Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. Find all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors.
I've tried checking the first 6-7 $n$'s on wolframalpha, but I don't see any patterns for even nor odd $n$'s. At first I thought for all odd $n$'s it was di... | This is just a sketch of a partial answer...
$$2^n+2^{2n}+2^{3n}=2^n(1+2^n+2^{2n})$$
so you need to check if $1+2^n+2^{2n}$ is of the form $p^k$. First note that, if $n$ is even
$$1+2^n+2^{2n}\equiv 1 + 1 + 1\equiv 0\bmod 3$$
so, if $n$ is even, you need
$$2^n+2^{2n}=3^a-1$$
if $a$ is odd, then $3^a-1\equiv -1-1\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of a Sequence involving 3rd root I'm not finding any way to simplify and solve the following limit:
$$
\lim_{n \to \infty} \sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}
$$
I've tried multiplying by the conjugate, but this give a more complex limit.
| Notice the $n^{th}$ term of the sequence can be rewritten as
$$
\sqrt{n^2+n+1} - \sqrt[3]{n^3 + n^2 + n + 1}
= \sqrt{\frac{n^3-1}{n-1}}-\sqrt[3]{\frac{n^4-1}{n-1}}
= n \left[\sqrt{\frac{1-n^{-3}}{1-n^{-1}}} - \sqrt[3]{\frac{1-n^{-4}}{1-n^{-1}}}\right]
$$
When one expand what's in the square/cubic roots in the rightmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int\limits_{\sqrt{2}}^{2}\frac1{t^3\sqrt{t^2-1}}$ This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$
Because of the $\sqrt{t^2-1} = \sqrt... | Your working looks alright.$$\sin \theta= \sin (\pi - \theta)$$
Is another identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I determine if the sequence $\frac{n^{2} + 3^{n}}{n^{4} + 2^{n}}$ converges or diverges? Determine whether the sequence $\frac{n^{2} + 3^{n}}{n^{4} + 2^{n}}$ converges. If it converges find the limit and if it diverges determine whether it has an infinite limit.
Proof: let $a_{n} = \frac{n^{2} + 3^{n}}{n^{4} + ... | your sequence is equivalent to $a_n=\frac{\frac{n^2}{2^n}+\left(\frac{3}{2}\right)^n}{1+\frac{n^4}{2^n }}$ this goes to infinity for $n$ tends to infinity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a... | Every non-zero element of $\mathbb Q[√5]$ is a unit. And other proof you have already shown. That's why we can conclude it's a field.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x$? How to evaluate the following integral
$$\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x $$
The numerical result is $= -1.38104$ and when I look at it, I have no idea how to work on it. Could you provide m... | Your integral can be rewritten as
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x(1-x)(1+x)}{\rm d}x
=\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x+\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x
\end{align}
The first integral is
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x
=&\chi_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Simplfy trigonometric functions by only considering integer inputs? I have the below function which only takes integer input,
$$ 2 \sqrt{3} \sin \left(\frac{\pi t}{3}\right)+\sqrt{3} \sin \left(\frac{2 \pi t}{3}\right)-\sqrt{3} \sin \left(\frac{4 \pi t}{3}\right)+6 \cos \left(\frac{\pi t}{3}\right)+\cos \left(\frac... | Notice that
$$\begin{align}\sin\left(\frac{4\pi t}{3}\right) &= \sin\left(\frac{6\pi t}{3} - \frac{2\pi t}{3}\right)\\
&= \sin\left(2\pi t - \frac{2\pi t}{3}\right)\\
&= - \sin \left(\frac{2\pi t}{3}\right)\end{align}$$
For the cosine part,
$$\begin{align}\cos\left(\frac{4\pi t}{3}\right) &= \cos\left(\frac{6\pi t}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$? From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$
For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-in... | In general, an integral representation of the beta function is $$B(x,y) = \int_{0}^{1} \frac{t^{x-1} + t^{y-1}}{(1+t)^{x+y}} \ dt \ , \ \text{Re} (x), \text{Re}(y) >0 . $$
This can be derived by expressing the more well-known integral representation $$ B(x,y) = \int_{0}^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}} \ dt \tag{1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of an infinite series of Fibonacci numbers divided by doubling numbers. How would I find the sum of an infinite number of fractions, where there are Fibonacci numbers as the numerators (increasing by one term each time) and numbers (starting at one) which double each time as the denominators?
I'm assuming ... | Using Moivre-Binet
$$
F_n = \frac{\phi^n - \psi^n}{\phi - \psi}
$$
with $\phi$ and $\psi$ being the positive and negative root of the golden ratio equation $x^2 - x - 1 = 0$,
we get
\begin{align}
\sum_{n=0}^\infty \frac{F_n}{2^n}
&=
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \frac{\phi^n-\psi^n}{2^n} \\
&=
\frac{1}{\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested w... | According to Brian Bóruma's suggestion, integral can be evaluated using the residue theorem
$$\int\limits_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,{d\theta}=\dfrac{1}{2}\int\limits_{|z|=1}{\dfrac{z+\frac{1}{z}}{\left[1-\frac{a}{2}\left(z+\frac{1}{z}\right)\right]^3} \frac{dz}{iz}}=\\
=-4i\int\limits_{|z|=1}{\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
If $f(x) = -3x^3 + 2x^2$, find $f(-1)$ and $f(1/2)$. I hope you can help me out
For number 8 and 9 find each value if $f(x) = -3x^3 + 2x^2$
8) $f(-1)$
9) $f(1/2)$
| For x=-1 we write $f(-1)=-3*(-1)^3+2*(-1)^2=3+2=5$
For x=1/2 we write $f\left(\frac{1}{2}\right)=-3\left(\frac{1}{2}\right)^3+2\left(\frac{1}{2}\right)^2=\frac{1}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
| Let (x, y, z) be the numbers showing on the 3 dice.
