Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× ... | There are a couple of well-known approaches. The first parallels the answer of @egreg and goes back to Gauss. For a prime $p$ find a primitive root $g$ and then for and divisor $d$ of $p-1$ and integer $0\le j<d$ let
$$\sigma_{dj}=\sum_{k=0}^{\frac{p-1}d-1}\omega^{g^{j+kd}}$$
Where $\omega=e^{2\pi i/p}$. In our case $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is... | Let $u=x^2+1\implies x^2=u-1\implies du=2x\,dx$. So $$\begin{align}\int(x^2+1)^7x^3\,dx&=\int (x^2+1)^7\cdot\frac{x^2}2\cdot2x\,dx\\&=\int u^7\cdot\frac{u-1}2\,du=\frac12\int(u^8-u^7)\,du\\&=\frac12\left(\frac{u^9}9-\frac{u^8}8\right)+C\\&=\boxed{\frac1{144}(x^2+1)^8(8x^2-1)}+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begi... | $(\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$
Um... that's not right.
$(\alpha^2-2)^3-3(\alpha^2-2)+1=$
$\alpha^6 +3(-2)\alpha^4 + 3(-2)^2\alpha^2+ (-2)^3 +$
$-3\alpha^2 + 6 + $
$1 =$
$\alpha^6- 6\alpha^4 + 9\alpha^2 -1$
And
$\alpha^6- 6\alpha^4 + 9\alpha^2 -1=$
$\alpha^6 - 3\alpha^4 + \alpha^3 -3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Common tangents of two ellipses I have two ellipses: $x^2/25+y^2/9=1$ and $x^2/16+y^2/25=1$. I have to find the equations of common tangents.
I understand that $xx_0/25+yy_0/9=1$ could be the equation of common tangents, ($x_0,y_0$) being the point of intersection. We will have 4 points of intersection thus 4 tangent... | There exists a line $y = mx + b$
$\frac {x^2}{25} + \frac {(mx+b)^2}{9} = 1\\
(9+ 25m^2) x^2 + 50mbx +25b^2- 225 = 0$
Since the line is tangent
$x = \frac {-50mb \pm \sqrt {(50mb)^2 - 4(9+25m^2)(25b^2 + 225)}}{2(9+25m^2)}$
Since the line is tangent (and not intersecting) the discriminant is $0$
$(50mb)^2 - 4(9+25m^2)(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n... | Also, you can use Stolz: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
$$\lim_{n\rightarrow+\infty}\frac{1^4+2^4+...+n^4}{n^5}=\lim_{n\rightarrow+\infty}\frac{n^4}{n^5-(n-1)^5}=\frac{1}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all the local extrema for $f(x,y)=x^3y^2+27xy+27y$ I have that $f'_x=3x^2y^2+27y$ and $f'_y=2x^3y+27x+27$. Setting $f'_x=0$ I get that
$$x^2=\frac{-27y}{3y^2}=-\frac{9}{y}\Leftrightarrow x=\pm\frac{3}{\sqrt{y}}i$$
Setting $x=\frac{3}{\sqrt{y}}i$ in the equation $f'_y=0$ gives
$$0=2\left(\frac{3}{\sqrt{y}}i\right)^... | Here a way to solve the system $\;\begin{cases}3x^2y^2+27y=0,\\[1ex]2x^3y+27x+27=0.\end{cases}$
The first equation is equivalent to $\;y(x^2y+9)=0$. So:
*
*either $y=0$, and the second equation yields $\;x=-1$;
*or $x^2y=-9$. Plugging this relation into the second equation, multiplied by $xy$ (note that $x$ has to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Origin Triangle Tetrahedron Volume I have a problem that goes like this:
Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?
I really can... | Maybe this will help you to visualize
$OA^2 + OB^2 = 5^2\\
OB^2 + OC^2 = 6^2\\
OA^2 + OC^2 = 7^2$
And even though it looks quadratic, it is really a system of linear equations.
If it feels strange to treat variables with squared terms as linear equations, rewrite it as:
$x + y = 25\\
y + z = 36\\
x + z = 49$
And onc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\f... |
$$\int { \frac { \cos ^{ 4 } x+\sin ^{ 4 } x }{ \sqrt { 1+\cos4x } } dx } =\frac { 1 }{ \sqrt { 2 } } \int { \frac { { \left( \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } \right) }^{ 2 }-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \sqrt { \cos ^{ 2 }{ 2x } } } dx } =\\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1-\frac { \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \... | $$r=sin(3\theta )$$
$$ r=3sin(\theta)-4(sin^3(\theta))$$
$$ r^2=3rsin(\theta)-4rsin(\theta)sin^2(\theta)$$
$$x^2+y^2=3y-4y(y^2/(x^2+y^2))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the ... | \begin{align*}
\lim_{x\rightarrow 0}\dfrac{x^{2}-\log(1+x^{2})}{x^{2}\sin^{2}x}&=\lim_{x\rightarrow 0}\dfrac{x^{2}-x^{2}+\dfrac{1}{2}x^{4}-\dfrac{1}{3}x^{6}\cdots}{x^{2}\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}-\dfrac{1}{3}x^{4}\cdots}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
P.M.F and expected value/expected payout The question is:
Let a random variable X be the number of days that a certain patient needs to be in the hospital. Suppose that X has the p.m.f.
$$\displaystyle f(x) = \frac{5 - x}{10}, \quad x = 1, 2, 3, 4 $$
If the patient is to receive 166 dollars from an insurance compan... | There are a number of minor mistakes.
Part of your computation looks like the computation for "expected number of days."
$$1 \cdot \frac{5-1}{10} + 2 \cdot \frac{5-2}{10} + 3 \cdot \frac{5-3}{10} + 4 \cdot \frac{5-4}{10}.$$
You also seem to be computing "expected payment on the last day in the hospital."
$$166 \cdot \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if div... | It is:
$$2^{5n+1}+5^{n+2}=2\cdot 32^n+25\cdot 5^n=$$
$$2\cdot (27+5)^n+25\cdot 5^n=2\cdot (27^n+\cdots +5^n)+25\cdot 5^n=$$
$$27m+2\cdot 5^n +25\cdot 5^n=27(m+5^n).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
If $f:[a,b]\to\mathbb{R}$ is a function of first class, does it mean that $f$ is continuous everywhere except countably many points in $[a,b]?$ Let $a<b.$
We say that $f:[a,b]\to\mathbb{R}$ is a function of first class if $f$ is a pointwise limit of some sequence $(f_n)_{n=1}^\infty$ of real-valued continuous functions... | Not a direct answer to your question, but this might be of interest: Let $C$ be the Cantor set, let $d(x,C)$ be the distance from $x$ to $C,$ and set $f_n(x) = (1-d(x,C))^n.$ Then each $f_n$ is continuous on $[0,1],$ and $f_n \to \chi_C$ pointwise on $[0,1].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | This is equivalent to multiplying $111111 \times 111111$. There is a principle in logic called "universal generalization". Since no property of the $10$ in $10^k$ the base of $11111$ is being used, because there are no carries, it can be generalized to $x^k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Is $\arccos\left (\frac{\sqrt 6 +1 }{2\sqrt3}\right)= \arctan\left(\frac{\sqrt3 - \sqrt 2}{1+\sqrt 6} \right)$?
