Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prob. 19, Chap. 1, in Baby Rudin: For what $\mathbf{c}$ and $r > 0$ does this equivalence hold? Here's Prob. 19 in Chap. 1 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $\mathbf{a} \in \mathbb{R}^k$, $\mathbf{b} \in \mathbb{R}^k$. Find $\mathbf{c} \in \mathbb{R}^k$ and $r > 0$ ... | First a comment. What this problem really says is that the subset of $\mathbb{R}^k$ defined by the equation $|x - a| = 2|x - b|$ is a sphere. So there is really not going to be any choice in determining $c$ and $r$. The point $c$ will need to be the centre of the sphere and $r$ its radius.
I imagine the main point of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\int \frac{2x}{9x^2+3}dx=?$ So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way:
\begin{align}
& \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx... | I'd have written $u = 9x^2+ 3$ so $du = 18x\,dx$ and $2x\,dx = \dfrac{du} 9$. Then you have
$$
\int \frac{2x}{9x^2+3} \, dx = \frac 1 9 \int \frac{du} u = \frac 1 9 \log|u|+C = \frac 1 9 \log ( 9x^2+3) + C.
$$
This is the same as
$$
\frac 1 9 \log ( 3( 3x^2+1)) + C = \frac 1 9 \left( \log(3x^2+1) + \log 3 \right) + C ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equi... | You have done the hard work
As $0<\tan A,\tan B,\tan C<\infty$
$0<A,B,C<\dfrac\pi2$
As $\tan A,\tan B,\tan C$ are distinct, so will be $A,B,C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of region under 2 curves. Find the area of the region R enclosed by the line $y=2x−1$ and the parabola $y^2=4x+141$
Here is what I have done:
*
*Write each equation in terms of x. ($x=\frac{y+1}{2}$ and $x=\frac{y^2-141}{4}$)
*Find intersections: y= -11 and 13
*Write the integral (I don't quite know).
*Calcu... | I checked $1$ and $2$ and it is correct.
Step 3:
If $f(y)>g(y)$ for $x$-values between $[-11,13]$ then the integral is
$\int_{-11}^{13}{\left(f(y)-g(y)\right)}{dy}$
Since $\frac{y+1}{2}>\frac{y^2-141}{4}$ from $[-11,13]$ we have
$\int_{-11}^{13}\left(\frac{y+1}{2}-\frac{y^2-141}{4}\right)dy$
$\frac{1}{4}\int_{-11}^{13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\lef... | I will assume you know that $\frac{\sin u}{u}\xrightarrow[u\to 0]{} \sin^\prime(0) = 1$.
*
*First, observe that $\frac{x+1}{4x^2+3} \xrightarrow[x\to\infty]{} 0$.
*Use this observation to rewrite
$$
\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)
= \frac{2x^3+x}{x^2+1}\frac{x+1}{4x^2+3} \frac{\sin\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\lim\limits _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{... | I'd set $x=1/t^3$ (with $x>0$, which is not restrictive) so the base becomes a fraction with
$$
\frac{1}{t^3}+\sqrt[3]{\frac{1}{t^{12}}+1}=
\frac{t+\sqrt[3]{1+t^{12}}}{t^4}
$$
at the numerator and
$$
\frac{1}{t^3}+\sqrt[6]{\frac{9}{t^{24}}+3}=
\frac{t+\sqrt[6]{9+3t^{24}}}{t^4}
$$
at the denominator. So, taking the loga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) +... | Using summation notation gives
$$3 + 9 + 15 + \cdots + 6n - 3 = \sum_{k = 1}^n (6k - 3)$$
Let $P(n)$ be the statement
$$\sum_{k = 1}^{n} (6k - 3) = 3n^2$$
To prove this statement by induction, we must show that $P(1)$ holds and that whenever $P(m)$ holds for some positive integer $m$, then $P(m + 1)$ holds since we th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finiteness of the sums of reciprocals of positive solutions of $\tan x = x$ and $x = \tan \sqrt x$ Let $a_n$ be the sequence of positive solutions of the equation $\tan x=x$ and $b_n$ be a sequence of positive solutions of the equation $x=\tan \sqrt x$.
Prove that $\sum \dfrac{1}{a_n}$ diverges but $\sum \dfrac{1}{b_n... | For each integer $n$, $\tan x$ is a bijection from $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$ to $\mathbb{R}$.
So for each positive integer $n$, there is exactly one positive solution to $x = \tan x$ in each interval $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$. Also, the only solution in $(-\tfrac{1}{2}\pi,\tfrac{1}{2}\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over... | \begin{align}
J&=\int_0^\pi \theta\ln\left(\tan\left(\frac{\theta}{2}\right)\right)d\theta\\
&\overset{u=\tan\left(\frac{\theta}{2}\right)}=4\underbrace{\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du}_{=K}\\
K&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ... | Let $\sqrt{x+2}=t\implies \frac{dx}{2\sqrt{x+2}}=dt$ or $dx=2t\ dt$
$$\int \frac{x+3}{\sqrt{x+2}}\ dx$$$$=\int \frac{t^2-2+3}{t}(2t\ dt)$$
$$=2\int (t^2+1)\ dt$$
$$=2\left(\frac{t^3}{3}+t\right)+C$$
$$=2\left(\frac{(x+2)^{3/2}}{3}+\sqrt{x+2}\right)+C$$
$$=\frac 23(x+5)\sqrt{x+2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
| $$a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2byxz$$
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2axcz+2byxz$$
$$a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2byxz+c^2y^2=0$$
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$$
From here the answer is clear. All three terms a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we h... | The solutions of a quadratic with integer coefficients are either both rational or both irrational. So if one of $a_1,a_2$ is an integer the other must be rational. Say $a_1\in \Bbb Z$ and $a_2=\frac nm\in\Bbb Q$ is a rational with denominator $m$. Then the denominator of $a_1+a_2$ is also $m$ and the denominator of $a... | {
"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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If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit. If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit.
So I am thinking that I should be able to do this by contradiction. So if I assume there is some element $z\in R$ ... | You have, $(a-b)(a^2+b^2)=a^3+ab^2-ba^2-b^3=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of plane perpendicular to given plane Find the equation of the plane which contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and which is perpendicular to the plane $5x+3y-6z+8=0$
By setting $z=0$ I found a point $(-\frac{14}{3}, \frac{13}{3}, 0)$ but not able to obtain normal vect... | In answer to your last question, yes: because our required plane, which we shall call $P$, must contain a vector, $\mathbf{l}$ say, parallel to our line of intersection and also the vector, $\mathbf{n} := (5,3,-6)$, normal to the plane $5x+3y-6y+8=0$.
Vector $\mathbf{l}$ is perpendicular to the normals: $(1,2,3)$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$ $$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$
This result seems to me digitally correct?
Can we prove that the equality is exact?
| \begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1... | Note that the given expression implies:
$\tan(A)=\frac{1-\cos(B)}{\sin(B)} \cdot \frac{1+\cos(B)}{1+\cos(B)}=\frac{\sin(B)}{1+\cos(B)}$
Therefore
$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}=\frac{2\frac{\sin(B)}{1+\cos(B)}}{1-\left(\frac{1-\cos(B)}{\sin(B)}\cdot \frac{\sin(B)}{1+\cos(B)}\right)}=\frac{2\sin(B)}{2\cos(B)}=\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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integrate $\int dx \frac{2x+1}{(x^2-9)^\frac{5}{2}}$
$$\int dx \frac{2x+1}{(x^2-9)^\frac{5}{2}}$$
$x=\frac{3}{\sin\theta}$
$dx=\frac{3\sin\theta}{\cos^2\theta}d\theta$
$$\int d\theta \frac{\left(\frac{6}{\sin\theta}+1\right)\cdot\frac{3\sin\theta}{\cos^2\theta}}{\left(\frac{9}{\sin^2\theta}-9\right)^\frac{5}{2}}=\int... | This is not a continuation of your method, but rather another technique.