We want x + y + z = 13.
Assuming the dice are distinguishable, the possibilities are:
(1, 6, 6)
(2, 5, 6), (2, 6, 5)
(3, 4, 6), (3, 5, 5), (3, 6, 4)
(4, 3, 6), (4, 4, 5), (4, 5, 4), (4, 6, 3)
(5, 2, 6), (5, 3, 5), (5, 4, 4), (5, 5, 3), (5, 6, 2)
(6, 1, 6), (6, 2, 5), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Error in solution for a separable differential equation I "solved" the differential equation $x'=x^2-1$ a couple of months ago, now I checked the solution with wolfram and it seem I was wrong...
According to Wolfram the solution should be $x(t)=\displaystyle\frac{1-e^{2(t+C)}}{1+e^{2(t+C)}}$, but mine is $x(t)=\display... | The solutions $x(t)=\displaystyle\frac{1-e^{2(t+C_1)}}{1+e^{2(t+C_1)}}$ and $x(t)=\displaystyle\frac{1+e^{2(t+C_2)}}{1-e^{2(t+C_2)}}$ are the same in fact, but with different $C$. Note that $C$ can be complex. With this form of writing we have the relationship $C_2=C_1+i\pi$
Another equivalent form is $x(t)=\displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove the lower bound of $\frac{x^2}{\sin^2x}$? How to prove $$1+\frac{x^2}{3}\leq \frac{x^2}{\sin^2x}, x\in (0,\pi/2)?$$
I do want to show it by intermediate value theorem as
$$\frac{1}{\sin^2x}-\frac{1}{x^2}
=\frac{2}{\xi^3}(x-\sin x)>\frac{2(x-\sin x)}{x^3}.$$
However, this may corrupt, since the rhs $<=\frac... | You might prefer to prove
$$ 1+\frac{x^2}{3} \le e^{x^2/3} \le \left(\frac{x}{\sin x}\right)^2 $$
For the second inequality, the slickest proof is perhaps
$$ \frac{\sin x}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2 n^2}\right)
\le \prod_{n=1}^\infty \exp\left(-\frac{x^2}{\pi^2 n^2}\right)
= \exp\left(-\frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit with Arctan Here's a hard limit I've been trying to answer for a while :
$$\lim_{x\rightarrow 1} \dfrac{-2x\arctan{x} + \dfrac{\pi}{2}}{x-1}$$
I've tried all the tricks that the teacher has taught us and still nothing, I even tried to factor the top by $(x-1)$ but still nothing. Can I get some help on how to eva... | \begin{align*}
\frac{\frac\pi2 - 2x\arctan x}{x-1}
&= 2 \frac{\frac\pi4 - x\arctan x}{x-1} \\
&= 2\left(\frac{\frac\pi4 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan 1 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan\left(\frac{1-x}{1+x}\right)}{x-1} - \arctan x\right) \\
&= 2\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How many distinct solutions are there? Suppose you put the numbers $1,2,\cdots ,10$ in each of the boxes below
such that every connected row and column sum to the same number. How many distinct solutions are there? (By distinct we mean disregarding cyclic permutations, reflections, etc.)
I tried this problem and found... | There are ten solutions
\begin{align*}
&1,4,5,9,7,3,2,6,8,10 \\
&1,4,7,6,10,2,3,5,8,9 \\
&1,5,6,7,9,3,2,4,10,8 \\
&1,5,7,6,9,4,2,3,10,8 \\
&2,3,6,9,7,4,1,5,10,8 \\
&2,3,8,5,9,4,1,7,6,10 \\
&3,1,10,5,8,6,2,4,7,9 \\
&3,2,5,10,6,4,1,7,8,9 \\
&6,1,5,10,4,8,2,3,9,7 \\
&6,3,5,8,4,10,1,2,9,7
\end{align*}
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed
| From what is given, $a,b,c$ are not multiples of $7$. From $(\pm1)^2\equiv 1\pmod 7$, $(\pm2)^2\equiv 4\pmod 7$, $(\pm3)^2\equiv 2\pmod 7$., we see that the only few values for $a^2,b^2,c^2\pmod7$ are possible and the only combinations leading to $a^2+b^2=c^2$ are $1+1\equiv 2$ and $2+2\equiv4$ and $4+4\equiv1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
I am having trouble with this integral from the 2012 MIT Integration Bee: $\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$ $$\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}} $$
Could someone explain to me how to integrate this integral.
Thank you and cheers.
| $\displaystyle{\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}}$
Let's put $x=\sin^{2}t$, then
\begin{align*}
\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t-\sin^4 t}}} \\
&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t(1-\sin^2 t)}}}\\
&=\int{\frac{2\sin t\cos t dt}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | Let$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}$$
$$g(a,b,c)=abc-1=0$$
Using Lagrange Multiplier
$$\large\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}=
\frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}=
\frac{\frac{\partial f}{\partial c}}{\frac{\part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
} |
solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6... | Well actually you have applied the quadratic formula wrong. The roots of the equation $$ax^2+bx+c=0$$ is given by
$$\alpha, \beta ={-b \pm \sqrt {b^2-4ac} \over 2a}$$
So for the equation $$t^2+ \frac12t - \frac32=0$$
the roots become $$\alpha, \beta={- \frac12 \pm \sqrt { (\frac12)^2+4(\frac32)} \over 2}$$
which gives ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Blackboard operation $x,y,z\rightarrow x,y,1/(zx+zy)$ The three numbers $2,3,6$ are written on the blackboard. In each move, we can pick any two numbers, say $x,y$, and replace the third number $z$ by $1/(zx+zy)$. Using finitely many operations, is it possible to obtain the three numbers $2,3,4$?
In the first move, we ... | The invariance is motivated by $ \frac{ x+y} { z ( x + y) } = \frac{ 1}{z} $.
