't need to compute both upper and lower Riemann sums and verify that those are equal. This would be required to prove the Riemann integrability.
Now regarding the computation itself. You have
$$U_n=\frac{s}{n}\left(0^2+\left(\frac{s}{n}\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" ... | A way is to find a polynomial satisfied by $z$. (For low degrees, this is likely to be a minimal polynomial. To show this requires computing elements of $X$ successively and searching for a linear relation among the powers of $z$ produced so far. The first such relation is the minimal polynomial. All we can actuall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3904061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to prove transitive relation for this one? $R=\{(x,y) \mid 3 |(x-y^2) \}$ while R is relation on set A=$\mathbb{Z}$
I need to check if it's reflective , symmetric , transitive .
I figured already that it's not reflective or symmetric with (2,1) or (10,2),(2,10)
and I don't how to check if it's transitive or not.
I ... | If $3|x-y^2$ and $3|y-z^2$ does it follow that $3|x-z^2$?
$3|x-y^2\implies z\equiv y^2 \pmod 3$ and $3|y-z^2\implies y\equiv z^2 \pmod 3$.
So if $x\equiv y^2$ and $y\equiv z^2$ then $x \equiv z^4\equiv \pmod 3$.
Must it follow that $z^4 \equiv z^2 \pmod 3$?
If $z\equiv 0$ then $z^2 \equiv z^4 \pmod 3$ and, yes, $x\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Guessing pattern of Picard's iteration of ODE Given the following initial value problem $x'(t) = x(t) + e^t$ with $x(0) = 0$. I've calculated its picard's iteration, but having trouble guessing the pattern: $$\begin{cases}
x_1 = e^t - 1, \\[3pt]
x_2 = 2e^t - t - 2, \\[3pt]
x_3 = 3e^t - 2t - 3 - \fr... | If you expand with Maclaurin $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+O\left(t^6\right)$$
and plug this in the, let's say, $6$th step, you get
$$6 \left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}\right)-\frac{t^5}{120}-\frac{t^4}{12}-\frac{t^3}{2}-2 t^2-5 t-6+O\left(t^6\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probably $x_1 \leq x_3 \leq x_2$ From set ${1,2...n}$ we choose three random different numbers, $x_1, x_2,x_3$. If $x_1 \le x_2$, find probability $x_1 \le x_3 \le x_2$.
I set A is event $x_1 \leq x_3 \leq x_2$, and B $x_1 \leq x_2$, and find P(AB)=$\frac{1}{6}$, P(B)=$\frac{1}{2}$. Is it okay?
| The problem has two different solutions, depending on whether you take them with replacement or without it (your question is quite ambiguous in that sense: the word 'different' suggests that they are taken without replacement, and the sign '$\leq$' - that they are taken with replacement):
Solution for numbers taken wit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried ... | We have following divisibility rules :
*
*$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod 7$
*$\overline{abcdef} \equiv a-b+c-d+e-f \pmod {11}$
*$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod {13}$
So one obtains
$$ \overline{xyz} \equiv 5 \pmod 7$$
$$ x+y+z \equiv 6 \pmod {11}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequal... | let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$
$$xy+yz+zx=18 \tag{given}$$
$$\color{red}{x+y+z\ge 3\sqrt{6}} \tag1$$
notice that $${(x-\sqrt{6})}^2\ge 0\Rightarrow \sqrt{3+x^2}\ge \frac{\sqrt{6}}{3}x+1$$
Thus $$\sum \sqrt{3+x^2}\ge \sum \frac{\sqrt{6}}{3}x+1\ge\frac{\sqrt{6}}{3}\color{red}{3\sqrt{6}}+3=9$$
Note the in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Consider a stick, cut twice, probability the smallest is 1/5 Consider a stick of unit length, you take two points (uniform independent), and then break the stick on these two points. You now got 3 segment. What is the probability the smallest segment is less than 1/5?
$X, Y$ be the two cuts.
$P(X < 1/5 \text{ and } Y... | Let $P(S)$ be the probability that the smallest segment is less than $\frac{1}{5}$; let $X$ be the location of the first cut, and $Y$ be the location of the second cut.
If $X < \frac{1}{5}$ or $X > \frac{4}{5}$, then you already know that one of the segments has length less than $\frac{1}{5}$, so the conditional probab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone ou... | Work mod 3, and put aside induction for now. Note that $4 \equiv_3 1$, and so for each integer $n$:
$$4^n \equiv_3 1^n = 1.$$
Thus $12^n -4^n -3^n +1 \equiv_3 (0 -1 -0 -1) = 0$.