$\int \frac{(2x+1) dx}{(x^2-9)^\frac{5}{2}}=\int \frac{2xdx}{(x^2-9)^\frac{5}{2}}+\int \frac{dx}{(x^2-9)^\frac{5}{2}}$
For the first term, let $u=x^2-9$, making $du=2x dx$, setting the integral up for a power rule.
For the second term, let $x=3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest positive value taken by $a^3+b^3+c^3-3abc$
Find the smallest positive value taken by $a^3+b^3+c^3-3abc$ for positive integers $a,b,c$. Find all integers $a,b,c$ which give the smallest value.
Since it is generally hard to find the minimum of a multivariate polynomial, I tried factoring it at first. ... | The answer is $4$, for $a = 1, b = 1, c = 2$ (or some permutation).
We have
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)[(a + b + c)^2 - 3(ab + bc + ca)].$$
This shows that any positive value taken by the expression must be at least as large as $a + b + c$, which leaves only the above cases and $a =b = c = 1$ to examine.
Ed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Is there an identity that says $|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}| \leq |\sqrt {x^2} - \sqrt {y^2}|$? Is there an identity that says $|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}| \leq |\sqrt {x^2} - \sqrt {y^2}|$?
Because of the nature of the square root function, its derivative monotonically decreases. so differences "further... | Yes.
$$
\left|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}\right| =\frac{\lvert x^2-y^2\rvert}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|}
= |\sqrt {x^2} - \sqrt {y^2}|\cdot \frac{|\sqrt {x^2} + \sqrt {y^2}|}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|}
\leq |\sqrt {x^2} - \sqrt {y^2}|
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find all primes $p,q$ and even $n > 2$ such that $p^n+p^{n-1}+\cdots+1 = q^2+q+1$
Find all primes $p,q$ and even $n > 2$ such that $p^n+p^{n-1}+\cdots+1 = q^2+q+1$.
Attempt
The first thing I would do is simplify the geometric series to $\dfrac{p^{n+1}-1}{p-1} = q^2+q+1$. I was thinking from here we could use a modula... | Suppose that $n=2k$ for some integer $k>1$. Then we have $q(q+1)=p\left(\frac{p^{2k}-1}{p-1}\right)$. As $k>1$, $p\neq q$, so that $q$ divides $p^{2k}-1=\left(p^k+1\right)\left(p^k-1\right)$. That is, $q$ divides either $p^k-1$ or $p^k+1$. In any case, $q\leq p^k+1$. Therefore, $$p^{2k}+p^{2k-1}+2< p^{2k}+p^{2k-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Discriminant of Elliptic Curves In the study of elliptic curves, specifically in Weierstrass form, you have the equation
$E : y^2 = x^3 +ax +b$.
However I have found the discriminant comes in two different forms:
$\Delta = -16(4a^3 + 27b^2) $ or $\Delta = 4a^3 + 27b^2$
I understand how to get the second equation, bu... | A cubic over $k$ in Weierstrass form (affine form) is given by $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ The discriminant is defined by $$\Delta = -b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6,$$ where $ b_2=a_1^2+4a_2$, $ b_4=2a_4+a_1a_3 $, $b_6=a_3^2+4a_6$ and $b_8 = a_{1}^{2} a_{6}+4 a_{2} a_{6}-a_{1} a_{3} a_{4}+a_{2} a_{3}^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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For which values of $x$ does the series converge? $$\sum_{k=0}^{\infty} \frac{3^k\cdot 3^2 \cdot x^k}{2^k}$$
I have to find the values of $x$ for which the series converges.
I used the root test,the limit of the absolute value of the kth root of the series must be smaller than 1 and I ended up with $|3x/2|<1 $ which le... | The geometric series
\begin{align*}
\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}\qquad\qquad |x|<1
\end{align*}
is convergent for $x\in(-1,1)$.
OPs example is a geometric series (somewhat in disguise), since
\begin{align*}
\sum_{k=0}^{\infty}\frac{3^k3^2x^k}{2^k}&=3^2\sum_{k=0}^{\infty}\left(\frac{3x}{2}\right)^k\\
&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An ellipse is drawn with the major and minor axes of lengths of $10$ and $8$, Then radius of circle which touches that ellipse.
An ellipse is drawn with the major and minor axes of lengths of $10$ and $8$ respectively
Using one focus as center is drawn such that it is tangent to the ellipse.
The radius of the circle i... | The circles of centers $(3,0)$ and radius $r = 5 \pm 3$ are clearly tangent to the ellipse. So why aren't the discriminant of your equation in $x$ equal to $0$ for these values of $x$? Because there are also other solutions in $x$. The problem is that the corresponding $y$'s are complex ( try solving the system obtaine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but... | Solving by factoring means reducing the polynomial to a lower degree by dividing it by factors $(x-x_i)$ for already known roots $x_i$.
The reason for this is that the lower degree polynomial might be easier to solve. Here this is not necessary, as a order $2$ polynomial is already solvable with moderate effort.
How to... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: L... | For the last induction step
$$\sqrt{k^2 + k}+1> k +1 \implies \frac{\sqrt{k^2 + k}}{\sqrt{k+1}}+ \frac1{\sqrt{k+1}}> \sqrt{k+1}\implies \sqrt{k} +\frac1{\sqrt{k+1}}> \sqrt{k+1}.$$
Hence
$$\sum_{j=1}^{k} \frac1{\sqrt{j}} > \sqrt{k} \implies \sum_{j=1}^{k+1} \frac1{\sqrt{j}} > \sqrt{k} + \frac1{\sqrt{k+1}} > \sqrt{k+1}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | Start off with the fact that
$$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca)
$$
Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term:
$$
x^3+kx+m = 0
$$ with $ab+bc+ca = k$. And by the way, that says that
$$a^2+b^2+c^2=-2k$$
Now start chaining upward, expressing $a^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 1
} |
Orthogonality relation as double sum of products of binomial coefficients I have stumbled upon the following sum over $x,y$ for non-negative integers $\kappa,\lambda$:
$$
\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{\displaystyle{\kappa+\lambda \choose x+y}}{... | Consider the integral
$$
I=\int_0^1 \left[\left(\sqrt{1-t^2}+it\right)^{2k}+\left(\sqrt{1-t^2}-it\right)^{2k}\right]\cdot \left[\left(\sqrt{1-t^2}+it\right)^{2\lambda}+\left(\sqrt{1-t^2}-it\right)^{2\lambda}\right]\frac{dt}{\sqrt{1-t^2}}.
$$
One can easily check that it equals
\begin{align}
I=&\ 4\sum_{x=0}^{\kappa}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| Note the fact that $2^8+2^{11}=48^2$
This implies that we are trying to find values of $n$ where $2^n=(x-48)(x+48)$.
Thus, we must find $k,l$ where $2^k-2^l=96$(where $x+48=2^k$, $x-48=2^l$)
Note the fact that $k \ge 7$.