This suggests that we want $ xz, yz, \frac{1}{z}$ as terms in the invariance. Of course, it needs to be cyclic, so the potential invariance has the form
$$ A ( xy+yz+zx) + B ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) $$
Hint: The invariance... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How pro... | I have been trying to find a more aesthetic answer, but until I do, I will post a variational approach.
I have simplified the argument a bit, but I am still looking for a simpler approach.
To minimize $x^3+y^3+z^3-2x^2-2y^2-2z^2$ given $xyz=1$, we need to find $x,y,z$ so that for all variations $\delta x,\delta y,\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Solving $T(n)= 2T(n/2) + \sqrt{n}$ without master theorem (algebraically & recurrence tree) $$T(n)= 2T(n/2) + \sqrt{n}$$
This recurrence was in a stackoverflow question, and I want to solve it without relying on the master method. The solution was given, but wolframAlpha gives a slightly different equation:
$$(1) \quad... | There is another closely related recurrence that admits an exact
solution. Suppose we have $T(0)=0$ and for $n\ge 1$ (this gives
$T(1)=1$)
$$T(n) = 2 T(\lfloor n/2 \rfloor) + \lfloor \sqrt{n} \rfloor.$$
These would seem to be reasonable assumptions since the work step of a recursive algorithm involves a di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to write a sum as an integral to solve it? I was wondering, for example,
Can:
$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$
Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.
But actually writing an integral form. Like
$$\displaystyle \sum_{n=1}^{\... | In such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.
Given that $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\sum_{n=1}^{\infty} T_{n}$$
Where, $T_{n}$ is the $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 7,
"answer_id": 2
} |
$10\sin(x)\cos(x) = 6\cos(x)$ In order to solve
$$10\sin x\cos x = 6\cos x$$
I can suppose: $\cos x\ne0$ and then:
$$10\sin x = 6\implies \sin x = \frac{6}{10}\implies x = \arcsin \frac{3}{5}$$
And then, for the case $\cos x = 0$, we have:
$$x = \frac{\pi}{2} + 2k\pi$$
Is my solution rigth? Because I'm getting another... | Clearly either $\sin x = \frac{3}{5}$ or $\cos x = 0$. Now you have to find out all the possible values for $x$.
*
*If $\sin x = \frac{3}{5}$, then $x=\arcsin \frac{3}{5} + 2k\pi$ ... or $x = \pi - \arcsin \frac{3}{5} + 2k\pi$.
*If $\cos x = 0$, then $x = \frac{\pi}{2} + 2k\pi$ ... or $x = -\frac{\pi}{2} + 2k\pi$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{... | $$\frac{\pi ^2}{9}=\frac{\pi ^2}{6}\times 6 \times \frac{1}{9}=(\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+
\cdots)\times 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Recurrence Fibonacci Sequence Proof I'm having troubles proving that in a fibonacci sequence if n is divisible by four, then Fn is divisible by three
So when Fn is 6, n is 8 and so on. I was thinking maybe I could use mod 3 or mod 4 but don't really know what to do with it.
| This follows from the matrix formulation, which is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
We will prove by induction that $3$ divides $F_{4k}$.
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Then $A^{4(k+1)}=A^{4k}A^4$ :
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Question about $\frac {\Gamma'(z+1)}{\Gamma(z+1)}$ If $\psi (z)= \log\Gamma(z+1)$
Prove that : $$\psi(n)+\gamma=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
My Proof :
$$\psi (z)= \frac {\Gamma'(z+1)}{\Gamma(z+1)}=-\frac{1}{z+1}-\gamma + \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+z+1} $$
$$= \sum_{n=0}^{\infty}\frac{1}{n+1}\... | One error:
From
$\psi(z) = \sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z+1}\right) -\gamma $
you set $z = n$
to get
$\psi(n)+\gamma = \sum_{n=0}^{\infty} \left(\frac{1}{n+1}-\frac{1}{2n+1}\right)$,
You use $n$ both as an argument of $\psi$ and as an index of summation.
You can't do that.
(added later)
From thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{... | $$ AM \ge GM $$
$$\frac{\frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)}}{3} \ge \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}$$
$$ \frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)} \ge \frac{3}{\sqrt[3]{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}}$$
$$AM \ge GM $$
$$ \frac{2abc+ab^2 + ba^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_... | We have
$$\sqrt{x^2 + x + 1} - x = \frac{x^2 + x + 1 - x^2}{\sqrt{x^2 + x + 1} + x} = \frac{x + 1}{\sqrt{x^2 + x + 1} - x}.$$
L'Hopital's rule then gives
$$\lim_{x\to\infty} \sqrt{x^2 + x + 1} - x = \lim_{x\to\infty}\left(1 + \frac{2x+1}{2\sqrt{x^2 + x + 1}}\right)^{-1} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
\begin{array}{l}
2x\equiv 3\pmod 5 \\
4x\equiv 2\pmod 6 \\
3x\equiv 2\pmod 7
\end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, gi... | You seem to have solved $$\left\{\begin{array}{c} x\equiv3 \pmod{5}\\x\equiv 2\pmod{6}\\x\equiv 2 \pmod{7}\end{array}\right.$$
But you need to take into account the coefficients on $x$ in each congruence.