So this quantity is divible by 3 for each positive integer $n$.
Lets now get back to mod 6 though. So $12^n -4^n -3^n +1$ is divisible by 3. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| $$(x^2 + 5x + 5-1)(x^2 + 5x + 5+1) - 48 = (x^2 + 5x + 5)^2-7^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \i... | If you are familiar with generalized harmonic numbers.
The expression you are considering is
$$a_n=\frac{1}{\sqrt{n}}\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}$$ and, for large $p$, we have
$${H_p^{\left(-\frac{1}{2}\right)}}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta
\left(-\frac{1}{2}\right)+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits... | Continuing from your method $$f'(x)=\lim_{h\to 0} \frac{f(h)+1 -(e^{-x}-1)(e^{-h}-1))}{h}$$ for derivative to exist $f(0)=-1$ hence by L-Hospitals rule $$f'(x)=\lim_{h\to 0}\frac{f(h)+1}{h}-\lim_{h\to 0}(e^{-x}-1)\frac{(e^{-h}-1)}{h}$$
$$f'(x)=f'(0)+e^{-x}-1=e^{-x}+C\Rightarrow f(x)=-e^{-x}+Cx$$
Now for the given limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920512",
"timestamp": "2023-03-29T00:00:00",
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We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90... | The given answer is wrong and you are correct. $ABC$ happens to be two consecutive sides of a regular octagon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate... | Another method, because it wasn't mentioned, is using Lagrange's trigonometric identities (which should not be difficult to prove using complex numbers, like here)
$$\sum _{k=1}^{n}\sin(k\theta )=
{\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(n+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
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Generalisation of IMO 1990/P3:For which $b $ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?
For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?
It was from my LTE/ Zsigmondy handout.
By taking examples, it loo... | We can construct b, we want:
$b^n+1\equiv 0 \mod (n^2)$
If $b=t.n^2-1$ and $n =2k+1$ is odd, then we have:
$(t.n^2-1)^n+1=M(n^2)-1+1=M(n^2)\equiv 0 \ mod (n^2)$
Where $M(n)$ is a multiple of $n^2$.
The condition is that n must be odd.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Perfect Square Number, prime I want to show that $p^{p+1}+(p+1)^p$ is not perfect square number. ($p$ is prime number)
This is what I do
Assume $p^{p+1}+(p+1)^p=k^2$.
$(p+1)^p=(k+p^{\frac {p+1} 2})(k-p^{\frac {p+1} 2})$.
Let $2ab=p+1, gcd(a,b)=1$.
$k+p^{\frac {p+1} 2}=2^{p-1}a^{p}, k-p^{\frac {p+1} 2}=2b^p$.
Therefore ... | Your proof is quite faulty (mathematically speaking, you omitted 2-3 cases i think), so I will reweite it entirely.
$$p^{p+1}+(p+1)^p=k^2\Leftrightarrow (p+1)^p=\big(k-p^{\frac{p+1}{2}}\big)\big(k+p^{\frac{p+1}{2}}\big)$$
Notice that $\text{gcd}(p+1;p)=1$ and $\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Drawing numbers from a set, quantile We have a set containing $20$ numbers from $1$ to $20$. Each time we draw only one number, and repeat it $15$ times (without replacement). Let's denote $X-$ the largest drawn number. Find the smallest ${{16} \choose {15}}/{20 \choose 15}$-fractile of a random variable $X$.
So, we l... | There are $\binom {20} {15}$ ways to choose 15 numbers from the set.
$\binom{16}{15}$ is the number of ways to choose 15 numbers only from $1$ to $16$ inclusive. The maximum of such subset is at most $16$. So the cumulative probability up to and including $t=16$ is the $\binom {16} {15} / \binom {20} {15}$.
Alternativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
I'm having trouble solving this equation. I've tried squaring both sides and got this
$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$
after simplifying
$$19 - 6\sqrt{... | Hint:
$11=(\sqrt7)^2+2^2$ and $\sqrt7-2>0$
$8=(\sqrt7)^2+1^2$ and $\sqrt7-1>0$
Finally
$$\sqrt{a^2+b^2-2ab}=|a-b|=a-b\text{ if } a-b\ge0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for continued fraction I am working with a recursive algorithm and I have realized that in each step it computes something equivalent to:
$A_{0}=\frac{a}{b}$
$A_{1}=a/(b-\frac{a^{2}}{b})$
$A_{2}=a/(b-\frac{a^{2}}{b-\frac{a^{2}}{b}})$
$\vdots$
$A_{n+1}=a/(b-aA_{n})$
And so on. I was wondering if there could ... | The closed form does exist, I found it with Mathematica and it's amazing, to use in a program I mean.