Is $\arccos\left(\dfrac{\sqrt 6 +1}{2\sqrt3}\right)= \arctan\left(\dfrac{\sqrt3 - \sqrt 2}{1+\sqrt 6}\right)?$
They are equal according to the calculator but how?
I made a triangle with base $\sqrt 6+ 1$ a... | All you did is fine. Now, notice that$$\left(\sqrt3-\sqrt2\right)^2=3+2-2\sqrt6=5-2\sqrt6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Deriving the closed form of $M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$ I have the sequence, let $M_0=1$
$$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$$
Which I would like first to study the convergence and fine the closed form.
I failed to show that $M_n$ is bound... | Answer thanks to @Masacroso comment.
$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
and
$$\lim_{n\to\infty}M_n = \lim_{n\to\infty} \frac{π}{4} \frac{\tan\left(\frac{π}{2^{n+2}}\right)}{\frac{π}{2^{n+2}}}=\frac{π}{4}\lim_{x\to 0} \frac{\tan\left(x\right)}{x} =\frac{π}{4}$$
One can show that $M_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| Using Lagrange multipliers, write
$$F(x,y,z) = \frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3y^2+1}-\lambda(x+y+z)$$ Taking derivatives, we get $$\frac{6x}{(3x^2+1)^2}=\frac{6y}{(3y^2+1)^2}=\frac{6z}{(3z^2+1)^2}=-\lambda$$
Now it's straightforward to show that $f(t) = \frac{6t}{(3t^2+1)^2}$ is one-to-one on $t>1,$ so $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that a polynomial is irreducible using Galois I need to show that the polynomial $X^4-3X^2+4$ is irreducible in $\mathbb{Q}[X]$. As Eisenstein's criterion fails and I hardly try to avoid something such as “Let's assume $X^4-3X^2+4$ is reducible, then $(X^2+aX+b)(X^2+cX+d) = X^4-3X^2+4$”.
My idea is that its roots ... | We see that our polynomial has no rational roots.
Now, since coefficients before $x^3$ and before $x$ they are $0$, we have two cases only:
$$x^4-3x^2+4=(x^2+px+2)(x^2-px+2)$$ or
$$x^4-3x^2+4=(x^2+px-2)(x^2-px-2)$$ for $p\in\mathbb Z$ and easy to check that they are impossible even for a rational $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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More Questions from Mathematical Analysis by Apostol I was solving the exercise questions of the book "Mathematical Analysis - 2nd Edition" by Tom Apostol and I came across the questions mentioned below. While I was able to solve a few questions, the others I did not even get any hint of!
1.
(a) By equating imaginary p... | With the 2nd question, 2nd part, you are asked to
$$z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right) \Rightarrow \prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$
Note that when $k=n$
$$e^{\frac{2ki\pi}{n}}=e^{2i\pi}=1$$
also
$$z^n-1=(z-1)(z^{n-1}+z^{n-2}+z^{n-3}+...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$ but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind ... | Elliptic integral of the first kind is
$$E_1(\varphi, k) = \int\limits_0^\varphi \dfrac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \int\limits_0^{\sin\varphi}\dfrac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt.$$
Elliptic integral of the second kind is
$$E_2(\varphi, k) = \int\limits_0^\varphi \sqrt{1-k^2\sin^2\theta}\,d\theta = \int\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x... | Notice that the RHS after simplification must have an identical denominator as with the LHS.
The LHS denominator has a cubic term, hence the RHS must also be cubic.
So the choice for the term under $C$ has to be $(x+1)^2$, and not the ones you have suggested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number ... | If $E_n$ is the expected number required when $n$ out of $5$ are already a six then clearly $E_5=0$ and you can write (dropping your final column)
$$\begin{pmatrix}E_0 \\ E_1 \\ E_2 \\ E_3 \\ E_4 \end{pmatrix} =\begin{pmatrix}1 \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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$\lim_{n \rightarrow \infty}\int_0^1f_n(x)$ For $n=1, 2,...,$ let $f_n(x)=\frac{2nx^{n-1}}{x+1}, x\in [0, 1]$. Then $\lim_{n \rightarrow \infty}\int_0^1f_n(x)$
Function is unbounded at $1$, How do I solve?
| \begin{align*}
\int_{0}^{1}f_{n}(x)dx&=\int_{0}^{1}\dfrac{2nx^{n-1}}{x+1}dx\\
&=\dfrac{2}{x+1}\cdot x^{n}\bigg|_{x=0}^{x=1}-\int_{0}^{1}-\dfrac{2}{(x+1)^{2}}\cdot x^{n}dx\\
&=1+2\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx,
\end{align*}
now
\begin{align*}
\dfrac{x^{n}}{(x+1)^{2}}\leq\dfrac{1}{(x+1)^{2}},~~~~x\in[0,1],
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$.
Here is my solution I am looking to see if it is correct or if there is another way to do it, thanks!
The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$ . Since there are ${... | Expressions like above can be evaluated by the multinomial Theorem:
$(a+b+c)^n = \sum_{k,l,m;k+l+m=n} \frac{n!}{k!l!m!}a^kb^lc^m$.
This sum means you sum over all possible $k,l,m$ with the condition that $k+l+m=n$.
Now you set $n=10,a=1,b=x^3,c=x^8$ and you obtain now that the $x^{25}$ Terms are obtained when $3l+8m=25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{... | Durgesh: to complement Jack D'Aurizio fantastic answer above.
If you use Euler's identity
$$e^{ix} = cos x + i\cdot sin x$$
and then apply for $-x$
$$e^{-ix} = cos (-x) + i\cdot sin(-x) = cos x - i\cdot sinx$$
and sum both equations, you get the well known
$$cos x={e^{ix}+e^{-ix}\over2}$$
so
$$\frac12+cos x={1+e^{ix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
How to simplify the trigonometric term?
Simplify $$\frac{2\tan\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)-1\right)^2} + \frac{\tan^2\left(\frac{x}{2}\right)+1}{2\tan\left(\frac{x}{2}\right)}$$
It should be like: $$\frac{2\sin(x) - 2\sin^2(x) - 2}{2(\sin(x)-1)\sin(x)}$$
I can use any trigonometric ide... | Let $y=\dfrac x2$.
Then using the identities $$\sec^2y=1+\tan^2y\tag{1}$$ $$\sin 2y=2\sin y\cos y\tag{2}$$ the expression becomes $$\begin{align}\frac{2\tan y}{(\tan y-1)^2}+\frac{\tan^2y+1}{2\tan y}&=\frac{2\tan y}{\tan^2y+1-2\tan y}+\frac{\sec^2y}{2\tan y}\tag{1}\\&=\frac{\frac{2\sin y}{\cos y}}{\frac1{\cos^2y}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1... | Let $m=2018n+1$ then $\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{m^4}$
=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{m^4}\frac{(m-1)^4}{(m-1)^4}}$
=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{(m-1)^4}{\frac{1}{(1-\frac{1}{m})^4}}}$
Known that $\lim _{m\to \infty }\big(1-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4... | Firstly, $z=x+yi$, where $x$ and $y$ are reals.