$2x\equiv 3\pmod{5}$ means $x\equiv 4\pmod{5}$
$4x\equiv 2\pmod{6}$ means $x\equiv 2 \mbox{ or } 5 \pmod{6}$
$3x\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
positive fractions, denominator 4, difference equals quotient (4,2) are the only positive integers whose difference is equal to their quotient. Find the sum of two positive fractions, in their lowest terms, whose denominators are 4 that also share this same property.
| I prove your first sentence below:
If $x, y\in\mathbb Z$ satisfy $x-y=x/y,$ then we have :
$$x(y-1)=y^2.$$
Thus, if a prime $p$ divides $y-1,$ then $p$ divides $y$ as well, a contradiction as $y$ and $y-1$ are coprime. Hence no prime divides $y-1,$ i.e. $y=2.$ Therefore $x(2-1)=4.$
As to the second sentence, if $(x-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \s... | $ \lim_{x\to 0} \frac { \sqrt{1+\tan(x)} - \sqrt{1+sin(x)}} { x^3 } = \lim_{x\to 0} \frac {tan(x) - sin(x)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {sin(x)(\frac {1} {cos(x)} - 1)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {1 - cos(x)} {cos(x)x^2(\sqrt{1+\tan(x)} + \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Simplfiying $x(5xy+2x-1)=y(5xy+2y-1)$ I want to simplify: $x(5xy+2x-1)=y(5xy+2y-1)$ to $(x-y)(5xy+something-1)=0$ but I can't figure out what to do with the $2x$ and $2y$ on both sides.
| \begin{eqnarray}
&&x(5xy+2x-1)=y(5xy+2y-1)\\
&\Rightarrow&(5x^2y-5xy^2)+(2x^2-2y^2)-(x-y)=0\\
&\Rightarrow&5xy(x-y)+2(x-y)(x+y)-(x-y)=0\\
&\Rightarrow&(x-y)(5xy+2x+2y-1)=0
\end{eqnarray}
Moreover, note that $5xy+2x+2y-1=5(x+\frac{2}{5})(y+\frac{2}{5})-\frac{9}{5}$, you can simplifying as following (if you want):
$(x-y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How many 10-letter words are there in which each of the letters e,n,r,s occur at most once? Solve with a generating function. My solution was
$$g(x)=\left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)^4 \left(\frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + ... \right)^{26-4}$$
$$g(x) = (1 + x)^4e^{22x}$$
$$g(x) =\left( \bi... | In your computation you are missing the factor $\frac{1}{i!}$ for $f_i$ coefficients.
Fixing this, what you should be looking for is: $P(n=10,r=4)$ (this should answer your second question)that is:
$$\sum_{i=0}^{r}h_i f_{n-i}x^n= \sum_{i=0}^{r} \binom{r}{i} \frac{22^{n-i}}{(n-i)!}x^n$$
substituting for $r=4$ and $n=10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$ One of the ways to compute the integral
$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$
is to m... | We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$
Then,
$\displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Mathematical Induction Proof - Exponent with n in denominator Use mathematical induction to prove the following:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}=2-\frac{n+2}{2^n}; n ∈ N $$
I am having trouble figuring out how to solve this with an exponent in the denominator.
$$2-\frac{n+2}{2^n}+\frac{n+1}... | $$\begin{align*}\underbrace{\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}\right)}_{=2-\frac{n+2}{2^n}}+\frac{n+1}{2^{n+1}}&=\left(2-\frac{n+2}{2^n}\right)+\frac{n+1}{2^{n+1}}\\&=2-\left(\frac{2(n+2)}{2^{n+1}}-\frac{n+1}{2^{n+1}}\right)\\\\&=2-\frac{(n+1)+2}{2^{n+1}}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of Sum to Infinity Is $ \Sigma_{n=0}^\infty (\sqrt[3]{n^3+1} - n)$ convergent or divergent?
For expressions of the form $\sqrt{n^2+1} - n$, I believe the common trick is to multiply by the "conjugate" $\frac{\sqrt{n^2+1} + n}{\sqrt{n^2+1} + n}$.
Is there a similar trick for other roots (i.e. not square roots) as... | $$\sqrt[3]{n^3+1}-n=\\\sqrt[3]{n^3+1}-n \cdot \frac{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\\\frac{n^3+1-n^3}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\frac{1}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\<\frac{1}{\sqrt[3]{(n^3)^2}+n\sqrt[3]{(n^3)^1}+n^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$ Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$. I’m new to the subject and the square roots are throwing me a bit off.
| You first have to multiply the top and bottom by the conjuage, which is $\sqrt{n^2+3} + \sqrt{n^2+1}$.
So you get $$\dfrac{(\sqrt{n^2+3}-\sqrt{n^2+1})(\sqrt{n^2+3}+\sqrt{n^2+1})}{\sqrt{n^2+3}+\sqrt{n^2+1}} = \dfrac{n^2+3 - (n^2+1)}{\sqrt{n^2+3} + \sqrt{n^2+1}} = \dfrac{2}{\sqrt{n^2+3} + \sqrt{n^2+1}}$$
Clearly, the $\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as... | Your approach looks fine, and it is explained well. Here is a variant of your solution, that uses similar ideas:
1) $\displaystyle\lim_{x\to0}(\sqrt[3]{ax+b}-2)=\lim_{x\to0}\bigg(\frac{\sqrt[3]{ax+b}-2}{x}\cdot x\bigg)=\frac{5}{12}\cdot 0=0,\;\;\;$ so $\sqrt[3]{b}=2$ and $b=8$.
2) Let $g(x)=\sqrt[3]{ax+8}.\;\;\;$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
prove equation equals $-1$ I was wondering if it was possible to prove that
$\left( \frac{( a^2 - c^2 + (\frac{d}{c}a)^2-d^2)}{ \sqrt{(a-c)^2+(\frac{d}{c}a)-d)^2} \sqrt{(a+c)^2+(\frac{d}{c}a)+d)^2}} \right) = -1$ when $a,d,c \in [-1,1]$ and $|a|<c$, and $|b|<d$?
Attempt: By Asdrul, I know that
$$\left[(a-c)^{2}+\bigg(... | Hint: note that
$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculating the value of $\frac{a-d}{b-c}$ If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of:
$$\frac{a-d}{b-c}$$
Where $a,b,c,d$ are real numbers.
Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this val... | You have $\displaystyle{\frac{a-b}{c-d}}=2$ and $\displaystyle{\frac{a-c}{b-d}}=3$, hence
$$a-b = 2c-2d\\a-c = 3b-3d$$
By subtracting,
$$c-b=2c-3b+d$$
Or, adding $2b-2c$ to both terms,
$$b-c=d-b$$
Then
$$2=\frac{a-b}{c-d}=\frac{a-d+d-b}{c-b+b-d}$$
In denominator, $c-b$ and $b-d$ are equal, so
$$2=\frac12\frac{a-d}{c-b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Complex numbers, polynomials Let $a$ be complex number such that $a^5 + a + 1 = 0$. What are possible values of $a^2(a - 1)$?
I have tried to find $a$. Is there any way to find it?
| $$
\begin{align}
a^5+a+1&=(a^2+a+1)(a^3-a^2+1)\\
&=(a^2+a+1)(a^2(a-1)+1)
\end{align}$$
If $a$ vanishes the second factor then $a^2(a-1)=-1$.
If $a$ vanishes the first factor, then divide $a^2(a-1)$ by $a^2+a+1$. We get $$a^3-a^2 = (a-2) × (a^2+a+1)+a+2$$
Therefore $a^2(a-1)=a+2$. We know that the roots of $a^2+a+1$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Showing two things are equal by Fourier series Given the Fourier series for the function:
$$f(x) = x+\frac14x^2 \quad -\pi\leq x \lt \pi$$
$$f(x)=f(x+2\pi) \quad -\infty \leq x \lt \infty$$
is
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(nx)}{n^2}-\frac{2\sin(nx)}{n}\right)$$
show that$$\sum \li... | Why not substitute $\;x=0\;$ to make things easier (though perhaps slightly longer)?
$$f(x)=x+\frac{x^2}4\implies f(0)=0=\frac{\pi^2}{12}+\sum_{n=1}^\infty (-1)^n\frac1{n^2}\implies$$
$$\implies\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi}{12}$$
But
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=1-\frac14+\frac19-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$ $$\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$$
I don't really know how to do, but I was trying to do like that:
$a=x$,
$b=y^2$
then I was trying to do this
$$\lim_{(x,y)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
then I don't know no more how to d... | Conversion into polar coordinates can be of help as well. Letting $x=r\cos\theta$ and $y=r\sin\theta$, we can write the limit as follows:
$$\lim_{r\to 0}\dfrac{r^3\cos^3\theta \cdot r^3\sin^3\theta}{r^2\cos^2\theta+r^6\sin^6\theta}=\lim_{r\to 0}\dfrac{r^3(\cos^3\theta\sin^3\theta)}{r^2(\cos^2\theta+r^4\sin\theta)}=\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to find the splitting field of $X^4-10X^2+1$? How to find the splitting field of $X^4-10X^2+1$ ?
I found the roots
\begin{align*}
X^4-10X^2+1=0&\iff (X^2-5)^2-24=0\\
&\iff X^2-5=\pm 2\sqrt 6\\
&\iff X^2=5\pm 2\sqrt 6\\
&\iff X\in\left\{\sqrt{5+2\sqrt 6},\sqrt{5-2\sqrt 6},-\sqrt{5+2\sqrt 6},-\sqrt{5-2\sqrt 6}\right... | Hint: $(\sqrt 3 + \sqrt 2)^2 = 5 + 2\sqrt6$ and $(\sqrt 3 - \sqrt 2)^2 = 5 - 2\sqrt 6$, so your splitting field contains $\sqrt 2$ and $\sqrt 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| Note that
$$(1^2+2^2+\ldots+n^2)^2\ge 1^4+2^4+\ldots+n^4$$
Now we can use Stolz–Cesàro theorem (kind of a discrete l'Hôpital's rule):
$$\lim_{n\to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\ldots+n(n+1)(n+2)}{1^4+2^4+\ldots+n^4}=\lim_{n\to\infty}\frac{n(n+1)(n+2)}{n^4}=0$$
And conclude the original limit was also $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Order of group from its presentation Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^2y = xy^3 = 1$.
Ok so it would be nice to know the order of $x$ and $y$ respectively. We can readily conclude from the equations above that $x^2 = y^{-1}$ and $x^{-1} = y^3$.
Next we can derive th... | $x^2 = y^{-1}, x = y^{-3} \Rightarrow x^6 = y^{-3} = x \Rightarrow x^5 = 1 \Rightarrow |x| = 5 \Rightarrow |x^2| = 5 \Rightarrow |y^{-1}| = 5 \Rightarrow |y| = 5.$ Now $y \in <x> \Rightarrow$ the group is the cyclic group of order $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the limit $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$ Find the limit if it exists
$$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$$
| Let's see the limit in the plane $z=0$. So,
\begin{align*}
\lim_{(x, y, z)\rightarrow (0,0,0)} \frac{xy + yz + xz}{x^2 + y^2 + z^2} &= \lim_{(x, y) \rightarrow (0,0)} \frac{xy} {x^2 + y^2}
\end{align*}
Now analyzing by paths, we have to take the path $y=mx$, with $m>0$.
\begin{align*}
\require{cancel}
\lim_{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Comparison between lebesgue integral and riemann integral of $f(x)=x^2$ in $[0,2]$ If we have an example $f(x)=x^2$ let's say for $[0,2]$.