This implies that $2^k$ is divisible by $32$, which implies that $2^l$ is also divisible by $32$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
} |
Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$? I was solving this equation:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=1$$
I solved it like this:
The given equation can be written as:
\begin{align*}
\sqrt{\cos{\the... | Hint:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1.$$
Multiply by $\cos(\theta)$ and take the square root. You get
$$\sqrt{\cos(\theta)}=\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}}...=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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How does this factoring work? $$ (z^2 - 2i ) = (z -1 -i)(z + 1 +i) $$
I see if you multiply out the right-hand side, you obtain the left-hand side, but how does one know to factor like that or this?
$$ (z^2 − 3iz − 3 + i) = (z − 1 − i)(z + 1 − 2i) $$
| If we are trying to factor $p(z) = z^2 - 2i$ we can determine the factorization by finding the zeros of the polynomial.
If $p(z) = 0$ then $z^2 = 2i$. Thus $|z| = \sqrt{2}$. Writing $z=\sqrt{2} e^{i\theta}$ we have $e^{i2\theta} = e^{i \frac{\pi}{2}}$. This yields $$\cos(2\theta) + i \sin(2\theta) = \cos(\pi/2) + i \si... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $b$ is an odd composite number and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$? (Note: An improved version of this question has been cross-posted to MO.)
Let $\sigma(X)$ be the sum of the divisors of $X$. For example, $\sigma(2) = 1 + 2 = 3$, and $\sigma(4) = 1 ... | This is only a partial answer to my initial question.
Let $I(x) = \sigma(x)/x$ be the abundancy index of $x$.
If $b$ is an odd composite and
$$\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r) = 2^{r+1} - 1,$$
then I know that
$$b^2 - 1 = \sigma(2^r)\sigma(b^2) - {b^2}\left(2^{r+1} - 1\right)$$
so that
$$\sigma\left({2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | Integrating by parts, $$\int\dfrac{2t^2}{(1+t^2)^2}dt=t\int\dfrac{2t}{(1+t^2)^2}dt-\int\left(\dfrac{dt}{dt}\cdot\int\dfrac{2t}{(1+t^2)^2}dt\right)dt$$
$$=-\dfrac t{(1+t^2)}-\int\dfrac{dt}{1+t^2}=?$$
Alternatively, choose $\sqrt x=\arctan u\implies x=(\arctan u)^2,dx=\dfrac{\arctan u}{1+u^2}du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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How Many Ways to Arrange These Boys and Girls?
There are $7$ boys and $3$ girls. In how many ways can they be arranged in a row such that the two ends are occupied by boys and no two girls are seated together?
The answer is $6 \cdot 5 \cdot 4 \cdot 7!$ and the book gave the same explanation is the same as the one h... | You have $20$ different seat-combinations for the girls:
*
*$2,4,6$
*$2,4,7$
*$2,4,8$
*$2,4,9$
*$2,5,7$
*$2,5,8$
*$2,5,9$
*$2,6,8$
*$2,6,9$
*$2,7,9$
*$3,5,7$
*$3,5,8$
*$3,5,9$
*$3,6,8$
*$3,6,9$
*$3,7,9$
*$4,6,8$
*$4,6,9$
*$4,7,9$
*$5,7,9$
Then, you can permute the girls in $3!$ different ways.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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steps to calculate the space surface area cut by a cylinder(see the picture) The space surface(in yellow) $ x^2+y^2 = 2az\ $ is cut by a cylinder(in green) $x^2+y^2=3a^2 (a>0)$
How to calculate the cut out part area $A$?
I think the part is between the two planes $z=0$ and $ z= \frac32a $, and I only have the knowledge... | Your work is corrrect.
Note that you can find the surface also as the surface of revolution of the parabola
$$
x^2=2az
$$
around the $z-$ axis for $0\le x\le a\sqrt{3}$. This is a bit easier and gives the integral:
$$
A=2 \pi\int_0^{a\sqrt{3}}x\sqrt{1+\frac{x^2}{a^2}}dx
$$
that can be solved with the substitution $1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and... | Hint: Use the substitution $u^2=x^7+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to find the roots of the polynomial We consider the irreducible polynomial $g=y^4+y+1 \in \mathbb{F}_2[y]$ and let $b$ be a root of $g$.
I want to find all the roots of $g$ and also three generators of $\mathbb{F}_{16}^{\ast}$ as for the basis $\{1, b, b^2, b^3 \}$.
I have tried the following:
I applied the euclid... | Observe that if $ b $ is a root of $ g $, then so are $ b+1, b^2, b^2 + 1 $. All of these are obviously pairwise distinct since $ g $ is an irreducible polynomial.
EDIT : $ b^2 $ is a root because we have $$ g(b^2) = (b^2)^4 + b^2 + 1 = b^8 + b^2 + 1 = (b+1)^2 + b^2 + 1 = 0 $$ From the fact that $ b $ is a root $ \impl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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On a remarkable system of fourth powers using $x^4+y^4+(x+y)^4=2z^4$ The problem is to find four integers $a,b,c,d$ such that,
$$a^4+b^4+(a+b)^4=2{x_1}^4\\a^4+c^4+(a+c)^4=2{x_2}^4\\a^4+d^4+(a+d)^4=2{\color{blue}{x_3}}^4\\b^4+c^4+(b+c)^4=2{x_4}^4\\b^4+d^4+(b+d)^4=2{x_5}^4\\c^4+d^4+(c+d)^4=2{x_6}^4$$
As W. Jagy pointed o... | After mulling over the problem, it turned out the same elliptic curve can make five of the $x_i$ as integers. To start, note that,
$$x^4+y^4+(x+y)^4 = 2(x^2+xy+y^2)^2$$
Thus, the system is reduced to finding,
$$\color{blue}{a^2+ab+b^2 = x_1^2}\tag1$$
$$\color{blue}{a^2+ac+c^2 = x_2^2}\tag2$$
$$b^2+bc+c^2 = x_4^2\tag3$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Determine all the roots of the equation given by $z^2(1-z^2)=16.$ For my third year Complex variable course, the question is
Determine all the roots of the equation given by
$$z^2(1-z^2)=16.$$
My attempt:
Let $z^2 = x$
$x(1-x) = 16$
$x-x^2 = 16$
$x^2-x-16 = 0$
$x = \frac{1 \pm \sqrt{1 -4(16)}}{2}$
$x = \frac{1 \... | Remember it is a fourth order equation, so the complete solution has four roots, in this example they occur in complex-conjugate pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$
My Try:: Using $\b... | The elementary estimate
$$\frac{m}{n+m} < \sum_{k = 0}^{m-1} \frac{1}{n+k} < \frac{m}{n}$$
for $m > 1$ (we have equality on the right for $m = 1$) gives
$$\frac{1}{61} = \frac{33}{2013} < \sum_{k = 1980}^{2012} \frac{1}{k} < \frac{33}{1980} = \frac{1}{60},$$
whence $\lfloor A\rfloor = 60$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2... | The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is a circle of radius $a$ in $(\hat x,y)$ coordinates, where $\hat x=\dfrac{a}{b}x$. This transformation multiplies areas by the constant $\dfrac{a}{b}$, so the problem is equivalent to finding the rectangle of maximum area in a circle, which is well-known to be a s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Moment Generating Function I am little unsure if I am doing the following question right:
Suppose Y is a discrete random variable such that
$P(Y=0)=\frac{1}{3}$; $P(Y=1)=\frac{1}{6}$; $P(Y=2)=\frac{1}{6}$;
$P(Y=3)=\frac{1}{3}$;
a) Write down the moment generating function of $Y$.