Thus, it should be $$|x^2-y^2|=\frac{1}{2}$$ and $$x^2+y^2=1.$$
Finally, we obtain $$\left\{\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,\pm\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2... | write it as $$\frac{n^2+\sqrt{n^4+4n^2+n}}{n^4-n^4-4n^2-n}$$ and this is $$\frac{n^2(1+\sqrt{1+1/n^2+1/n^3})}{n^2(-4-1/n^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)... | The roots of $\Phi_{10}(x)=1-x+x^2-x^3+x^4$ are the primitive tenth roots of unity, $\xi,\xi^3,\xi^7,\xi^9$ with $\xi=\exp\left(\frac{2\pi i}{10}\right)$. The roots of $\Phi_{20}(x)=\Phi_{10}(x^2)=1-x^2+x^4-x^6+x^8$ are the primitive $20$-th roots of unity, $\zeta,\zeta^3,\zeta^7,\zeta^9,\zeta^{11},\zeta^{13},\zeta^{17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\fr... | $$\lim_{x\rightarrow0}\frac{3^x+2^x-2}{4^x+2^x-2}=\lim_{x\rightarrow0}\frac{\frac{3^x-1}{x}+\frac{2^x-1}{x}}{\frac{4^x-1}{x}+\frac{2^x-1}{x}}=\frac{\ln3+\ln2}{\ln4+\ln2}=\log_86$$
I used $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\lim_{x\rightarrow0}\left(\frac{e^{x\ln{a}}-1}{x\ln{a}}\cdot\ln{a}\right)=\ln{a}$$ for all $a>0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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decomposing and the summing a sum.
Prove:
$$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) \quad \text{ where } \{x\} = x-\lfloor x \rfloor $$
We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n} n -\sum_{k\leq n} k \l... | We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n} n -\sum_{k\leq n} k \lfloor \frac {n}{k} \rfloor $$
$$= n^2 - \sum_{k \leq n} \sigma (k) $$
Since $ \sum_{k \leq n} \sigma (k)= (\frac{\pi^2}{12}n^2 + O(n \log n) )$
Hence we have that $$= n^2 - \sum_{k \leq n} \sigma (k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\m... | Assume $s>0$ and notice that for $s\ne\sqrt3$
$$\frac12\left(s+\frac3s\right)>\sqrt3$$ as can be established by computing the minium.
So for $n>1$, we have
$$s_n>\sqrt3\implies s_n^2>3\implies s_n>\frac3{s_n}\implies s_n>\frac12\left(s_n+\frac3{s_n}\right)=s_{n+1}.$$
At the same time this establishes that the sequence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+2 \cos2m\theta$ I am struggling with a part of my textbook can anyone help me? It states:
$$ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+ 2\cos2m\theta$$
Why is this obvious? Do I have to perfo... | You can use the standard formula:
$$1+\cos x+\cos 2x+\dots +\cos mx=\frac{\sin\dfrac{(m+1)x}2}{\sin\dfrac{x}2}\,\cos\dfrac{mx}2.$$
Here, with $x=2\theta$, one obtains
\begin{align}
1 +{} & 2 \cos2\theta + \dots+2 \cos2m\theta =2(1 + \cos2\theta + \dots+\cos2m\theta)-1 \cr
={} &\frac{2\sin(m+1)\theta}{\sin\theta}\,\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is a proof using modular arithmetic in a question like this valid? It's been two years or so since I've finished my math undergrad (and I'm doing something non-math related now, unfortunately), so I apologize if what is to follow isn't a very good question!
Prove that for all Integers $n$, $n(n + 1)(2n + 1)$ will alway... | It's correct, but you can go further.
Fermat’s little theorem
If $p$ is a prime number, then, for every integer $n$,
$$n^p\equiv n \pmod{p}$$
There are several proofs. A simple one considers the fact that, if $n$ is coprime with $p$, then $n,2n,3n,\dots,(p-1)n$ are all distinct modulo $p$ and not congruent to $0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 0
} |
$\sin x$ and $\cos x$ - linearly independent There is a lot of posts about that subject, but I'm not sure.
Task:
Check if
$$f(x)=\sin(x) $$
and
$$g(x)=\cos(x) $$
are linearly independent in the space of functions.
That's what I have done:
$a \sin(x) + b \cos(x) = 0$,
so $a=0$ and $b=0$.
for $x=0 $,
$a \sin(0) ... | The Wronskian of $\sin$ and $\cos$ is not zero:
$$W(\sin, \cos)(x) = \begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x\end{vmatrix} = -\sin^2 x - \cos^2 x = -1 \ne 0$$
Therefore $\{\sin, \cos\}$ is linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the mistake in this? What is wrong in this proof?
In $\triangle ABC$ with right angle at $B$, $BD$ is drawn as an altitude on $AC$. Let $AB=a$ and $BC=b$. So by similarity, $AD=\frac{a^2}{\sqrt{a^2+b^2}}$ and similarly $CD=\frac{b^2}{\sqrt{a^2+b^2}}$. Now, $\triangle ABD,BDC$ are also right angled at $D$. So a... | $$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$ is obviously wrong. Are you assuming a=b?
Also when using Pythagorean Theorem you forgot to square AD and CD.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | So you have done some good work. Now the function is always positive for $x \not \in (0,1)$. Now lets analyse in this interval.
In $(0,1)$, $x^2+1$ lies in $(1,2)$ and $x(x-1)$ has a minimum of only $-1/4$ at $x=1/2$. Thats enough to conclude that $x(x-1)(x^2+1) > -1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
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Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
| $$e^{ix} = i\sin{x}+\cos{x}$$
$$e^{4ix} = (i\sin{x}+\cos{x})^4$$
$$i\sin{4x}+\cos{4x} = \sin^4{x}-4i\sin^3x\cos{x}-6\sin^2x\cos^2{x}+4i\sin x\cos^3{x} + \cos^4x$$
As all coefficients are real, we can say the imaginary parts of each side are equal:
$$i\sin{4x} = -4i\sin^3x\cos{x}+4i\sin x\cos^3{x}$$
$$\sin{4x} = 4\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Polynomial Doubt Question :
$x^4 + px^3 + qx^2 + px + 1 =0 $ has real roots. Then what is the minimum value of $ p^2 +q^2 $ .
How I started ?
I started by dividing the whole equation by $x^2$ then we get
$ (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0 $
Then put $(x + \frac{1}{x} ) = t$. Then discriminant... | I don't know how much this is going to help, but if you use this you can find the roots
\begin{eqnarray}
x_1&=& -\frac{1}{4} \sqrt{p^2-4 q+8}-\frac{1}{2}
\sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4
q+8}}-q-2}-\frac{p}{4},\\
x_2&=& -\frac{1}{4} \sqrt{p^2-4
q+8}+\frac{1}{2} \sqrt{\frac{p^2}{2}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving three alike problems $a_{n+1} = \frac{1+a_n^2}{1+a_n}$ I have two problems alike:
*
*If $a_1=a$, $a>0$, $a_{n+1} = \dfrac{1+a_n^2}{1+a_n}$, study its convergence.
*If $a_0 >0$, $a_1>0$ and $a_{n+2} = \dfrac{2+a_{n+1}}{2+a_n}$, study this one's convergence.