In lebesgue integral, I already use a sequence of function $f_n(x)$ as approximation to $f(x)$ ($f_n(x)$ converges to $f(x)$) which is stated by $f_n(x)=\sum_{k=1}^{2.2^n}\frac{k-1... | Since $f(2) = 2^2 = 4$, the upper limit for the sum should be $4 \cdot 2^n$:
$$f_n(x) = \sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}1_{E_n},$$
where
$$E_n = \left\{x : \frac{k-1}{2^n} \leqslant f(x)\leqslant \frac{k}{2^n}\right\}=\left\{x : \sqrt{\frac{k-1}{2^n}} \leqslant x\leqslant \sqrt{\frac{k}{2^n}}\right\}.$$
Then the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First Order Ordinary Differential Equation by Any Method (1R-24) I just need for someone to check my work and suggest a better way to solve this if one exists. I can use any method but not numerical or any other iterative series approximation. The following documents my process of attempting to solve this problem. It d... | You could check by yourself your result :
$$y+2\ln|y|-\frac{1}{y}(x+1)=c$$
$$x=y^2-2y\ln|y|-cy-1$$
$$dx=(2y-2\ln|y|-2-c)dy$$
Bringing it back into the ODE :
$$-ydx + (x + y^2 + 2y + 1)dy = 0$$
leads to :
$$-y\big((2y-2\ln|y|-2-c)dy\big) + \big((y^2-2y\ln|y|-cy-1) + y^2 + 2y + 1\big)dy = $$
after simplification $$=0$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in... | There is nowhere a converging series in sight. From the data given it seems that $N$ is large and
$$p:=Nk$$
is very small. Therefore we may write
$$S={1+k\over k}\bigl((1+k)^N-1\bigr)={1+k\over k}\left(\bigl(1+{p\over N}\bigr)^N-1\right)\doteq{1+k\over k}(e^p-1)\ .$$
If $N$ is not in the thousands use the first few t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute
$$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$
What I am doing is take
$$f(z)=\frac{(\log z)^2}{1+z^2}$$
and calculating
$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$
which came out to be $\dfrac{\pi}{2}-\df... | I would consider the following contour integral
$$\oint_C dz \frac{\log^3{z}}{1+z^2} $$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, about the positive real axis. Without explicitly showing that the integrals over the outer and inner circles vanish as $R\to\infty$ and $\epsilon \to 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Help with Linear Transformation of a multivariate normal Given X ~ $N_2$ (μ, Σ)$
Find the Distribution of
$$
\begin{pmatrix}
X+Y \\
X-Y
\end{pmatrix}
$$
Show independence if $Var(X) = Var(Y)$
Attempt:
Given proper of Multitvariate Normal Transformations $N_m$ (Aμ, $AΣA^t$ )
Using $$ ... | *
*Yes, for a Gaussian random vector $(X,Y)$, it is know that $X$ and $Y$ are independent if and only if the covariance $\text{cov}(X,Y)$ equals $0$. Since the entries on the off-diagonal are exactly the covariance, this means that $X$ and $Y$ are indeed independent.
*Well, if $\delta_y^2=0$ (which is equivalent to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Induction Proof that $\sum_{i=0}^n 3^{n-i} {n \choose i} = \sum_{i=0}^n (-1)^i 5^{n-i} {n \choose i}$ Show that for all $n\geq0$
$$\binom{n}{0}3^n+\binom{n}{1}3^{n-1}+\dotsc+ \binom{n}{n-1}3^{1}+\binom{n}{n} $$
$$= \binom{n}{0}5^n-\binom{n}{1}5^{n-1}+\binom{n}{2}5^{n-2}-\binom{n}{3}5^{n-3}+\dotsc (-1)^n\binom{n}{n}$$
I... | We have
$$(x+a)^n=\sum_{k=0}^{n}\binom{n}{k}x^ka^{n-k}$$
Now by putting $x=3$ and $a=1$ we get
$$\begin{align}(3+1)^n&=\sum_{k=0}^{n}\binom{n}{k}(3)^k(1)^{n-k}\\
&=\binom{n}{0}3^n+\binom{n}{1}3^{n-1}+\dotsc+ \binom{n}{n-1}3^{1}+\binom{n}{n}\\
\end{align}$$
And by putting $x=5$ and $a=-1$ we get
$$\begin{align}(5+(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\r... | Let us consider your last integral $$I=\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$ Changing variable $y=a z$, it becomes $$I=\frac 1 a \int_0^1 \frac{dz}{\left(z^2+1\right) \sqrt{z^2+2}}$$ Now, make a change of variable such that $$z=\frac{\sqrt{2} t}{\sqrt{1-t^2}}$$ (it did not come immediately to my mind, I must c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/... | Though I suspect this may be pushing common functions..
$2^2 + 2 + \Gamma(2) = 7$
$22/{\sqrt{2}^2} = 11$
$\int_{2/2}^{22}dx =21$
$\int_{2-2}^{22}dx =22$
$(2+2)! + \sqrt{2+2} = 26$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 1
} |
Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{... | $$ a_n = 3a_{n-1} + n^2 - 3 = 3^ka_{n-k} + n^2 + 3(n - 1)^2 + \ldots + 3^{k - 1}(n - k + 1)^2 - (3 + \ldots + 3^k) $$
$$ a_n = 3^ka_{n - k} + \sum_{m=0}^{k-1}3^m(n - m)^2 - \sum_{m = 0}^{k-1}3^{m+1}$$
Setting $k = n$, we have
$$a_n = 3^n + \sum_{m=0}^{n-1}3^m(n - m)^2 - \sum_{m = 0}^{n-1}3^{m+1} = \frac{1}{2}(-n^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solve this equation if we can't use Chinese remainder theorem. Let consider:
$$\begin{cases}6x \equiv 2 \mod 8\\ 5x \equiv 5\mod 6 \end{cases}$$
We can't use Chinese remainder theorem because $\gcd(8,6) = 2 > 1$
Help me.
| $$
\begin{array}{l}
\left\{ \begin{array}{l}
6x \equiv 2\left[ 8 \right] \\
5x \equiv 5\left[ 6 \right] \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x \equiv 1\left[ 4 \right] \\
x \equiv 1\left[ 6 \right] \\
\end{array} \right. \\
\Rightarrow \left\{ \begin{array}{l}
x \equiv - 1\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Infimum of $\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \mathbb N\right\}$ with a proof Consider $A=\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \:N\right\}$. I need to find and prove $\inf(A)$.