b) Find the mean and variance of $... | Let $Z=X_1+X_2$. Since they are independent random variables with the same distribution as $Y$
\begin{align}
M_Z(t) &= E[e^{t(X_1+X_2)}]
\\
&= E[e^{tX_1}] \:E[e^{tX_2}]
\\
&=(\frac{1}{3} + \frac{1}{6}e^{t} + \frac{1}{6}e^{2t} + \frac{1}{3}e^{3t})^2
\\
&=\frac1{36}(4+4e^t+5e^{2t}+10e^{3t}+5e^{4t}+4e^{5t}+4e^{6t})
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $f(x, y) \le 3 $ for $x \ge 0, y > 0$
Let $x \ge 0, y>0$ and
\begin{align*}
f(x,y)&=\sqrt{\dfrac{y}{y+x^2}}+4\sqrt{\dfrac{y}{(y+(x+1)^2)(y+(x+3)^2)}}\\[6pt]
&\qquad +4\sqrt{\dfrac{y}{(y+(x-1)^2)(y+(x-3)^2)}}.
\end{align*}
Prove that $f(x,y) \le 3$.
I can prove when $x\ge 2, f(x,y) < 3$, but $f(x,y)=3$ when... | No clue without an overview. With help of restrictions as found already by the OP, a contour plot / isoline chart has been produced for the function at hand:
The $y$-axis is in $\color{green}{\mbox{green}}$. Our viewport is:
xmin := -6 ; xmax := 6;
ymin := 0 ; ymax := 12;
The are 27 contour levels nivo, defined ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 0
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Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where:
$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
and $x$, $y$, $z$ are angles of a triangle.
I tried to use sine law but got nothing.
| My solution involves finding the size of angle x. Then getting it's sine.
From the sine law we can tell that the triangle has sides 3, 4 and 2. And that angle x is opposite to the side with length 4.
We can employ cosine law to find the size of angle x.
$c^2 = a^2 + b^2 + 2ab\cos\theta$
$4^2 = 2^2 + 3 ^2 - 2(3*2)\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$ How can I show that
$$
\lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2}
$$
without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
| You just need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$, since:
$$\begin{eqnarray*} \frac{1-\cos^n x}{x^2} &=& \frac{1-\cos x}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x) = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x)\\ &=& \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \sum_{k=0}^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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Find x for $\sqrt{(5x-1)}+\sqrt{(x-1)}=2$ Solve:
$$\sqrt{(5x-1)}+\sqrt{(x-1)}=2$$
When $x=1$, we get the following equation to equal to $2$
I've been trying to solve this problem but when I square both sides and simplify I end up with:
$$x^2+6x+2=0$$ and of course $x=1$ cannot be a solution. So im not sure what im doin... | Perhaps a slightly tidier solution:
$$\begin{align}
\sqrt{5x-1}+\sqrt{x-1}&=2 \\
\sqrt{5x-1} &= 2-\sqrt{x-1} &\text{from regrouping}\\
5x-1&=4-4\sqrt{x-1}+(x-1) & \text{after squaring}\\
4x-4&=4\sqrt{x-1} &\text{after regrouping again}\\
(x-1)^2&=x-1 &\text{divide by 4 and square again}\\
(x-1)^2-(x-1)&=0 \\
(x-1)(x)&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Focus of the Parabola Find the Focus of $$(2x+y-1)^2=5(x-2y-3)$$.
Clearly its a Parabola whose axis is $2x+y-1=0$ and since $x-2y-3=0$ is perpendicular to $2x+y-1=0$ Tangent at the vertex is $x-2y-3=0$.Also the Vertex is $(3,-1)$, but now how to find its focus?
| More generally if we have a parabola in the form (which all parabolas can be put into) $$(l_{\rm axis})^2=e\cdot l_{\text{tangent at vertex}}$$ or $$(ax+by+c)^2=e(-bx+ay+d),$$ and use the orthonormal change of basis $x'=\frac{ax+by}{\sqrt{a^2+b^2}},y'=\frac{-bx+ay}{\sqrt{a^2+b^2}}$, we get $$(\sqrt{a^2+b^2}x'+c)^2=(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$
Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$
My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\fra... | Since $\sin\left(x\right)/x
$ is a monotone decreasing function on $\left[\frac{\pi}{4},\frac{\pi}{3}\right]
$ we have $$\int_{\pi/4}^{\pi/3}\frac{\sin\left(x\right)}{x}dx\geq\frac{\sin\left(\pi/3\right)}{\pi/3}\int_{\pi/4}^{\pi/3}dx=\frac{\sqrt{3}}{8}.$$ For the upper bound, if we want more precision, we can use for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's me... | Let's use Bill Dubuque's AC method on this cubic. Start by factoring $228 = 2^3 \cdot 3 \cdot 19$ and $1116 = 2^2 \cdot 3^2 \cdot 31$. Their greatest common divisor is $2 \cdot 3 = 6$, so it is an ideal candidate to reach the form $f(6x) = a(6x)^3 + b(6x)^2 + c(6x) + d$.
To do this, transform the roots $k \mapsto 1/x$ ... | {
"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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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| $$I= \int \frac { x^2dx} {x^4 + x^2 -2}$$
Divide by $x^4$.
$$I=\int \frac{\frac{1}{x^2}dx}{1+\frac{1}{x^2}-\frac{2}{x^4}}$$
Let $\frac1x=u$. Then, $du=-\frac{1}{x^2}dx$.
$$I=\int\frac{du}{2u^4-u^2-1}=\int\frac{du}{2u^4-2u^2+u^2-1}=\int\frac{du}{(2u^2+1)(u^2-1)}=\int\frac{du}{(2u^2+1)(u+1)(u-1)}$$
Now this can be solve... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$ $C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$
My try:
I applied $|z^3-z+2|\leq|z|^3+|-z|+|2|$,i got $f(z)\leq 4$ but book says my answer is wro... | Let $\cos\theta=c,\sin\theta=s$ and $z=c+si$.
Then, since
$$\begin{align}z^3-z+2&=(c+si)^3-(c+si)+2\\&=c^3+3c^2si-3cs^2-s^3i-c-si+2\\&=(c^3-3cs^2-c+2)+(3c^2s-s^3-s)i\end{align}$$
we have
$$\begin{align}f(z)&=\sqrt{(c^3-3cs^2-c+2)^2+(3c^2s-s^3-s)^2}\\&=\sqrt{c^6+3 c^4 s^2-2 c^4+4 c^3+3 c^2 s^4+c^2-12 c s^2-4 c+s^6+2 s^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Identifying units in a polynomial ring
Problem Statement: Let $R$ be a domain. Identify the units in $R[x]$.
I am trying to identify the units in a domain $R$ by considering an arbitrary element $a=a_{n}x^{n}+\cdots+a_{1}x+a_{0}\in R[x]$ with an inverse $b=b_{m}x^{m}+\cdots+b_{1}x+b_{0}\in R[x]$.
If $b=a^{-1}$, then ... | Hint: if $R$ is a domain, then $\deg{fg} = \deg{f} + \deg{g}$ for all polynomials $f, g\in R[x]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of common roots of $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ The equations $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ have
*
*exactly one common root;
*no common root;
*exactly three common roots;
*exactly two common roots.