I can solve this using two different situations: ... | Hint with the 2nd one, it is easy to show by induction that $a_n>0$, then
$$\color{red}{0<a_{n+2}}=\frac{2+a_{n+1}}{2+a_{n}}<
\frac{2+a_{n+1}}{2}=
1+\frac{a_{n+1}}{2}<
1+\frac{1+\frac{a_{n}}{2}}{2}=
1+\frac{1}{2}+\frac{a_n}{2^2}<\\
1+\frac{1}{2}+\frac{1}{2^2}+\frac{a_{n-1}}{2^3}<...<\\
1+\frac{1}{2}+\frac{1}{2^2}+...+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Powers for the sum of three squares Say $n$ can be represented as a sum of three non-zero squares. (i.e. $n = a^2 +b^2+ c^2$, for some $n,a,b,c \in \mathbb{N}$)
Is it possible that every natural power of $n$ is also a sum of three non-zero
squares? (i.e. $n^k = x^2+y^2+z^2$ for $x,y,z,k \in \mathbb{N}$)
| Yes, more generally, if,
$$N = x_1^2+x_2^2+\dots+x_m^2$$
then one can always find integer $y_i$ such that,
$$N^k = y_1^2+y_2^2+\dots+y_m^2$$
Proof: Use the expansion of
$$(a+i\sqrt{w})^k$$
To get,
$$a^2 + w = N,\\ (a^2 - w)^2 + (2 a)^2 w = (a^2 +
w)^2,\\ (a^3 - 3 a w)^2 + (3 a^2 - w)^2 w = (a^2 +
w)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) ... | $$\frac{n(n+5)+4}{4(n+4)(n+5)}=\frac{n^2+5n+4}{4(n+4)(n+5)}=\frac{(n+4)(n+1)}{4(n+4)(n+5)}=\frac{n+1}{4(n+5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2... | $\displaystyle \sum_{k=1}^nk^2(2k-1)=2\sum_{k=1}^nk^3-\sum_{k=1}^nk^2=2\times\frac{1}{4}n^2(n+1)^2-\frac{1}{6}n(n+1)(2n+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
elimination part of a mathematical question During solving a math question I got stuck in the elimination part and I couldn't proceed! Can someone help me, please?
$$2(x+y)-3^\frac 1{2}(x-y)=0$$
$$4(x+y)+2(3)^{\frac 1{2}}(x-y)+9z=1$$
$$x+y+z=0$$
I want to get the values for $x$, $y$, and $z$.
| I would like to change the variables here.
let:
$u = x+y\\
v = \frac {\sqrt 3}{2} (x-y)$
then
$2u-2v = 0\\
4u + 4v + 9z = 1\\
u+z = 0$
$u=-1, v = -1, z = 1$
Now back to $x,y$
$x+y = u\\
x-y = \frac {2}{\sqrt 3}v$
$x+y = -1\\
x-y = -\frac {2}{\sqrt 3}$
$2x = -1-\frac {2}{\sqrt 3}\\
2y = -1 + \frac {2}{\sqrt 3}\\
x = -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Since$$x^4+5x^2+3=(x^2+1)^2+3(x^2+1)-1,$$ $P(x)=x^2+3x-1$ and therefore $P(x^2-1)=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Show that $\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right).$
Show that $$\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right),$$
where $d(n)$ is the number of divisors of $n$ and $E$ is Euler-Mascheroni constant.
I f... | If we write $D(x)=\sum_{n\leq x}d(n)$ then by the Dirichlet hyperbola method we have:
$$D(x)=x\text{log}(x)+x(2\gamma -1)+\mathcal{O}(x^{1/2})$$
Moreover applying Abel's summation formula to your partial sum gives us:
$$\sum_{2\leq n\leq x}\frac{d(n)}{\text{log}(n)}=\frac{D(x)}{\text{log}(x)}+\int_{2}^x\frac{D(t)}{t\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the integral of $\int\frac{dx}{\sqrt{x^3+a^3}}$?
What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$
I came across this integration in a physics problem. I suspect role of complex numbers here.
'$a$' is a constant
| As the other users said, this integral is unlikely to have an elementary form, however, it can be expressed in terms of the well known Gauss hypergeometric function, which can be easily evaluated by most CAS or even Wolfram Alpha.
First, let's consider the case $|x|<|a|$, then we can substitute:
$$x=at, \qquad |t|<1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$ Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$. We can rewrite n-th term as $a_n=\frac{(n-1)(n-3)\cdot...\cdot1}{n(n-2)\cdot..\cdot2}z^n=\frac{(2n)!}{2^{2n}\cdot(n!)^2}z^n$. But I don't know what should I do now. Cons... | Another approach (same origin as the other answer, but starting from a well-known formula): we have, for a real $z$ s. t. $|z|<1$,
$$\arcsin z=z+\frac12\frac{z^3}3+\frac{1\cdot 3}{2\cdot 4}\frac{z^5}5+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{z^7}7+\dotsm, $$
whence
$$(\arcsin z)'=\frac1{\sqrt{1-z^2\strut}}=1+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Approximate $\sqrt{(1.02)^3+(1.97)^3}$ using differential So $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
So the differential will be $$\sqrt{1^3+2^3}+\frac{3(1)^2}{2\sqrt{1^3+2^3}}\Delta x+\frac{3(2)^2}{2\sqrt{1^3+2^3}}\Delta y=3+\frac{1}{2}\Delta x+2\Delta y$$
is $\Delta x=0.02 \text{ and } \Delta y=-0.0... | Note that $$\sqrt{1.02^3+1.97^3}=\sqrt{(x+\Delta x)^3+(y+\Delta y)^3}$$ where $\Delta x=0.02$ and $\Delta y= -0.03$ with $x=1$ and $y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A question about how to factorise a binomial. If $x=\frac{-1}{3}$
$(x+1)^3-{3(x+1)}^2+3(x+1)$ =?
Using the binomial theorem coefficients I'm going to add 1 at the end
[$(a-b)^3=a^3-3a^2b+3ab^2-b^3]
So we now have
$(x+1)^3-{3(x+1)}^2+3(x+1)+1$
How do I continue from here?
| Note that by
$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$
with
*
*$a=x+1$
*$b=-1$
we have
$$(x+1)^3-{3(x+1)}^2+3(x+1)=(x+1)^3-{3(x+1)}^2+3(x+1)-1+1=[(x+1)-1)]^3+1=x^3+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $? $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=\sum_{n=1}^\infty\left(\frac{2n+1}{n^2}\frac1{{(n+1)}^2}\right)$$
I assume that I should get a telescoping sum in some way, but I'm couldn't find it yet.
| Since
\begin{align*}
\frac{2n+1}{n^4+2n^3+n^2}&=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\\
&=\frac 1{n^2}-\frac 1{(n+1)^2}
\end{align*}
then we have, for an integer $N>1$
$$\sum_{n=1}^N\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=1-\frac1{(N+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find Jordan basis of a matrix Assume matrix
$$A=
\begin{bmatrix}
-1&0&0&0&0\\
-1&1&-2&0&1\\
-1&0&-1&0&1\\
0&1&-1&1&0\\
0&0&0&0&-1
\end{bmatrix}
$$
Its Jordan Canonical Form is
$$J=
\begin{bmatrix}
-1&1&0&0&0\\
0&-1&0&0&0\\
0&0&-1&0&0\\
0&0&0&1&1\\
0&0&0&0&1
\end{bmatrix}
$$
I am trying to find a nonsingular $P$... | I got the Jordan blocks in slightly different order.