So I know that I need to prove that for every $\epsilon > 0$ exists some $a\:\in A$ such that $a\:=\:\frac{n_0}{n_0^2\:+1}$ for some $n_0$, s... | Revised to match corrected question and the OP’s background.
Let $a_n=\dfrac{n}{n^2+1}$. Then
$$\begin{align*}
\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}&=\frac{n+1}{n^2+2n+2}-\frac{n}{n^2+1}\\\\
&=\frac{(n+1)(n^2+1)-n(n^2+2n+2)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{n^3+n^2+n+1-(n^3+2n^2+2n)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{-n^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using an Integral to Solve for a Variable a I am struggling to use the following equation:
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0
$$
where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.
I integrate the integrand, via
$$
\... | Your question is now: If $\alpha =2\sqrt{1-\frac{1}{a^{2}}}$ how can one
prove that $\cos \alpha =\alpha ?$
Your question in the title is :'' solve for a variable $a$ ''
it means that: the problem is: solve for the variable $a$
the equation $\cos \alpha =\alpha ,$ where $a>1.$ So, lets go:
From $\frac{\alpha }{2}=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\disp... | You are just a baby step away from the answer...: $\dfrac{1}{(x-1)(x^2+1)} = \dfrac{1}{2}\left(\dfrac{1}{x-1} - \dfrac{x}{x^2+1} - \dfrac{1}{x^2+1}\right)$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the limit of a series
I got the n'th term but clueless about what to do next
| We have
$$t_r = \dfrac{r(r+1)(2r+1)/6}{r^2(r+1)^2/4} = \dfrac23 \dfrac{2r+1}{r(r+1)} = \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right)$$
Hence,
\begin{align}
S_n & = \sum_{r=1}^n (-1)^r t_r = \sum_{r=1}^n (-1)^r \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right) = \dfrac23 \left(\sum_{r=1}^n \dfrac{(-1)^r}r + \sum_{r=1}^n \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an in... | $x^2-x\ge0\iff x\not\in(0,1)$. Thus, we need to choose whether $x\ge1$ or $x\le0$.
For $x\ge1$, let $x=u+\frac12$ and $u=\frac12\sec(\theta)$. We can assume that $\theta\in[0,\frac\pi2)$ since that gives a full range for $x\ge1$. As is shown in this answer, $\int\sec(\theta)\,\mathrm{d}\theta=\log(\sec(\theta)+\tan(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \f... | A generalized version of the result might be easier to prove:
For every positive integers $k$ and $m$, $$\sum_{i=k+1}^{k(m+1)}\frac1{a+ib}\geqslant\frac{m}{a+(m+1)b}.$$
The question asks about the case $k=2^{n-1}$ and $m=1$.
To prove the claim, note that $a+ib\leqslant a+k(m+1)b$ for every $i$ used in the sum $S$ of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Need help setting this up: find $I_z$ the given lamina with uniform density of 1 Find $I_z$ for the given lamina with $\rho=1$,
$z = x^2 + y^2 \;\text{ and }\; 0 ≤ z ≤ h$.
I tried to set it up the following, but I am not sure if this is correct:
I know $\int_{0}^{h}dz, z-x^2-y^2=0$ and $I_z=\int_{s}\int x^2+y^2ds$
but... | Hint:
Since
\begin{align*}
I_z&=\int\int_S\left(x^2+y^2\right)\mathrm ds
\end{align*}
and the shape $S$ of the lamina is given by $z=x^2+y^2$ with $0\leq z\leq h$, so
\begin{align*}\mathrm ds &=\left[\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1\right]^{1/2}\mathrm d A\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
| Yes you can. Base case is easy. After it, you have to prove following implication.
$$\left(\underbrace{\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}}_{\mathrm{Assumption}}
\right) \Longrightarrow \left(
\underbrace{\sum_{k=0}^{n}(k^2)= \binom{n+1}{3} + \binom{n+2}{3}}_{\mathrm{Thesis}}
\right)
$$
Now, using assu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Gradient of $x^T B^T B x - x^T B^T b - b^T Bx$ I want to compute the gradient $\nabla_x f(x)$ of $f(x) = x^T B^T B x - x^T B^T b - b^T Bx$ with respect to the vector $x$. So far I have tried below. But when I try to add them together, I couldn't see they come together.
Edits
Now I think I got it. They does come togethe... | Hint: For a scalar $\epsilon$ and appropriate sized vectors $x,y$ we have:
$f(x+\epsilon y) = (x+\epsilon y)^TB^TB(x+\epsilon y) - (x+\epsilon y)^TB^Tb - b^TB(x+\epsilon y)$
$= x^TB^TBx + \epsilon y^TB^TBx + \epsilon x^TB^TBy + \epsilon^2 y^TB^TBy - x^TB^Tb - \epsilon y^TB^Tb - b^TBx - \epsilon b^TBy$
$= (x^TB^TBx - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int \frac{\mathrm{d}x}{49x^2+1}$ So I tried solving this by taking a substitute for the integrand, $t=49x$, so its derivative is $dx = \frac {dt} {49}$. Then you insert it into the integrand and get $$\int \frac{\mathrm dt}{49(t^2 +1)} = \frac{1}{49}\int\frac{1}{t^2 +1}dt = \frac{1}{49}\arctan t + c = \frac... | $$\begin{align}
F(x) &= \int \left(\frac{1}{49x^2 + 1} \right)dx \\
&= \int \left(\frac{1}{(7x)^2 + 1} \right)dx \\
t &= 7x \\
\left(7x \right)dx &= dt \\
\left(\frac{d}{dx}(7x) \right)dx &= dt \\
\left(7*\frac{d}{dx}(x) \right)dx &= dt \\
\left(7*\frac{dx}{dx} \right)dx &= dt \\
\left(7 \right)dx &= d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Summation notation for divided factorial. I have the following sum
$$5\cdot4\cdot3+5\cdot4\cdot2+5\cdot4\cdot1+5\cdot3\cdot2+5\cdot3\cdot1+$$$$5\cdot2\cdot1+4\cdot3\cdot2+4\cdot3\cdot1+4\cdot2\cdot1+3\cdot2\cdot1$$
It is basically $5!$ divided by two of the numbers in the factorial. So
$$\frac{5!}{1\cdot2}+\frac{5!}{1... | You can write it as a single sum as follows
$$\frac{5!}{1\cdot2}+\frac{5!}{1\cdot3}+\frac{5!}{1\cdot4}+...+\frac{5!}{3\cdot5}+\frac{5!}{4\cdot5}=5!\sum_{1\le i <j\le 5}\frac{1}{i\cdot j}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra - Find inverse of $A$ I have this problem :
$$A = \left(\begin{array}{ccc}
3 & -1 & 1 \\
2 & 0 & 1 \\
1 & -1 & 2 \end{array}\right) $$
1) Show that $A^3-5A^2+8A-4I=0$.