I found it with this function
RSolve[{A[n + 1] == a/(b - a A[n]), A[0] == a/b}, A[n], n]
Set:
$$
\begin{cases}
D=\sqrt{b^2-4 a^2}\\
x=\frac{b-D}{a}\\
y=\frac{b+D}{a}\\
\end{cases}
$$
Then
$$A_n=\frac{a (b+D ) y^n-a (b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$... | From linear independence of the rows we see that $A$ is invertible and so $Q = A^{-1}B$. You can use adjugate matrix and determinant to compute $A^{-1}$
$$
A^{-1} = \frac{{\rm adj} A}{\det A} = \frac{\begin{pmatrix}
1/3 & 1/3 & 0 \\
-1/3 & 1/3 & 1/3 \\
0 & -1 & 0
\end{pmatrix}}{\frac{1}{3}} =
\begin{pmatrix}
1 & 1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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USAJMO 2017 P4: triples $(a,b,c)$ such that $(a-2)(b-2)(c-2)+12$ is a prime number that divides the positive number $a^2+b^2+c^2+abc-2017$?
Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2)+12$ is a prime number that properly divides the positive number $a^2+b^2+c^2+abc-2017$?
My progres... | If we continue from here: $\;\;\;xyz+12 \mid x+y+z-41\;\;\;$ or $\;\;\;xyz+12|x+y+z+49$.
We can assume $x\leq y\leq z$. We see that $x,y,z\notin \{0,2,3,4,6,8,9\}$ since $xyz+12$ is prime.
*
*If $x>1$ then $x,y\geq 5$ and so $xyz+12\geq 25z+12$ so in
*
*first case: $$25z+1\leq |x+y+z-41|\leq 3z+41\implies z\le 1$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I need to solve $12x-30y+24z=18$ equation as a diophantine equation where $x,y,z$ is whole number I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended e... | Experiments in a spreadsheet show that $x$ $y$ $z$ combinations only add up to a given value if we increment $y$ by $4$ and $z$ by $5$ for a given value of $x$. Further experimentation shows offsets of $1$ and $2$, respectively to make the result zero.
There are infinite solutions starting with $x\in\mathbb{Z}$ and
$$y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$ I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4} }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sq... | Have you tried multiplying top and bottom of $\frac{\sqrt[3]{2+h}-\sqrt[3]{2}}{h}$ by $(2+h)^{\frac{2}{3}}+\sqrt[3]{2}\cdot\sqrt[3]{2+h}+2^{\frac{2}{3}}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from the... | Induction:
\begin{align*}2^{2^{6(k+1)}\cdot 4}=2^{2^{6k+6}\cdot 4}=\left(2^{2^{6k}\cdot 4}\right)^{64}&\equiv (2^4) ^{64}\mod 19\\&\equiv 2^{256}\mod19\end{align*}
By Fermat's little theorem, $2^{18}\equiv 1\mod 19$. Since $256=18*14+4$, then
\begin{align*}2^{2^{6(k+1)}\cdot 4}&\equiv (2^{18})^{14}2^4\mod19\\&\equiv 1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a pe... | The denominator of $1+e\cos\theta$ occurs in inverse $r^2$ force problems and when $e\gt1$ as here, corresponds to a hyperbolic orbit, like Rutherford scattering or ʻOumuamua. We can zap this denominator depending on whether the orbit is elliptical, parabolic, or hyperbolic. In the latter case we may let
$$\sin y=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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How do I find a plane that is tangent to two given spheres and passes through a given point? My problem is the following:
Given two spheres: $$(x-6)^2+(y-1)^2+(z+1)^2=1$$ and $$x^2+(y-5)^2+(z-4)^2=9$$ find a plane that is tangent to both of those spheres and passes through the point $$(5;2;0)$$
I tried plugging all the... | $C_1(6,1,-1);\;r_1=1$
$C_2(0,5,4);\;r_2=3$
A generic plane $\pi$ has equation
$$\pi:ax+by+cz+d=0$$
The distance from the plane to the center of the spheres must be equal to the respective radius. Furthermore the plane must pass through $P(5,2,0)$
$$\begin{cases}
\frac{|6a+b-c+d|}{\sqrt{a^2+b^2+c^2}}=1\\
\frac{|5b+4c+d|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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2D random walk - first hits A 2D symmetric random walk $(X_k,Y_k)$ $k\ge 0$ (Markov chain) where $(X_{k+1},Y_{k+1})$ takes one value of the next ones: $(X_k,Y_k+1),(X_k,Y_k-1),(X_k+1,Y_k),(X_k-1,Y_k)$, all have the same probability and the initial values of X and Y are zero.