What you seem to be missing is the consistency part: in my
$$ P =
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0
\end{array}
\right)
$$
we have a place where we do have $(A+I)^2 p_3 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
| Looking at the equation mod $7$ we see that $2^x = 1 \pmod 7$ which means $3 \mid x$ by Lagrange's theorem. Looking at the equation mod $3$ we see that $(-1)^x = 1 \pmod 3$ so $x$ is even.
Hence there is a positive integer $k$ such that $x = 6k$.
Rewrite the equation as $2^{6k} = 3^y7^z + 1$, which can also be written ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s method
Ferrari’s method for solving a quartic equation
$$x^4-15x^2-10x+ 24 = 0$$
begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of
the form:$$-2bx^2+b^2$$to both sides.
(a) Explain why this is good idea and what it accomplishes.
(b) Use $b= 7$ to find t... | $$(x^2-7)^2=x^2+ 10x+25\implies (x^2-7)^2-(x+5)^2=0$$
$$(x^2-x-12)(x^2+x-2)=0$$
Now solve each quadratic equation for $ x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let a and b be such that $-2A^2 + 4A -3I_2 = aA + bI_2$. To find $a + b$. $A \in M_2$ with characteristic polynomial $p(x) = x^2 -3x - 5$. Let a and b be such that $-2A^2 + 4A -3I_2 = aA + bI_2$. To find $a + b$.
| A matrix satisfies its characteristic polynomial.
$$-2A^2+4A-3I=aA+bI$$
$$-2A^2+(4-a)A+(-3-b)I=0$$
$$-2\left(A^2+\frac{a-4}{2}A+\frac{3+b}{2}I\right)=0$$
$$A^2+\frac{a-4}{2}A+\frac{3+b}{2}I=0$$
From this we need:
$$\frac{a-4}{2}=-3$$
$$a-4=-6$$
$$a=-2$$
and
$$\frac{3+b}{2}=-5$$
$$3+b=-10$$
$$b=-13$$
So $a+b=-15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Quadratic equation formula help / simplification I have this quadratic equation,
$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $
I use the quadratic formula to solve and simplify
$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$
my book says it should simplify to
$ \frac{1}{3} ( -5 \pm \sqrt{... | Your solution seems correct, indeed
$$x^{2} + \frac{10}{3}x -\frac{80}{3} = 0\iff3x^2+10x-80=0$$
$$ \frac{-10 \pm\sqrt{100-(4)(3)(-80)}}{6}=-\frac53\pm\frac{\sqrt{265}}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Power Representation of A Series I am asked to find the power representation of the series $ f(x)=xln(2+3x)+1/(2-x)^2 $ about a = 0. I took the power series of $ xln(2+3x) $ and $ 1/(2-x)^2 $ separately and added them together to give me: $ xln(2)+\sum_1^\infty((-1)^{n-1}3^nx^{n+1}) /n2^n + 1/4\sum_1^\infty(n-1)(x/2)^{... | Hint: The power series expansion of $\frac{1}{(2-x)^2}$ at $x=0$ is
\begin{align*}
\frac{1}{(2-x)^2}&=\frac{1}{4}\cdot\frac{1}{\left(1-\frac{x}{2}\right)^2}\\
&=\frac{1}{4}\sum_{n=0}^\infty\binom{-2}{n}\left(-\frac{x}{2}\right)^n
=\frac{1}{4}\sum_{n=0}^\infty\binom{n+1}{n}\left(\frac{x}{2}\right)^n\\
&=\frac{1}{4}\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Inequality $\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$ Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\l... | The presented proof relies alone on the Cauchy–Bunyakovsky–Schwarz inequality (CBS).
We exploit equivalent formulations of the given constraint
\begin{align*}
& \sum_{\text{cyc}}{1\over a+1} =1 \tag{1} \\[1ex]
\iff\quad & \sum_{\text{cyc}}{a\over a+1}=2 \tag{2} \\[1ex]
\iff\quad & abc = a+b+c+2 \tag{3}
\end{align*}
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
n-th element of recurrence relation I want to find formula for n-th element of recursively entered sequence using generating functions. It goes like this
$$a_{1} = 0, a_{2} = 1, a_{n+2} = a_{n+1} + a_{n} + 2$$
I rewrite this into generating functions like this:
$$a(x) = x(a(x)) + x^{2}(a(x)) + \frac{2}{1-x}-x-2$$
The... | As it almost always happens, the problem is in being very precise with the indexes. Start with
$$f(x)=\sum\limits_{n=1}a_nx^{n-1}=a_1+a_2x+\sum\limits_{n=3}a_nx^{n-1}=\\
x+\sum\limits_{n=3}(a_{n-1}+a_{n-2}+2)x^{n-1}=
x+\sum\limits_{n=3}a_{n-1}x^{n-1}+\sum\limits_{n=3}a_{n-2}x^{n-1}+2\sum\limits_{n=3}x^{n-1}=\\
x+\sum\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Arrange 6 adults and 12 children in 5 rooms, with at least 1 adult in each room Problem: How many ways can you arrange 6 adults and 12 children in 5 rooms of max 4 people such that there is at least 1 adult per room. Every person and room is distinguishable.
My take from the problem is the following:
*
*There are 2... | You could drive your consideration under 1. even further: you would have the following patterns (using a = adult, c = child):
$A = \{aacc, accc, accc, accc, ac\}$,
$B = \{accc, accc, accc, accc, aa\}$,
$C = \{aacc, accc, accc, acc, acc\}$,
$D = \{accc, accc, accc, aac, acc\}$.
Now let us consider first the rooms to be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x... | $$\sqrt {x^2-3}=x-3$$
Since, by definition, $\sqrt {x^2-3} \ge 0$, we must have $x-3\ge 0$. That is we must have $x \ge 3$.
If $x \ge 3$, then $x^2-3 \ge 6$. So there will be no complications caused by squaring both sides of the equation.
\begin{align}
x^2-3 &= x^2-6x + 9 \\
6x &= 12 \\
x &= 2
\end{align}
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Integral of $\sqrt{x-\sqrt{x^2+1}}dx$ On Wolfram Alpha's website, I find
$$\int\sqrt{x-\sqrt{x^2+1}}dx=\frac23 \sqrt{x - \sqrt{1 + x^2}}( {2 x + \sqrt{1 + x^2}})+C$$
I am trying to get this answer by substitution by $x=\tan(\theta), y=\sqrt{x-\sqrt{x^2+1}},$ or $y={x-\sqrt{x^2+1}}$. But the integrated converted to mo... | Leonardo says this will not be easy to solve. With
$$
t=\sqrt{x-\sqrt{x^2+1}}
$$
I get
$$
\int \sqrt{x-\sqrt{x^2+1}} \;dx =
-2\int\sqrt{\frac{(t^4-1)^2}{4t^2}+1}\;dt
=-\int \left(t^2+\frac{1}{t^2}\right)\;dt
$$
and that integral is easy.
$$\begin{align*}
-\int \left(t^2+\frac{1}{t^2}\right)\;dt &= \frac{-t^3}{3}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing a matrix inequality Can someone help me with showing the following
Let $X$, $B$, and $C$ all be positive definite matrices. Show that the following inequality is true:
$$ (X + C)^{-1} (X B X' + C ) ( X + C)^{-1} \ge (B^{-1} + C)^{-1}$$
for all $X$, where $A \ge B$ means $x' A x \ge x' B x$ for all column vect... | Put $D = B^{1/2}CB^{1/2}$ and $Y = B^{1/2}XB^{1/2}$, the inequality becomes
\begin{align}
(X+C)^{-1}(XBX+C)(X+C)^{-1} &\ge (B^{-1}+C)^{-1},\\
B^{1/2}(Y+D)^{-1}(Y^2+D)(Y+D)^{-1}B^{1/2} &\ge (B^{-1}+B^{-1/2}D^{-1}B^{-1/2})^{-1},\\
(Y+D)^{-1}(Y^2+D)(Y+D)^{-1} &\ge (I+D^{-1})^{-1},\\
(Y^2 + D) &\ge (Y+D)(I+D)^{-1}(Y+D).