I factored the first equation. I think the root... | $\gcd (x^3 + 2 x^2 +2x +1 , x^{200} + x^{130} + 1 ) = x^2+x+1$, so two roots.
Note that you don't need to know the two roots, just that the gcd has degree $2$ and is coprime with its derivative (so no repeated roots).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Unit Tangent Vector and Unit Normal Vector for a given value of t
To Find - unit tangent vector $T$ and the unit normal vector $N$ for $t=0$; I
know the following
$\mathbf r(t)=\mathbf i(t)+\frac{1}{2}\mathbf j(t^2)+\frac{1}{3}\mathbf k(t^3)$
$T=\frac{r'(t)}{|r'(t)|}$
and
$N=\frac{T'(t)}{|T'(t)|}$
now,
$\math... | \begin{align*}
\mathbf{v} &= \dot{s} \mathbf{T} \\
&=(1,t,t^{2}) \\
\dot{s} &= \sqrt{1+t^{2}+t^{4}} \\
\ddot{s} &= \frac{t(1+2t^{2})}{\sqrt{1+t^{2}+t^{4}}} \\
\mathbf{v}(0) &= (1,0,0) \\
\dot{s}(0) &= 1 \\
\mathbf{T}(0) &= \mathbf{i} \\
\mathbf{a} &= \ddot{s} \mathbf{T}+\kappa \dot{s}^{2}\mathbf{N} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Which two digit number when you find the product of the digits yields a number that is half the original? Which two digit number when you find the product of the digits yields a number that is half the original?
Let x=$ab$ be the $2$-digit number. So $x=10a+b$.
Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies ... | $2ab-10a-b=0$ is a simple quadratic Diophantine equation. The canonical approach in this case is the following.
You add $5$ to both sides and you obtain
$$
2ab-10a-b+5=5,\ \ \Rightarrow (2a-1)(b-5)=5
$$
so $b-5$ or $2a-1$ must be equal to $5$ (or $-5$, but it is not possible) and the other must be equal to $1$.
But $0<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$ If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$
Let $y=\frac{x^2+ax+b}{x^2+2x+3}$
$$x^2y+2xy+3y=x^2+ax+b$$
$$x^2(y-1)+x(2y-a)+3y-b=0$... | $$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4 $$
As the denominator is positive, this is equivalent to
$$-5x^2-10x-15 \le x^2+ax+b \le 4x^2+8x + 12$$
which can be considered as two quadratic inequalities,
$$6x^2+(a+10)x+(b+15) \ge 0, \quad 3x^2+(8-a)x+(12-b) \ge 0$$
Note that equality must happen as well for some $x$. For t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Deriving the formula for the height of a trapezoid The bases of a trapezoid have lengths $a$ and $b$, and its legs have lengths $c$ and $d$. A formula for the height is
\begin{equation*}
h = \frac{
\sqrt{(-a + b + c + d)(a - b + c + d)(a - b - c + d)(a - b + c - d)}
}
{2\vert a - b \vert} .
\end{equation*}
The formula ... | Your formula "is reminiscent of Heron's Formula" because it is based on Heron's Formula.
One formula for the area of a triangle is
$$A=\frac 12bh$$
and Heron's formula gives
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter given by
$$s=\frac{a+b+c}2$$
Here is a diagram for the derivation for the height of y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to express $f(n\alpha)$ in terms of $f(\alpha)$
Original question: Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x)=\dfrac{a^x-a^{-x}}{2}$, where $a>0$ and $a\ne 1$, and $\alpha$ be a real number such that $f(\alpha)=1$. Find $f(2\alpha)$.$^1$
A few years ago, I was a high school student and solved ... | From $f(\alpha)=\frac{a^\alpha - a^{-\alpha}}{2}$, we can induce the quadratic equation for $a^{\alpha}$:
$$
a^{2\alpha}-2f(\alpha)a^{\alpha}-1=0.
$$
By the quadratic formula, we get
$$
a^{\alpha}=f(\alpha)\pm \sqrt{(f(\alpha))^2+1}
$$
and using $a^{\alpha}>0$ we eliminate one possibility. Thus
$$
a^{n\alpha}=(a^{\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Determine all integers $x $ and $ y$ such that $|2^x − 3^y| =1$ I am having trouble solving this problem:
Determine all integers x and y such that $|2^x − 3^y| =1$
I would think that the only solutions to it is $x = y = 1$.
How can I show that there is no other solutions?
If there are other solutions, how can I find... | Hint: If $2^x - 3^y = 1$, then $2^x = 1 + 3^y$. Take both sides modulo $16$:
$$2^x \equiv 1 + 3^y \pmod{16}.$$
If $x\ge 4$, then the left-hand side is $0$. The sequence of $3^y \pmod{16}$ is $3,9,27=11,81=1,3,9,11,\ldots$, so the right-hand side is $1+3=4$, $1+9=10$, $1+11=12$, or $1+1=2$, none of which is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^{2}\sqrt{4x + 1} \text{d}x$ Evaluate $\displaystyle\int_0^{2}\sqrt{4x + 1}~\text{d}x$
This becomes:
$\displaystyle\int_0^{2}(4x + 1)^\frac{1}2~\text{d}x$
I am not sure where to go from here, I suspect it might use the chain rule or reverse chain rule.
| $$\int_{0}^{2}\sqrt{4x+1}\space\text{d}x=$$
Substitute $u=4x+1$ and $\text{d}u=4\space\text{d}x$.
This gives a new lower bound $u=4\cdot0+1=1$ and upper bound $u=4\cdot2+1=9$:
$$\frac{1}{4}\int_{1}^{9}\sqrt{u}\space\text{d}u=\frac{1}{4}\int_{1}^{9}u^{\frac{1}{2}}\space\text{d}u=$$
Use $\int y^{n}\space\text{d}y=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the general solution to differential equation $x(x+1)^2(y'-\sqrt x)=(3x^2+4x+1)y$ Equation can be transformed to linear differential equation:
$$LHS=x^3y'+2x^2y'+xy'-x^{7/2}-2x^{5/2}-x^{3/2}$$
$$RHS=3x^2y+4xy+y$$
$$\Rightarrow y'(x^3+2x^2+x)+y(-3x^2-4x-1)=x^{7/2}+2x^{5/2}+x^{3/2}$$
After dividing by $(x^3+2x^2+x),... | Starting from the equation in the title, we have
$$x(x+1)^2y'-(3x^2+4x+1)y=x(x+1)^2\sqrt x$$
It just so happens we have
$$[x(x+1)^2]'=3x^2+4x+1$$
This suggests use of the quotient rule.
$$\dfrac{x(x+1)^2y'-(3x^2+4x+1)y}{x^2(x+1)^4}=\left[\dfrac{y}{x(x+1)^2}\right]'=\dfrac{\sqrt x}{x(x+1)^2}$$
One messy integral left an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Through the point $A(4,5)$ a line is drawn. Through the point $A(4,5)$ a line is drawn inclined at $45°$ with the $+ve$ X - axis. It meets $x+y=6$ at the point $B$. Find the equation of $AB$.
My solution..
Equation of $AB$
$$(y-y_1)=m(x-x_1)$$
$$(y-5)=1(x-4)$$
$$x-y+1=0$$.
But the answer in my book is $3x-y=7$.