a) Find the value of $P(X_T=3,Y_T=0)$ and $E... | I am not sure my answer for part a) is right, just for your reference.
a) First, it is important to observe that the point $(X_{T-2}, Y_{T-2})$ must be one of the four:
\begin{align}
(1,0),\,(-1,0),\,(0,1),\,(0,-1).\tag{1}
\end{align}
Now, if $(X_T, Y_T)=(3,0)$, then $(X_{T-2}, Y_{T-2})=(1,0)$. Consider (1), and by sym... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$
Attempt. Using ratio test for $a... | Let us use Gauss's test.
$$\frac{a_n}{a_{n+1}} = \frac{n+\frac52}{n+\frac53} = 1 + \frac{\frac56}{n(1+\frac{5}{3n})} = 1 + \frac{\alpha}{n} + \underline{O}(n^{-2})$$
with $\alpha = \frac{5}{6} < 1$. Hence $\sum_n a_n$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$? I did
\begin{align}
& \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt]
= {} & \fra... | $(\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$
$\frac {1}{(t^2 + 2)^2} ((-2(t^2+2)+4t^2) \mathbf i + (4t^2+t^2(t^2+2))\mathbf j + (-2t^3-4t) \mathbf k)$
$\frac {1}{(t^2 + 2)^2} ((2t^2 - 4) \mathbf i + (t^4+6t^2) \mathbf j -(2t^3+4t) \mathbf k)$
$\frac {2t^2 - 4}{(t^2 +2)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\frac{a + b}{\gcd(a,b)^2}$ is a Fibonacci number when $ \frac{a+1}{b}+\frac{b+1}{a}$ is an integer?
Let $a, b$ be positive integers such that the number $ \dfrac{a + 1}{b} + \dfrac{b + 1}{a}$ is also integer. Then, show that $\dfrac{a + b}{\gcd{(a, b)^{2}}}$ is a Fibonacci number.
I prove that : $ \... | First of all we assume that $gcd(a,b)=d $ so we may assume that there exist coprime integers $x,y$ such that $a=dx , b=dy$ then from the hypothesis of the problem: $ab|a^2+b^2+a+b
\rightarrow ab|(a+b)(a+b+1)
\rightarrow dxy|(x+y)(dx+dy+1)$
because $d$ is coprime to $dx+dy+1$ , $d|x+y$ (first result) and because $x,y$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proof by induction for a sequence, given $a_1$ and $a_2$ Question: prove by induction that this statement holds.
Define the sequence $(a_{k}), k∈\mathbb N ,$
by $a_1 = 2, a_2 = 5 $
and
$a_{k+2} = 5a_{k+1} − 6a_{k} $ for all $k≥1$. Then $a_{k} =2^{n−1} +3^{n-1}$ for all $n≥1 $.
What I did was find the base case -- $ ... | Say $P(n)$ is $a_n=2^{n-1}+3^{n-1}$.
Base: $P(1): a_1=2=2^0+3^0$, $P(2): a_2=5=2^1+3^1$.
Now assuming $P(k)$ and $P(k+1)$, we have
$P(k+2): a_{k+2}=5a_{k+1}-6a_k=5(2^k+3^k)-6(2^{k-1}+3^{k-1})$
$=(5-3)2^{k}+(5-2)3^{k}=2^{k+1}+3^{k+1}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$
What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?
All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am stuck and would like to know how w... | We'll find sums at each place (units, tens, etc.) and add them.
At thousands place each of $1,2,3$ occurs $6$ times. Sum is $6(1+2+3)\cdot 1000=36000$.
For hundreds place, we first look at numbers with thousands digit $1$. These are six numbers in which each of $2,3,0$ occurs twice. Hence in the $18$ nos, each of $1,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very ... | $x^2 - 1 = (x+1)(x-1)$.
So you have $\sqrt{(x+1)(x-1)} =(x+5) \sqrt\frac {x+1}{x-1}$.
Now my first thoughts are dividing both sides by $\sqrt{x+1}$ and multiplying both sides by $\sqrt{x-1}$ to get $x-1 = x+5$.
BUT BE CAREFUL!
First off. When we divide both sides by $\sqrt{x+1}$ we are assuming that $\sqrt{x+1}\ne 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank y... | From the equation solve for $\sqrt{x^2-1}$ and get
$$\sqrt{x^2-1} = \frac{x}{\frac{35}{12}-x}$$
now raise to the square and get an equation of degree $4$. Only $2$ of its roots will be roots of the initial equation, since they also include the roots of $x-\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$.