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is
Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So... | $a^2+ab+b^2=9 \implies a^2 = 9 - ab - b^2 $
$b^2+bc+c^2=52 \implies b^2 = 52 -bc - c^2$
$c^2+ac+a^2=49 \implies c^2 = 49 - ac - a^2$
Substituting $a^2$ in $(iii)$
$(c-b)(a+b+c) = 40 --(iv)$
Similarly,
$(b-a)(a+b+c) = 3 -- (v)$
$(c-a)(a+b+c)= 43 -- (vi)$
Now adding $(i),(ii),(iii)$
$2(a^2 + b^2 + c^2) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I ... | Given that $b^2-4c<0$,
$$x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge-\frac{b^2}{4}+c=-\frac{1}{4}\left(b^2-4c\right)>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
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Expectation and variance of 4th generation Let $z_1,z_2,...,z_n$ describe a branching process in which each parent has j offspring with probability $p_j$, for
$ p_0 = \frac{1}{12}, p_1 = \frac{1}{6}, p_2 = \frac{3}{4}$.
Find the expectation and variance of the $4$th generation.
$H(s) = \frac{1}{12} + \frac{1}{6} s + ... | Let $$P(s) = \sum_{k=0}^\infty \mathbb P(Z_1=k)s^k=\frac1{12}+\frac16s+\frac34s^2$$ be the generating function of the offspring distribution. Then
$$
m:=\mathbb E[Z_1] = \lim_{s\uparrow 1}P'(s) = \frac53
$$
and
$$
\mathbb E[Z_1(Z_1-1)] =\lim_{s\uparrow 1}P''(s)=\frac32,
$$
so
$$
\sigma^2:= \mathsf{Var}(Z_1) = P''(1) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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GCD of the elements of a set
Let $S$ be the set of numbers of the form
$n(n + 1)(n + 2)(n + 3)(n + 4),$
where $n$ is any positive integer. The first few terms of $S$ are
\begin{align*}
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\
3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2... | For $n\ge 1$, the number $$\frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ is just the binomial coefficient $\binom{n+4}{5}$ which is always an integer. Hence all the numbers are divisible by $120$.
Because $120$ is the smallest , the $\gcd$ of all the numbers is $120$
| {
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"url": "https://math.stackexchange.com/questions/2711456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the method of "changing variables" to solve the following recurrence: $T(n) = 2\cdot T(n-2) + n$
I tried doing this, but I don't know how to continue from here. I think it doesn't work and is not correct.
Please help, thanks in advance!
| Let $f(x)=T_0+T_1x+T_2x^2+....+T_nx^n+...$. Thus, $x^2f=T_0x^2+T_1x^3+T_2x^4+...+T_{n-2}x^n+....$. Also, note that $\frac{x}{(1-x)^2}=x+2x^2+3x^3+...nx^n+...$
Thus,
$$f-2x^2f-\frac{x}{(1-x)^2}=T_0+(T_1-1)x+(T_2-2T_0-2)x^2+...+(T_n-2T_{n-2}-n)x^n+...$$
$$=T_0+(T_1-1)x$$
Thus,
$$f=\frac{T_0}{1-2x^2}+\frac{x}{(1-2x^2)(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$? Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ ... | $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\left.\frac12\log\frac{1+x}{1-x}\right|_{\sin\left(-\frac{5\pi}{12}\right)}^{\sin\left(\frac{5\pi}{12}\right)}=\frac12\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}-\frac12\log\frac{1+\sin\left(-\frac{5\pi}{12}\right)}{1-\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION
Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION.
Vowels are IIIEEAO. Conson... | Another approach worth mentioning is that of exponential generating functions (egfs). Let $x$ enumerate vowels, $y$ enumerate consonants and $z$ enumerate total letters. Then the egf factors for vowels break down as follows
III
$$1+xz+\frac{1}{2!}(xz)^2+\frac{1}{3!}(xz)^3$$
EE
$$1+xz+\frac{1}{2!}(xz)^2$$
A, O
$$1+xz$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4... | You seem to be going in the right direction. There is a mistake somewhere, but your algebra is a bit convoluted, so I cannot find it exactly. Here is what I got following your procedure:
Solve $4x^2+9y^2=3600$ for $y$ to get $$y=\frac23\sqrt{(30-x)(x+30)}.$$ Then, we want to maximize $$A=4xy=\frac{8x}{3}\sqrt{(30-x)(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+... |
Given $\;x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}\,$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+\color{red}{3}$
That's actually a good approach, and it does in fact work out nicely, but:
*
*in order to reverse the fractions, you have to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| \begin{align*}
\sum_{k=1}^\infty\frac{1}{k^3}&<\sum_{k=2}^\infty\frac{1}{(k-1)(k)(k+1)}+1\\
&=\frac{1}{2}\sum_{k=2}^\infty\left[\frac{1}{k-1}-\frac{2}{k}+\frac{1}{k+1}\right]+1\\
&=\frac{1}{2}\left(1-\frac{1}{2}\right)+1\\
&=\frac{5}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
The difference between the radii of the largest and smallest circles having centres on the circumference of $x^2+2x+y^2+4y=4$ Owing to restriction of 150 characters in the title section I include the latter part of the problem here below in bold and italics
Also given that both the circles(largest and smallest) pass th... | Any point on the circle $$(x+1)^2+(y+2)^2=3^2$$ can be $P(3\cos t-1,3\sin t -2)$
So, if $r$ is the radius of the new circle $$r^2=(a+1-3\cos t)^2+(b+2-3\sin t)^2$$
$$=(a+1)^2+(b+2)^2+9-6\{(b+2)\sin t+(a+1)\cos t\}$$
$$=(a+1)^2+(b+2)^2+9-6\sqrt{(b+2)^2+(a+1)^2}\sin\left(t+\arctan\dfrac{a+1}{b+2}\right)$$
$$r\ge|\sqrt{(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Is there a closed form of this sequence? The sequence is $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
and I need to find n th..
Here's my approach..
I bound them with who has same s... | The $n$-th term can be expressed as ${\large{\frac{a}{b}}}$, where
\begin{align*}
a&=n-\frac{k(k+1)}{2}\\[4pt]
b&=k+2-a\\[4pt]
k&=
\left\lfloor
{
\frac
{
\left\lfloor
\sqrt{8n-7}
\right\rfloor
-1
}
{2}
}
\right\rfloor\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
sum (difference) of polynomials to the power n Is there a way to simplify the expression:
$D = (f_1(\omega)+f_2(\omega))^n-(f_1(\omega)-f_2(\omega))^n$
where $n$ is a positive integer.