Can an... | The equation of the first line is: $y - 5 = \tan(45^{\circ})(x - 4)= 1(x-4) = x-4\Rightarrow y = 5+x-4 = x+1$. Thus the intersection is found by: $x + 1 = 6- x \Rightarrow x = \dfrac{5}{2}\Rightarrow y = 6 - \dfrac{5}{2} = \dfrac{7}{2}\Rightarrow B = (\frac{5}{2}, \frac{7}{2})$. Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$ The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.
if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?
Since $x^3=x^2+x+1$ from the graphs of $x^3$... | Isn't this a trick question?
Let $$A_{n}=\frac{a^n-b^n}{a-b}$$
Then note $$a^{n+3}=a^{n+2}+a^{n+1}+a^{n}$$$$b^{n+3}=b^{n+2}+b^{n+1}+b^{n}$$
Thus subtract the two, and divide by $a-b$.
This gives us $$A_{n+3}=A_{n+2}+A_{n+1}+A_{n}$$
In a similar fashion, if $$B_{n}=\frac{b^n-c^n}{b-c}$$$$C_{n}=\frac{c^n-a^n}{c-a}$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How many base $10$ numbers are there with $n$ digits and an even number of zeros?
How many base $10$ numbers are there with
$n$ digits and an even number of zeros?
Solution:
Lets call this number $a_n$.
This is the number of $n-1$ digits that have an even number of zeros
times $9$ possibilities for the $n$th digit... | Here's another solution: clearly the number of $n$-digit numbers with $k$ zeroes is $f(n,k) = {n \choose k} 9^k$. Then we have
$$f(n,0) + f(n,1) + \cdots + f(n,n) = \sum_{k=0}^n {n \choose k} 9^k$$
and by the binomial theorem this is $(9+1)^n = 10^n$. On the other hand,
$$f(n,0) - f(n,1) + \cdots + (-1)^n f(n,n) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719341",
"timestamp": "2023-03-29T00:00:00",
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proof $ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}- \frac{1}{\binom{n+k+1}{k}} \right)=\frac{k}{k+1} $ I would appreciate if somebody could help me with the following problem
Q: How to proof (by combinatorial proof)
$$ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}
- \frac{1}{\binom{n+k+1}{k}... | It’s a bit involved, but here is a combinatorial argument.
As usual, for $m\in\Bbb Z^+$ let $[m]=\{1,\ldots,m\}$. Let $\mathscr{A}$ be the family of $k$-element subsets of $[n+k+1]$, $\mathscr{B}$ the family of $(k+1)$-element subsets of $[n+k+1]$, $\mathscr{C}$ the family of $(k+1)$-element subsets of $[n+k+1]$ that c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | You can approach this problem with the substitution
$$
\begin{cases}
X=\cos x\\[4px]
Y=\sin x
\end{cases}
$$
that transforms the equation into
$$
\begin{cases}
X+Y=\dfrac{1}{3} \\[6px]
X^2+Y^2=1
\end{cases}
$$
Rewriting the second equation as $(X+Y)^2-2XY=1$, we can substitute and get
$$
\begin{cases}
X+Y=\dfrac{1}{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
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Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1... | Method$\#1:$
$$33^2=(30+3)^2\equiv3^2+2\cdot3\cdot30\equiv190-1\pmod{100}$$
$$\implies33^{100}\equiv(190-1)^{50}$$
$$(190-1)^{50}=(1-190)^{50}\equiv1-\binom{50}1\cdot190\pmod{100}\equiv?$$
Method$\#2:$
Alternatively, Carmichael Function $\lambda(100)=\cdots=20\implies33^{20}\equiv1\pmod{100}$
As $100\equiv0\pmod{20},33... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes when $b_1<b_2\le 4$, where $b_1,b_2$ are the numbers of objects in box... |
Three boxes have no restrictions with respect to the objects which results in $e^{3x}$. The restriction $b_1<b_2\leq 4$ of the other objects together with the objects of the three boxes is encoded as
\begin{align*}
e^{3x}&\left(\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)+x\left(\frac{x^2}{2!}+\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Greatest $n<1000$ such that $\left \lfloor{\sqrt{n}}\right \rfloor-2 \mid n-4$ and $\left \lfloor{\sqrt{n}}\right \rfloor+2 \mid n+4$? My first attempt was incorrect, and it is supposed to be a middle school problem.
So, if $n=k^2$
Then $k-2 \mid n-4$ and $k+2 \mid n+4$, so $n-4 \mid (n-4)(n+4)$. I then assumed the ans... | Let $k=[\sqrt n]$. Then we know that
$$
k^2\le n<k^2+2k+1.
$$
So $n-4$ is in the range $[k^2-4,k^2+2k-4]$. Of the numbers in that range
only $k^2-4$, $k^2+k-6$ and $k^2+2k-8$ are divisible by $k-2$, so $n$ has to be one of $k^2,k^2+k-2, k^2+2k-4$.
We also see that $n+4$ is in the range $[k^2+4,k^2+2k+4]$. Of those num... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
How to solve $a = \cos x - b\sin x$ where $a$ and $b$ are real numbers? I found this equation when solving a physics problem related to finding an angle when entering a river, that has a known current, and trying to get to a specific point on the other side. I'm not sure how to solve this equation explicitly for the a... | Another way to solve this to use $x=y+z$ and expand the trig into
$$ a = \cos(y) \left( \cos(z) - b \sin(z) \right) - \sin(y) \left( b \cos(z) + \sin(z) \right) $$
Solve $\cos(z) - b \sin(z)=0$ with $z =\tan^{-1}(\frac{1}{b})$ and simplify the above into $$ a = - \sin(y) \left( b \cos(z) + \sin(z) \right) = - \sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Determinant of $N \times\ N$ matrix So the question asks:
For $n \geq 2$, compute the determinant of the following matrix:
$$
B =
\begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdo... | To add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\cdots r_n) where the $r_k$ are the roots of p(X)" see https://en.wikipedia.org/wiki/Vandermonde_matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$ I have to find number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$. I guess it would be $[x^n](1+x+x^2 \dots)(1 + x^2 + x^4 \dots)^2$, but how to compute it? I know only that $\frac{1}{1-x} = 1+x+x^2 \dots$.
| If $n$ is odd, then $x_1$ must be odd, and if $n$ is even, then $x_1$ must be even. Thus there is an integer $y$ such that $x_1=2y+1$ for former and $x_1=2y$ for latter. Thus
$$
y+x_2+x_3=\frac{n-1}{2}\text{ or }y+x_2+x_3=\frac{n}{2}.
$$
Therefore, the answer is
\begin{align}
\binom{\frac{n+3}{2}}{\frac{n-1}{2}} &= \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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What can we do to solve the following equation with $6$ variables with some information provided?
Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3... | 5xum has already pointed out your error.
Multiplying the both sides by $6!$ gives
$$\text{(integer)}+\frac{a_7}{7}=\frac{5\cdot 6!}{7}\quad\Rightarrow\quad a_7\equiv 5\cdot 6!\pmod 7\quad\Rightarrow\quad a_7=2$$
Multiplying the both sides by $7!/6$ gives
$$\text{(integer)}+\frac{7a_6+a_7}{6}=5!\times 5\quad\Rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
Th... | If $x=y=-1$, then $x+y+2=xy-1=0$. Therefore, some of the solutions below need to be prefaced with "except for $x=y=-1$".
If $x=y$, then $\frac{x+y+2}{xy-1}=\frac{2}{x-1}$, which says that
$$
x=y\in\{0,2,3\}\quad\text{are the only solutions with $x=y$}\tag{1}
$$
Furthermore,
$$
x+y=-2\quad\text{is a solution}\tag{2}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
What is the simplified average rate of change What is the simplified average rate of change between $x = 3$ and $x = 3 + h$ for the function $f(x) = -x^2 + 5$?