Worth plotting the grap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
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Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \... | For $n\ge2$,$$[x^{2n}]\sqrt{1-x^2}=(-1)^n\binom{1/2}{n}=(-1)^n\frac{1}{n!}\frac12\prod_{k=1}^{n-1}(\tfrac12-k)=-\frac{(2n-3)!!}{n!2^n},$$so$$\sqrt{1-x^2}=1-\tfrac12x^2-\sum_{n\ge2}\frac{(2n-3)!!}{n!2^n}x^{2n}.$$See how that looks in Desmos.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substitute... | Wouldn't be integration by parts much easier? Start with
$$\frac12\int x^2\cdot2x\sqrt{x^2-9}\,dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$ Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Originally I thought to rephrase it in geometric terms, see... | Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$
Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like:
$$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denomin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ This problem $5b$ from Chapter $14$ of Spivak's Calculus:
Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ (Note $g$ is not assumed to be continuous at $0$).
I think this question is impossible? First of all there's no restrictions on $x$, so ... | The function $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3}$$ seems to satisfy the condition. Below is how to find it:
Let $G(x) = \int xg(x)dx$. Then, we have
$$\int_0^{x^2}tg(t)dt \overset{\text{(FTC)}}= G(x^2) - G(0) \overset{*}= x + x^2
$$
Now, differentiate both sides of $(*)$ to get
$$\begin{align} 2x[x^2g(x^2)] &=... | {
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How many rectangles can be found in this shape?
Question: How many rectangles can be found in the following shape?
My first solution:
Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1... | Searching ${n+3 \choose 2}$ at OEIS you can find in A000332 the reference to A004320 and there the document Counting the lattice rectangles inside Aztec diamonds and square biscuits by Teofil Bogdan and Mircea Dan Rus and the solution to problem 3, on page 3. Basically, for each rectangle you have to choose four coordi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many N-digit numbers have the sum of odd places digits equal the sum of even places digits? We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers ... | This answer is intended to be simple and self-contained but please ask if it has not been expressed sufficiently clearly.
The expansion of $(1+x+x^2+...+x^9)^n$ as $\sum a_ix^i$
Consider multiplying out $(1+x+x^2+...+x^9)(1+x+x^2+...+x^9)...(1+x+x^2+...+x^9).$
For a given integer $M$, a term $a_Mx^M$ in the product is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$ If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$
I attempted to solve this question using binomial coefficients, saying:
$(\sqrt{m}+\sq... | This is a USAMTS problem. The official solution can be found here. You are actually on the right track.
Below is my solution. Notice that I used the same notation as in the original problem.
My proof:
We prove by induction that: If $m$ is odd then
$$x^m = a_m\sqrt n + b_m \sqrt{n+1}, na_m^2+1 = (n+1) b_m^2 $$
If $m$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3971804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin... | By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$
$\cos 2x =2\cos^2x-1=1-2\sin^2x$
$\cos 2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$
Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does the Minus-1 trick to get particular and general solutions work? Minus-1 trick: A practical trick where the reduced row-echelon form of a system of equations is augmented with -1 rows. It is used to get particular and general solutions.
While it is a handy trick, I was wondering why does it work.
The following ... | In the example, set the non-pivot variables to $x_2=-\lambda_1$, $x_5=-\lambda_2$ and solve the system for other variables. You will get the exact same result.
The minus-1 trick is just a schematic way to achieve the same with less writing.
Another way to look at this: The columns $1$, $3$ and $4$ of $A$ are the standa... | {
"language": "en",
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"source": "stackexchange",
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}... | There is no need for induction, use a straight proof.