In this particular problem:
$f_1(\omega)=-\omega^2+2$
$f_2(\omega)=\omega \sqrt{\omega^2-4}$
Expanding $D$ for some values of $n$:
$n=1... | By the binomial theorem
$$
(f_1+f_2)^n -(f_1-f_2)^n = \sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}f_2^k -\sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}(-f_2)^k.
$$
Notice that since we have $(-f_2)^k$ the second sum is alternating, so every other term cancels, leaving
$$
2\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2j+1}f_1^{n-(2j+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int x \, \tan^{-1} (x) \, dx$ Integrate $\int x \, \tan^{-1} (x) \, dx$
My Attempt:
\begin{align}
\int x \, \tan^{-1} (x) \, dx &= \int x \, \tan^{-1} (x) \, dx \\
&= \dfrac {x^2}{2} \tan^{-1} (x) - \int \dfrac {x^2}{x^2+1} \, dx.
\end{align}
| Write the integrand as $\dfrac{x^2+1-1}{x^2+1}$ and rewrite it as $1-\dfrac{1}{x^2+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$ I have encountered this integral and I am stuck evaluating it:$I=\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$
My try was to expand the numerator into power series, indeed: $$x\cos^2x=\frac{x}{2}(1+\cos(2x)) =\frac{x}{2} +\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1} x^{2n... | Hint. Another way is to use the residue theorem. Since, for a suitable function, we have $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ we get $$\sum_{n\geq1}\frac{1}{2n^{2}}+\sum_{n\geq1}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Ho... | $$\frac{1}{x^2}-\frac{1}{\sin^2(x)}=\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$
You have that $\sin(x)=x-\frac{x^3}{3!}+o(x^3)$,
and thus $$\sin^2(x)=x^2-\frac{x^4}{3}+o(x^4).$$
Finally,
$$\frac{1}{x^2}-\frac{1}{\sin^2(x)}=\frac{-x^4+o(x^4)}{3x^2\sin^2(x)}\underset{x\to 0}{\longrightarrow }-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$
Given the limit:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \alpha$$
Find the value of $\alpha$
I know the series does not converge (it is equivalent to the harmonic series. Correct me, please, if I am wrong).
D... | Note that
$$\frac{2k}{k^2-2k-3}=\frac{2k}{(k+1)(k-3)}= \frac{1}{2}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)$$
So, with $H_n = \sum_{k=1}^{n}\frac{1}{n}$
$$\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \frac{1}{2}\sum_{k=4}^{n}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)\leq \frac{1}{2}H_{n+1} + \frac{3}{2}H_{n+1} = 2H_{n+1}\leq 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$ Find Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$
My try:
First of all $x$ belongs to First quadrant
Also $$\sqrt{\sin x}+\sqrt{\cos x}-1 \ge 0$$ Squaring both sides we get
$$\sin x+\cos x+2\sqrt{\sin x\cos x} \ge 1$$
Any clue here?
| For
$f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$,
you first of all have to
restrict $x$ to the reals.
Then,
since $\sin$ and $\cos$
are periodic
with period $2\pi$,
you can restrict $x \in [0, 2 \pi)$.
Any restrictions there
are replicated in
$[2k\pi, 2(k+1)\pi)$.
Since you need
$\sin(x) \ge 0$,
this gives
$x \in [0, \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with r... | $$\sum_{n\ge 1}\left(\frac1n - \frac1{2n-1} - \frac1{2n+1}\right) = \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n-1}\right) + \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n+1}\right) = \sum_{n \ge 1} \frac{(-1)^n}{n} + \sum_{n\ge 2}\frac{(-1)^n}{n} = 2\sum_{n\ge 1} \frac{(-1)^n}{n} + 1 = -2\log(1-(-1)) + 1 = 1 - 2\log2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to properly find a critical point for $xy^2$? Let $f(x,y) = xy^2$ and the domain $D = \lbrace (x,y)| x,y\geq0, x^2 + y^2 \leq 3 \rbrace$
$f_x(x,y) = y^2$ and $f_y(x,y) = 2xy$
Therefore, the critical points should be $\lbrace (x,y)| y = 0, \sqrt{3} \geq x \geq 0 \rbrace$.
The determinant of the Hessian is
$$\det(... | On the circle of radius $r$, we have $x^2+y^2=r^2$. Therefore,
$$
xy^2=r^2x-x^3\tag1
$$
This implies that the interior critical points are at
$$
(x,y)=\frac r{\sqrt3}\left(\pm1,\pm\sqrt2\right)\tag2
$$
with the corresponding values of
$$
xy^2=\pm r^3\frac2{3\sqrt3}\tag3
$$
At the endponts of $x=\pm r$, we get the value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many $m$ such that : $\sum\limits_{k=1}^m \left\lfloor\frac{m}{k}\right\rfloor$ be even? Find how many $m \le 1000$ such that :
$$\sum\limits_{k=1}^m \left\lfloor \frac{m}{k}\right\rfloor$$
be even ( $\lfloor x\rfloor$ is the largest integer smaller than $x$.)
I think that one case is $\sum\limits_{k=1}^m \left\l... | Hint, further to the comments
$$\sum\limits_{k=1}^{m} \left \lfloor \frac{m}{k} \right \rfloor=\sum\limits_{k=1}^{m}d(k) \tag{1}$$
*
*$m=1 \Rightarrow \sum\limits_{k=1}^{1} \left \lfloor \frac{1}{k} \right \rfloor=\sum\limits_{k=1}^{1}d(k)=\color{blue}{1}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Inequality with Sum of Binomial Coefficients My friend noticed that for $n>12$, we have the following pattern in the binomial coefficients.
$$\sum_{i=0}^{\lfloor n/3 \rfloor}\binom{n}{i} < \binom{n}{\lfloor n/3 \rfloor + 1}$$
$$\sum_{i=0}^{\lfloor n/3 \rfloor + 1}\binom{n}{i} > \binom{n}{\lfloor n/3 \rfloor + 2}$$
We... | This is partially a comment that is slightly too long. In the Math Overflow article, we want to bound
$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}}
= {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$
The author chooses to use a geometric series as an upper bound s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Analytical solution of Laplace's equation with robin/third boundary condition I'm new to PDE solution, and feel difficult to solve the Laplace's equation with robin boundary conditions.
The equation $\Delta u=0$ with boundary conditions shown in the picture:
picture of problem.
Does this problem has an analytical solut... | Using separation of variables $u = X(x)Y(y)$, we obtain the ODEs
\begin{align}
X'' &= -\lambda X \\
Y'' &= \lambda Y
\end{align}
where we used $-\lambda$ for our separation constant, with associated boundary conditions
\begin{align}
u(L,y) &= 0 \implies X(L) = 0 \\
u_{x}(0,y) &= 0 \implies X'(0) = 0 \\
u_{y}(x,0) - hu(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in ad... | First we simplify the numerator of $$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
$$(5+6)(5^2+6^2)=(6-5)(5+6)(5^2+6^2)=(6^4-5^4)$$
$$(5+6)(5^2+6^2)(5^4+6^4)=(6^4-5^4)((5^4+6^4)=(6^8-5^8)
\\.\\.\\ (5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})=(6^{2048}-5^{2048})$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$
| You can decompose $n^3-n = (n-1)n(n+1)$. So $n^3-n$ is a product of three consecutive integers, and in such a sequence you always have an even number and a multiple of $3$. Hence $n^3-n$ is a multiple of $6$.