I know you start off with $\frac{f(x+h)-f(x)}{h}$ but after substitution I got $-15+h+x^2$. What is the correct solution?
| $$
\begin{eqnarray}
\frac{f(x+h)-f(x)}{h} &=& \frac{(-(x+h)^2 + 5) - (-x^2 + 5)}{h} \\
&=& \frac{(-x^2 - 2xh - h^2 + 5) - (-x^2 + 5)}{h} \\
&=& \frac{-2xh - h^2}{h} \\
&=& -2x - h
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find a maximum of: $x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $ $x,y,z \ge 0 $ , $ x+y+z =1$
Find a maximum of:
$$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $$
and when it is reached.
my attempt:
1) $$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x \le $$
$$\le \left( \frac{2016x+y}{2017}\right... | *
*$x\geq y\geq z$
Hence, $x^{2016}y+y^{2016}z+z^{2016}x\leq(x+z)^{2016}y=2016^{2016}\left(\frac{x+z}{2016}\right)^{2016}y\leq$
$\leq2016^{2016}\left(\frac{2016\cdot\frac{x+z}{2016}+y}{2017}\right)^{2017}=\frac{2016^{2016}}{2017^{2017}}$.
*$x\geq z\geq y$.
In this case $x^{2016}y+y^{2016}z+z^{2016}x\leq x^{2016}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find a formula for a periodic sequence? I would like to find the formula for a periodic sequence such as 4, 1, 1/4, 1/4, 1, 4... with a period of 6
|
Here is a variation which utilizes the special structure of the sequence.
The sequence can be written as
\begin{align*}
(a_n)_{n\geq 0}&=\left(4,1,\frac{1}{4},\frac{1}{4},1,4,4,1,\frac{1}{4},\ldots\right)\\
&=\left(4^1,4^0,4^{-1},4^{-1},4^0,4^1,4^1,4^0,4^{-1},\ldots\right)
\end{align*}
Since
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$
Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$
My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}... | You want to show that
$$
\frac{\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}}{\sin\frac{x}{2}}
+\cos((n+1)x)=
\frac{\sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2}}{\sin\frac{x}{2}}
$$
which is the same as
$$
\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}+\sin\frac{x}{2}\cos((n+1)x)=
\sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2}
$$
Set $s=nx/2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$ Evaluate the following integral:
$$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$
Could some give me hint as how to approach this question?
I tried to use the fact that $\sin ^4x+\cos ^4x=1-\fra... | Dividing by $\cos^2x$ on the top and bottom gives $\displaystyle\int\frac{\sqrt{\tan^4x+1}}{\frac{\sin^3x}{\cos x}}dx=\int\frac{\sqrt{\tan^4x+1}}{\tan^3x}\sec^2x \;dx$.
Now let $u=\tan x$ to get $\displaystyle\int\frac{\sqrt{u^4+1}}{u^3}du,\;$ and then let $t=u^2$ to get $\displaystyle\frac{1}{2}\int\frac{\sqrt{t^2+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can you prove this by strong induction? The sequence $b_1,b_2,...$ is defined recursively as:\begin{align} b_1&=0;\\ b_2&=1;\\ b_n&=2b_{n-1}-2b_{n-2}-1 \ \text{for} \ n\geq3. \end{align} Prove that this means: $$\forall n\geq1: b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$$
Edit:
I have tried to prove th... | Since you have verified the base case, we proceed to the inductive case. Fix $n >2$ and suppose that for all $k <n$, $P(k)$ is true. We want to show that $P(n)$ is true.
$$
\begin{aligned}
b_{n}&=2b_{n-1}-2b_{n-2}-1\\
&=2\left((\sqrt{2})^{n-1}\sin\left(\frac{1}{4}\pi (n-1)\right)-1\right)-2\left((\sqrt{2})^{n-2}\sin\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $a, b, c$ values from function $y=ax^2-bx+c$ and minimum value $D$ The problem reads like this:
The quadratic function which takes the value $41$ at $x = -2$ and $20$ at $x = 5$, is: $y = Ax^2-Bx+C$
The minimum value for this function is: $D$
I order it like this:
$$
\left\{
\begin{array}{1}
A(-2)^2-B(-2)+C=41\\... | The question as given in the post does not make sense as the value can be as small as you want, according to this Desmos graph.
However, if we add in the restriction that $A, B, C$ are all positive integers, then we can solve the problem. Following on from egreg's solution, we need $A>0, B = 3A+3 ≥ 0$, which gives $A >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) =... | Yes it is correct. It would do harm dividing out $\sqrt{x^2 + 9}.$ Perhaps an easier example of why this is so... Consider
\begin{equation}
x^2 = x.
\end{equation}
Obviously the two roots are 0,1, but do you see what happens when you divide by $x?$ You reduce the order of the polynomial, hence throwing away a root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| Let $z=e^{i\theta}$, then
\begin{align}
\int_0^{2\pi}\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}&=\int_C \frac{1}{a^2\left(\frac{z+\frac{1}{z}}{2}\right)^2+b^2\left(\frac{z-\frac{1}{z}}{2i}\right)^2}\frac{dz}{iz}\\
&=\int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz,
\end{align}
where $C:|z|=1$. If $a=b$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$ I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\fr... | I will give you a hint:
For second integral take $-1/x^2$ common becase if you take $t= 1/x$ then $dt=-1/x^2dx $ Now substitute t in place of $1/x$ so it will be converted into simple integral.
so next step becomes
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Diophantine equations $x^n-y^n=2016$
Solve equation
$$x^n-y^n=2016,$$
where $x,y,n \in \mathbb N$
My work so far:
If $n=1$, then $y=k, x=k+2016, k\in \mathbb N$
If $n=2$, then $2016=2^5\cdot 3^2 \cdot 7$
$x-y=1; x+y=2016$
$x-y=2; x+y=1008$
...
If $n=3$, then $x^3-y^3=(x-y)(x^2+xy+y^2)=2016$
$n\ge4 $. I need help... | You found the solutions for $n=1$ and your method for $n=2$ works (with Hagen von Eitzen's caveat) to give $(x,y)=(45,3),(46,10),$ $(50,22),(54,30),(65,47),$ $(71,55),(79,65),(90,78),$ $(130,122),(171,165),$ $(254,250),(505,503)$.
For $n=3$ we need $(x-y)(x^2+xy+y^2)=2016$. Since $x^2+xy+y^2>x^2-2xy+y^2>(x-y)^2$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If one angle of a triangle be $60^\circ,$ the area $10\sqrt3$ sq cm.,and the perimeter $20$ cm,find the lengths of the sides. If one angle of a triangle be $60^\circ$, the area $10\sqrt3\ \mbox{cm}^2$, and the perimeter $20\ \mbox{cm}$, find the lengths of the sides.
Let $\angle A=60^\circ$ and $\frac{1}{2}bc\sin A=10... | You have $(b+c)^2=(20-a)^2$ and so $b^2+c^2-bc=a^2-40a+400-120$. But by the cosine formula that is $a^2$, so you have $40a=280$ and hence $a=7$, so $b+c=13,bc=40$. Hence $b,c=5,8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$? while doing the Integration problem using Limit of a sum approach i have a doubt how
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln2$$
by infinite geometric series we have
$$1-x+x^2-x^3+x^4-x^5+\cdots =\frac{1}{1+x}$$ for $|x| \lt 1$ Integrting both sides... | Abel's Theorem shows that you can substitute $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$... | The analysis in the OP works. Here, we address the question regarding the necessity of using complex analysis.