The equation is
$$f(n+1)-g(n+1)=3(f(n)-g(n))+8n-12$$ with $$f(0)-g(0)=4$$ and it does verify the solution
$$f(n)-g(n)=4-4n$$ as is shown by substitution,
$$4-4(n+1)=3(4-4n)+8n-12,$$ equivalent to $$-4n=-4n.$$
Furthermore, $$4=4-4\cdot 0.$$
| {
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"source": "stackexchange",
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Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$ Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$
where $a$ and $b$ are real numbers and not equal. Find the ... | Note that the log function in the integrand can be decomposed as a sum of odd and even ones
\begin{align}
\ln\frac{1+ax}{1+bx}& =\frac12\ln\frac{1+ax}{1-ax}-\frac12\ln\frac{1+bx }{1-bx} + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2} \\
& =\tanh^{-1}(ax)-\tanh^{-1}(bx) + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2}
\end{align}
where th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$? The problem is as follows:
Find the value of:
$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$
Where $A$ is given by:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\ri... | Alternative approach
Using the diagram and half-angle formulas:
$$\cot(A/2) = \frac{1 + \cos(A)}{\sin(A)}
~~\text{and}~~ \cot(B/2) = \frac{1 + \cos(B)}{\sin(B)} = \frac{1 + \sin(A)}{\cos(A)} \implies $$
$$\frac{\sin(A)}{\cos(A)} = \frac{2 + \frac{1 + \sin(A)}{\cos(A)}}{4 + \frac{1 + \cos(A)}{\sin(A)}} =
\frac{2\cos(A) ... | {
"language": "en",
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Find real solution $x$ in order to $x + \sqrt{2020}$ and $\dfrac{5}{x} -\sqrt{2020}$ are integers $x + \sqrt{2020}$ and $\dfrac{5}{x} - \sqrt{2020}$ are integers
$\Rightarrow x + \dfrac{5}{x}$ is an integer
$\Rightarrow \dfrac{x^2 + 5}{x}$ is an integer
$\Rightarrow x^2 + 5\ \vdots\ x$
$\Rightarrow x^2 + 5 - x^2\ \vdot... | As Edward H's question comment hint suggests, let $a$ be the integer where
$$a = x + \sqrt{2020} \implies x = a - \sqrt{2020} \tag{1}\label{eq1A}$$
Then let $b$ be the other integer, and use \eqref{eq1A}, to get
$$\begin{equation}\begin{aligned}
b & = \frac{5}{a - \sqrt{2020}} - \sqrt{2020} \\
b + \sqrt{2020} & = \frac... | {
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Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with t... | Just another solution (similar to Albus Dumbledore's).
$$x^4+4x^3+4x^2-4x+3=x^2(x+2)^2-4x+3 = x^4+4x^3+(2x-1)^2 +2$$
For $x<0$ we have $-4x+3>0$ and therefore $x^2(x+2)^2-4x+3>0$.
For $x>0$ it follows from $x^3>0$ that $x^4+4x^3+(2x-1)^2 +2>0$.
| {
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Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$ Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$
Attempt:
We can write $x^5+x+1 = (x^2+x+1)(x^3-x^2+1)$ which could be a hint to be reducible, if we look clously we see that $(x^2+x+1),(x^3-x^2... | Let $x=w,w^2$, where $w$ cube root of unity. The $f(x)=x^5+x+1 \implies f(w)=w^2+w+1=0, f(w^2)=w+w^2+1=0$. So $g(x)=(x-w)(x-w^2)=x^2+x+1$ is a factor of $f(x)$ Further $\frac{f(x)}{g(x)}=x^3-x^2+1.$
So $f(x)$ is reducible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the general form of the $n$th derivative of $\ln^2(x+1)$ $f(x)=\ln^2(x+1)$.
I tried to write $f(x)=\ln (x+1)\cdot\ln (x+1)$ and then apply Leibniz but i think maybe i did something wrong.
Could you explain me the right method?
| Ok we talked a bit about it in the comments: we can start with the chain rule to find $f'$. Then we calculate a few more derivatives, and eventually we find a nice form that looks like a good guess. That guess can be proven by induction. I will leave the guessing and the proof by induction for you to do. Alright, here ... | {
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theory of equation let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
( c+d)^{-1}+ (c+d)^{-1}$$
my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck
| Notice that
$$\frac{1}{a+b} + \frac{1}{c+d} = \frac{a+b+c+d}{(a+b)(c+d)} = \frac{3}{(a+b)(c+d)}.$$
Repeat for two more sets of two fractions. Then add the three results.
| {
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Show that $(\mathbf{a}·\mathbf{a})(\mathbf{b}·\mathbf{b})-(\mathbf{a}·\mathbf{b})^2=(\mathbf{a}\times\mathbf{b})·(\mathbf{a}\times\mathbf{b})$? I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,
$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf... | The left hand side is the determinant of the Gramian matrix of $\mathbf a$ and $\mathbf b$:
$$
\det\begin{pmatrix}\mathbf a\cdot\mathbf a & \mathbf a\cdot\mathbf b \\
\mathbf b\cdot\mathbf a & \mathbf b\cdot\mathbf b\end{pmatrix}.
$$
In general, the determinant of a Gramian matrix is the square of the $n$-dimensional v... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction: for every integer $n > 0$, $a_n = 2 \cdot 3^n + n^2-6$ I am working on what should be a simple induction proof exercise. However, I'm not sure if I am failing to simplify an expression after substitution correctly, or whether a prior step is incorrect.