By induction: the base case is $n = 0$, which is trivial. Now suppose $6$ divides $n^3-n$. Claim: $6$ divides $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find variance of $X$ given its cumulative distribution function A rv $X$ has the cumulative distribution function
$$ F(x) = \begin{cases} 0 \; \; \; \; x<1 \\ \frac{x^2-2x+2}{2} \; \; \; \; 1 \leq x < 2 \\ 1 \; \; \; \; x \geq 2 \end{cases} $$
Calculate the variance of $X$
attempt
First since $F'(x) = f(x)$, then
$$ f... | It doesn't have a density. Note that $F(1)=\frac12$. If you assume that there is a density and compute according to your formula, you will notice that $E(X) <1$, which imply that something went wrong.
Let $y \in [1,4)$,$$P(X^2 \le y)=P(X \le \sqrt{y})=\frac{y-2\sqrt{y}+2}{2}$$
$$E[X^2]=\int_0^1 \, dx+\int_1^4 \frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $P(X + 10/X > 7 )$ where $X$ is uniform
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?
Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \le... | Note: $\mathsf P(X+\tfrac {10}X>7) ~{=~ \mathsf P(X^2+10>7X)\\ =~ \mathsf P((X-5)(X-2)>0) \\=~ \mathsf P(X<2~\cup~5<X)}$
Plot $~y=(x-5)(x-2)~$ and look at where it is above the x-axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integrate $\int \ln(x^2 +1)\ dx$ $$\int \ln(x^2 +1)\ dx$$
I done it using integration by parts where
$\int u\ dv = uv - \int v\ du$
Let $u$ = $\ln(x^2 +1)$
$du = \frac{2x}{x^2+1} dx $
Let $dv = dx$ so $v=x$
$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $
I integrate $2\int \frac{x^2}{x^2+1} $ sepa... | Your integration by parts in the beginning is correct. Truly :
$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x =x\ln\left(x^2+1\right)-{\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x$$
Now, handling the second integral, first of all let's factor out the constant and write $x^2$ as $x^2 + 1 - 1$ to split it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is the range correct? What is the range, using interval notation, of:
$$
f(x) = \frac{2x-4}{x^2+x-2}
$$
The answer I have is
$$
(-\infty, 2/9] \cup [2, \infty)
$$
but i am not sure if it is correct.
| First, you decompose your function w.r.t. the poles
$$
f(x)=\frac{2x-4}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
$$
you get $A+B=2, 2A-B=-4$ hence $3A=-2, A=-\frac{2}{3}, B=\frac{8}{3}$. The derivative of $f$ is
$$
\frac{-2 x^2 + 8 x}{x^4 + 2 x^3 - 3 x^2 - 4 x + 4}
$$
Now, as the denominator is $((x-1)(x+2))^2$ the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The smallest distance between any point on a curve and the parabola $y=x^2$ is 1. What is the equation of the curve? Find the equation of the function $f(x)$, where:
*
*the optimal distance between any one point $P$ on the curve $y=f(x)$ and the parabola $y=x^2$ is always equal to $1$
*$f(x)>x^2$ for all $x$ (looki... | Let a point on the curve be $(a,f(a)$ and the tangent point on the parabola be $(b,b^2)$. Two conditions must be true for all $a \in \mathbb R$
First, the distance is $1$
$$ (b-a)^2 + (b^2-f(a))^2 = 1 $$
Second, the slope between the two points must be normal to the tangent slope, which is $2b$
$$ \frac{b^2-f(a)}{b-a} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
throwing a two-dice 7 with in total 6 dice We have been stuck on the following questions regarding the game Qwixx:
What is the probability that there is a combination of two dice which add up to exactly 7 when throwing with a total of 6 dice once?
Thank you in advance!
| For $k=1,2,3$ let's say that we are in status $S_k$ if exactly $k$ distinct faces have shown up and no pair exists that adds up to $7$.
Now throw a die.
It does not matter what outcome it has and we have landed in status $S_1$.
From here $5$ steps are made giving a total of $6^5=7776$ possible routes.
Note that:
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral involving the Lambert W or product log function. How to calculate the integral
$\displaystyle \int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx$ ?
I tried making a substitution but it doesn't seem to work.
Does this integral have a symbolic solution or a series expansion?
| If you change variable $$\frac{1-x}x=t\implies x=\frac{1}{t+1}\implies dx=-\frac{dt}{(t+1)^2}$$ you end with
$$I=\int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx=-\int \frac {t}{(1+t^2) \, W(t)} \,dt$$ You can expand the integrand around $t=0$ using composition of Taylor series. This would give
$$ \frac {t}{(1+t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Distribution of similar objects of one type and distinct object of other type
$4$ identical oranges and $6$ distinct apples(each of different variety) are to be distributed into $5$ distinct boxes. Then probability that each box get a total of $2$ objects is
Try: Number of ways in which $4$ identical object of one t... | Number of distributions of four identical oranges and six distinct apples to five distinct boxes: Since the oranges are identical, what matters is how many are placed in each box. Let $x_k$ be the number of oranges that are placed in the $k$th box. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 = 4$$
is an equation in the nonnega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Logarithm simultaneous equation
Let $(x_1, y_1,z_1)$ and $(x_2, y_2, z_2)$ - where $x_1\ge y_1\ge z_1$ and $x_2\le y_2\le z_2$ - be two triplets satisfying the following simultaneous equations:
$$
\begin{align}
\log_{10}(2xy)&=\log_{10}x\cdot\log_{10}y\\
\log_{10}(xy)&=\log_{10}z\cdot\log_{10}y\\
\log_{10}(2xz)&=\log... | Too long for a comment.
I think that you did a good job and that the solution $b=c=\color{red}{x}$ is the one to consider (I changed notation on purpose).
You ended with the cubic equation $x^3-2x^2-k=0$ which you do not know how to solve.
Let us follow the steps given here for $a=1$, $b=-2$, $c=0$, $d=-k=$. So, we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find a line equation that divides a triangle in half Let $(0,0), (2,2), (4,0)$ be the vertices of a triangle. Then find the line $l$ that passes through $(1,-1)$ and divides the triangle in half (equal area).
I think can solve this problem by
Letting $P$ and $Q$ be the intersection of the line and the triangle and $m$ ... | Just an alternative solution. Not very nice.
Let $A=(0,0)$, $B=(4,0)$, $C=(2,2)$, $P=(1,-1)$ and $D=(2,0)$.
Suppose that $l$ meets $AB$ at $H$, $CD$ at $K$ and $BC$ at $L$. Let$ DK=a$.
Then $m=a+1$, $K=(0,a)$, $\displaystyle H=\left(2-\frac{a}{a+1},0\right)$ and $L=(4-b,b)$, where $\displaystyle \frac{b-a}{2-b}=a+1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:
approaching by the y-axis:
$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$
approaching by $y=x$
$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$
So I think,that th... | Hint:
Use polar coordinates: $(x,y)\to (0,0)\iff r\to 0$. In polar coordinates, thexpression becomes
$$\frac{x^3+y^3}{x^2+y^2}=r(\cos^3\theta+\sin^3\theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.