To that end, we first note that using standard trigonometric identities, we can evaluate the sum
$$\begin{align}\sum_{n=1}^N \cos(nx)&=\frac{\sin(x)}{\sin(x)}\sum_{n=1}^N \cos(nx)\\\\
&=\csc(x)\sum_{n=1}^N\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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System of equations that can be solved by inequalities: $(x^3+y^3)(y^3+z^3)(z^3+x^3)=8$ and $\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32$
S367. Solve in positive real numbers the system of equations:
\begin{gather*}
(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\
\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32.... | Note that
$$
\begin{align}\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}&=\left(\frac{x^2}{x+y}+y-x\right)+\left(\frac{y^2}{y+z}+z-y\right)+\left(\frac{z^2}{z+x}+x-z\right)
\\
&=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32\,.
\end{align}$$
Therefore,
$$\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Problem in solving a question of vector space. The question is :
Let, $V$ be the subspace of all real $n \times n$ matrices such that the entries in every row add up to zero and the entries in every column also add up to zero. What is the dimension of $V$?
Now,I find that for $n=2$ the given condition yields a symmetri... | Actually when $n=2$, you should get 1. Suppose we decide to make the top entry $a$. Then the top right and lower left entries must be $-a$. And once we see that, the bottom right entry must be $a$. So $$V = \textrm{span}\left[\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix} \right].$$
If $n=3$, you can fill in the the to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Algebra, finding the elements of the field and solving irreducible polynomials I'm trying to do this problem from a practice final but there are no solutions.
I honestly am pretty stumped.
My thought was since it has 7 elements, then the degree of the polynomial must be one and we must be working with the field $F_7$
1... | The squares in $\mathbb{F}_7$ are indeed
$$
0^2=0,\quad
1^2=1,\quad
2^2=4,\quad
3^2=2,\quad
4^2=2,\quad
5^2=4,\quad
6^2=1,
$$
so you get $\{0,1,2,4\}$.
The cubes are
$$
0^3=0,\quad
1^3=1,\quad
2^3=1,\quad
3^3=6,\quad
4^3=1,\quad
5^3=6,\quad
6^3=6,
$$
so you get $\{0,1,6\}$.
In order to find a field with $7^2$ elements,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve for unknown matrix Let $A = \begin{bmatrix}
2 & 3 \\
4 & 5
\end{bmatrix}$
and let $B =
\begin{bmatrix}
3 & 4 \\
5 & 6
\end{bmatrix}$
Solve $A X = B$ for a matrix $X$
My guess is that i:
let $X =
\begin{bmatrix}
x \\
y
\end{bmatrix}$
Then solve it using a linear equation but i'm not too ... | Let
$$X=\left[\begin{matrix} a &b\\ c &d\end{matrix}\right].$$
Expanding the matrix product, you need to solve
$$\begin{cases}2a+3c=3\\4a+5c=5\\2b+3d=4\\4b+5d=6\end{cases}$$
Hint:
$$\begin{cases}2a+3c=3\\4a+5c=5\end{cases}$$ $$\begin{cases}2b+3d=4\\4b+5d=6\end{cases}$$
The real, efficient method is by solving the sys... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$
Find all real roots of the equation
$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$
I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn't work ... | Let $\sqrt{x+2}=a$ and $\sqrt{x+7}= b$.
So, $x + 1 = a^2 -1$
$x+6 = b^2 -1$
$ x^2 + 7x + 12= (x+3)(x+4)= (b^2-4)(a^2+2)$
Substituting and solving the system of equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Find triples $(a,b,c)$ of positive integers such that... Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$
I found this on a local question paper, and I am unable to solve it.
Any help will be appreciated.
| We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.
Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Hurwitz Zeta in terms of Bernoulli polynomials. @Raymond Manzoni showed nicely in
this post how the Riemann zeta function is related to the Bernoulli numbers using the Euler-Maclaurin sum. The result is :
\begin{eqnarray}
\zeta(1-k) = -\frac{B_k}{k}.
\end{eqnarray}
(see his equation (4) )
Following a similar process b... | I found an answer:
We will need the following easy to prove (I can expand here if necessary)
relations:
\begin{eqnarray*}
B_{k+1}(x) = \sum_{i=0}^{k+1} B_i
\binom{k+1}{i}
x^{k+1-i} \quad \quad (1),
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{i=0}^{n-1} (i+m)^{k-1} =
\frac{B_{k}(n+m) - B_{k}(m)}{k} \quad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | Hint:
Write your matrix $A$ as $I+N$ where $I$ is the identity matrix and $N$ is a nilpotent matrix. Then use the definition of $e^{At}$ as a power series, noting that $N^k=0$ for some $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 0
} |
Question on Indefinite Integration: $\int\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$ Give me some hints to start with this problem: $${\displaystyle\int}\dfrac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$$
| Because of the cube in denominator, you can say that $$\int \frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,dx=\frac{P_n(x)}{(x^5+x^3+1)^2}$$ where $P_n(x)$ is a polynomial fo degree $n$.
Now differentiate both sides to get $$\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}=\frac{\left(x^5+x^3+1\right) P_n'(x)-2 x^2 \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integral of the function $\frac{\cos ^2 x}{1+\tan x}$ Evaluate $$\int \frac{\cos ^2 x}{1+\tan x}dx$$ I tried converting in double angle and making the derivative of the denominator in the numerator. But, it didn't work out. Some help please. Thanks.
| Another possibility is a phase shift in the argument.
$$\begin{align}\int\frac{\cos^2x}{1+\tan x}dx&=\int\frac{\cos^3x}{\cos x+\sin x}dx=\int\frac{\cos^3x}{\sqrt2\cos\left(x-\frac{\pi}4\right)}dx\\
&=\frac1{\sqrt2}\int\frac{\cos^3\left(u+\frac{\pi}4\right)}{\cos u}du\\
&=\frac1{\sqrt2}\int\frac{\left(\frac1{\sqrt2}\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y... | You can do this if $x,y,z$ are integers, otherwise there are infinitely many solutions.
First suppose that $x \geq y \geq z$, this is possible because of symmetry.
Suppose that $x \geq 26$, then $3^x\geq3^{26}$ and hence $3^x+3^y+3^z>3^{26}$. Hence $x \leq 25$.
If $z \leq 24$, then $3^x+3^y+3^z \leq 3^{25}+3^{25}+3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Eigenvalues of matrices of order $n$ How to find eigenvalues of following matrices of order $n$?
$$A=\begin{bmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & 1 & \cdots & 1 \\
1 & 1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & 0 \... | Using row reduction,
$$
\det A=\begin{vmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & 1 & \cdots & 1 \\
1 & 1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & 0 \\
\end{vmatrix}_n
=\begin{vmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 ... | Note that $y$ is positive, and $$y^2=3+2\sqrt{x(3-x)}$$ It's clear from this that a minimum value for $y$ is $\sqrt{3}$.
Meanwhile, the largest that $x(3-x)$ can get is $\frac32\left(3-\frac32\right)$, since the expression $x(3-x)$ is quadratic with negative leading coefficient, and $x=\frac32$ is the axis of symmetry... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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.