The exercise:
Let $a_1 = 1$ and for every integ... | It is the needed result if you notice $$(2\cdot 3^{k+1}) + k^2 + 2k - 5 = (2\cdot 3^{k+1}) + k^2 + 2k +1 - 6 = (2\cdot 3^{k+1}) + (k+1)^2 - 6$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty wei... | Hint:
Let $\sqrt{x^2+12y}-\sqrt{y^2+12x}=1+2a$
$$\implies\sqrt{x^2+12y}=\dfrac{23+1+2a}2,\sqrt{y^2+12x}=11-a$$
$\implies11\ge a\ge-12\ \ \ \ (1)$
$$(12+a)^2-(11-a)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
Replace the value of $x+y$ to find $x-y$
Consequently we find $x,y$ in terms of $a$
But we have $$x^2+12y=(12+a)^2$$
Repl... | {
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Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$ Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$.
I drew the picture and found two surfaces: $S_1$ and $S_2$,
where $S_1$ is $\{ 0 \leq y \leq 1 \land -\sqrt{(... | In cylindrical coordinates, the cylinder $x^2 + y^2 = y \ $ is
$r^2 = r \sin \theta \implies r = \sin \theta, \ 0 \leq \theta \leq \pi$.
This is a cylinder of radius $\frac{1}{2}$ with center at $(0,\frac{1}{2})$.
Now parametrization of cylinder is
$r(\theta, z) = (r\cos\theta, r\sin\theta, z) = (\frac{1}{2} \sin 2\the... | {
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Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degre... | $K$ has $2^4$ elements:
$$ax^3+bx^2+cx+d, \ (a,b,c,d) \in {\mathbb{Z}_2}^4$$
We have $$g(0)^2=g(1)^2=1$$
So the only possible candidates for $g$ are
$$1\\
x^3+x^2+1\\
x^3+x+1\\
x^2+x+1
$$
First we calculate
$$x^4\equiv x^2+x$$
$$x^6\equiv (x^4+x^2+x)x^2 +x^4+x^3\equiv x^3+x^2+x$$
We can skip the constant polynomial $1$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
S... | Note
\begin{align}
\frac{(-a^3 + a^2)^6 - 1 }{(-a^3 + a^2)^3 - 1 }&
=(-a^3 + a^2)^3 +1 = (-a^3 + a^2 +1 )[ (a^3 -a^2)^2 + (a^3 - a^2)+ 1]\tag1\\
\end{align}
where
\begin{align}
& (a^3 -a^2)^2 +(a^3 - a^2)+ 1 \\
= & (a^3 -a^2+a-a)^2 +(a^3 - a^2+a -a)+ 1 \\
=& (a^3 -a^2+a)^2 -2a (a^3 -a^2+a) +(a^3 - a^2+a ) +a^2-a+ 1 ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $p$ is an odd prime (different than $5$) prove that $10 \mid p^{2}-1$ or $10 \mid p^{2}+1$ I was reading some properties on prime numbers in a book and I’ve found this problem.
Using Modular Arithmetic isn’t allowed
Using the division algorithm I have found:
$p \in \{5m+1, 5m+2, 5m+3, 5m+4\}$
I’ve tried to plug thos... | Why on earth would modular arithmetic not be allowed. As I am of the theory all arithmetic is modular arithmetic that may be hard.
But as $p$ is an odd prime $p^2 -1$ and $p^2 +1$ are both even and $2|p^2 \pm 1$ so it is sufficient to show that $5|p^2 -1$ or $5|p^2 + 1$. .... Oh, wait, that was modular arithmetic! W... | {
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Proving combinatorial formula by induction and combinatorially I need to prove this equation by induction and by combinatorial explanation. $$\sum_{k=0}^{n} \left( \begin{array}{c} 4n \\ 4k \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1}$$
My induction proof:
Base case: p(1): $$\left( \begin{array}{c} 4 \\ 0 \en... | The problem is that the $n+1$ case isn't actually obtained from the $n$ case by adding an extra term to the end, the way you're thinking of it.
Here are the first few cases of this formula:
*
*For $n=1$, $\binom 40 + \binom 44 = 2^2 - 2^1$.
*For $n=2$, $\binom 80 + \binom 84 + \binom 88 = 2^6 + 2^3$.
*For $n=3$, $\... | {
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"timestamp": "2023-03-29T00:00:00",
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The matrix of $T(x_1,x_2)=(x_1+x_2,x_1-x_2)$ with respect to a basis Consider the linear map $T:\mathbb{R}^2\to \mathbb{R}^2, T(x_1, x_2)=(x_1+x_2,x_1-x_2)$. Let $B_1$ be the canonical base of $\mathbb{R}^2$ and consider another basis $B_2=\{f_1,f_2\}$, where $f_1=(1,1)$ and $f_2=(1,2)$.
So, according to my computation... | The second column of your matrix of $T$ with respect to $B_2$ is wrong.
Indeed, $T(f_2) = T(1,2) = (3,-1) \neq 3f_1 - f_2 = (2,1).$
| {
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Factorize polynomial of degree 4 given statements If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?
Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+... | Let $$f(x) = x^4-x^3-13x^2+26x-8$$
Observe that $f(2) = 0$ and $f(-4) = 0$
Hence $(x-2)$ and $(x+4)$ are two factors of $f(x)$
Also $f(x)$ is divisible by $(x-2)(x+4) = x^2 + 2x - 8$
Now divide $f(x)$ by $x^2 + 2x - 8$ to obtain the other quadratic factor $x^2 - 3x +1$
If you solve $x^2 - 3x +1 = 0$, you'll find $x = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the last part of a diferential equation I'm trying to solve this problem but I've some doubts about the final steps.
Solve the initial value problem using matrix functions:
$$
\frac {dy}{dx} =
\begin{pmatrix}
1 & a & 1 \\
0 & 1 & a \\
0 & 0 & 1 \\
\end{pmatrix}
y
\text{ ; }
y(0) =
\begin{pmatrix}
0 \\
0 \\
-... | A simple solution:
$$
\frac {dy}{dx} =
\begin{pmatrix}
1 & a & 1 \\
0 & 1 & a \\
0 & 0 & 1 \\
\end{pmatrix}
y
\text{ ; }
y(0) =
\begin{pmatrix}
0 \\
0 \\
-1 \\
\end{pmatrix}
$$
Let $y=\begin{pmatrix} y_1 \\ y_2 \\y_3 \end{pmatrix}$
Then you have three couples ODEs as
$\frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse of fourth root function I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$
but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quad... | The equation is easier to work into the quadratic equation if we first get rid of radicals and avoid trying to "simplify" by using the $y^4/a^4$ rational. Expansion gives us the best view of a,b,c.
$$y=a\sqrt[^4]{b^2-(x-c)^2}\implies y^4=a^4\big(b^2-(c^2 - 2 c x + x^2)^2\big)\\
\implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$ Calculate
$\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$.
The only idea that I have is to substitute $t=x+1$, but I do not think it is a good one.
| As @Tito Eliatron commented, integrating by parts
$$I=\int\frac{\tan ^{-1}(x)}{x^2-4 x-1}dx$$
$$I=\tan ^{-1}(x)\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{2 \sqrt{5}}+\frac J{2 \sqrt{5}}$$
$$J=\int\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{x^2+1}dx$$
$$\frac 1{x^2+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just expo... | I came up with a weird proof, but the idea behind is kind of simple: the fact is that I expect that for $x$ outside an interval the terms $x^8+x^7+x^6$ will be greater than the negative terms, and in this range the remaining terms will be positive. On the other side, for small $x$ the -5 will paly his role and make thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a ... | The standard substitution is $x=b+c-a , y=c+a-b , z=a+b-c $ ($x,y,z \geq 0$ by the triangle inequality.) These invert to give
\begin{eqnarray*}
2a&=&y+z \\ 2b&=&z+x \\ 2c&=&x+y.
\end{eqnarray*}
Subbing these gives
\begin{eqnarray*}
x^3z+y^3x+z^3x-xyz(x+y+z) \geq 0
\end{eqnarray*}
which you could $\color{red}{\text{not}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$? If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?
I need to solve it using combinatorics.
I've started with cal... | The straightforward way to get the result is to compute the coefficient of $X^{12}$ in $(1+X+X^2+X^4+X^5+X^6)^4$. Though computing that coefficient basically amounts to counting solutions explicitly in a systematic manner, it is not so hard to do even by hand: $$(1+X+X^2+X^4+X^5+X^6)^2=\\1+2X+3X^2+2X^3+3X^4+4X^5+6X^6+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $ If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $
Prove $ abc \geq 8 $
I have tried AM-GM and substituting the 1, without results.
| Let
$$x = \frac{1}{a+1}, \quad y = \frac{1}{b+1}, \quad z = \frac{1}{c+1},$$
then $x+y+z=1$ and
$$a = \frac{1-x}{x} = \frac{y+z}{x}.$$
The inequality become
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant 8.$$
Which is true because
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant \frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{xyz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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