2) Using (1) To find $A^{-1}$.
I did (1) show that is correct, Usually to find inverse I use the Identity Matrix.
Anyhow for this proble... | You have
$A^3 - 5 A^2 + 8A = 4I $
Left multiply both sides by $A^{-1}$. Use the fact that matrix multiplication is associative, and $A^{-1}A=I$,
$$
A^2 - 5A + 8I = 4A^{-1}
$$
which gives you an $A^{-1}$ of
$$
\frac{1}{4}
\begin{bmatrix}
1 & 1 & -1 \\
-3 & 5 & -1 \\
-2 & 2 & 2
\end{bmatrix}
$$
You can now check that $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that... | Since $a_n=c\left(\frac{1}{3}\right)^n+dn\left(\frac{1}{3}\right)^n$, $a_0=2\implies c=6$ and $a_1=5\implies d=9$, so
$\displaystyle a_n=\frac{6}{3^n}+\frac{9n}{3^n}=\frac{2+3n}{3^{n-1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | Call $S = 1 + 4 + \ldots + [3n-2]$.
Add the numbers in reverse direction: $S = [3n-2] + [3n-5] + \ldots + 1$.
Add the two equations term by term: $2S = (1 + [3n-2]) + (4 + [3n - 5]) + \ldots + ([3n-2]+1) = n (3n-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 0
} |
Show $\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$. How to show that
$$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$
?
My try:
We have
$$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$
so
$$\frac{1}{n^2+3n+1}=... | I think we can make some use of the residue theorem. Write $n^2+3 n+1 = (n+3/2)^2-5/4$ and the sum is
$$\sum_{n=1}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} = \frac12 \sum_{n=-\infty}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} - 1 +1$$
(To get the doubly infinite sum, I had to add b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I... | We have:
$$\lim_{x\to 0} \frac{\sin^2 x -x^2}{x^2\sin^2 x}$$
Since direct substitution yields $\frac{0}{0}$, we use L'Hospital's rule:
$$\lim_{x\to 0} \frac{2\sin x \cos x - 2x}{2x\sin^2 x + 2x^2\sin x \cos x }$$
Using it again:
$$\lim_{x\to 0} \frac{2\cos2x - 2}{x^2(2\cos^2 x-2\sin^2 x) + 2\sin^2 x +8x\sin x \cos x }$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that for a $2\times 2$ matrix $A^2=0$ Assuming that given a $2\times2$ matrix $A$ with the property $A^3=0$ show that $A^2=0$.
Okay so this question is messing with me. When it says $A^3=0$ do you think it means the determinant? How would one approach this? Any help would be much appreciated
| It does not mean the determinant. It means that if you multiply $A$ with itself three times, you get the zero matrix. The nullspace of $A^2$ is contained in the nullspace of $A^3$, i.e. if $A^2 x = 0$, then clearly $A^3 x = A(A^2 x) = 0$. Particularly, this means that the nullspace can only increase as you apply more p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $n^5+n^4-3=x^2\pmod p$ Prove that for every odd prime number $p$ there is a natural number $n$ such that the equation $n^5+n^4-3=x^2\pmod p$ has no solutions.
So we have to understand that for each $p$ we can find $n$ such that the Legendre symbol $\left(\dfrac{n^5+n^4-3}{p}\right) = -1$. For $p=4k+3$ we can t... | The curve $x^2=n^5+n^4-3$ has genus $2$, and so the Hasse-Weil bound implies that the number of ordered pairs $(x,n)$ modulo $p$ satisfying the equation is between $p-4\sqrt p$ and $p+4\sqrt p$. There can be at most five values of $n$ for which the right-hand side is congruent to $0\pmod p$; for all other values of $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$? What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$?
I'm not good at algebra so please explain in easy to understand steps.
Thanks
| Subbing in $x=2$ and $y=-4$ into the equation gives:
$2(2^2) +3(2)(-4)-4((-4)^2)=2(4) + (-24) -4(16)=8-24-64=8-88=-80$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Hint:- $$y^3-1=y^2(y-1)+y(y-1)+(y-1)=(y-1)\left(y^2+y+1\right)$$
Solution:-
$y=x^2\implies x^6-1=\left(x^2-1\right)\left(x^4+x^2+1\right)=(x-1)(x+1)\left(x^4+x^2+1\right)$ $$\boxed{\therefore\dfrac{x^6-1}{x-1}=(x+1)\left(x^4+x^2+1\right)=x^5+x^4+x^3+x^2+x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 2
} |
Use Laplace Transform to solve the following IVP: I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace.
The problem is:
$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$
Attempt at Solution
$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + ... | We can take inverse Laplace transform by using the Bromwich integral.
That is,
\begin{align}
\mathcal{L}^{-1}\{Y(s)\} &= \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma -iT}^{\gamma +iT}Y(s)e^{st}ds\\
&= \sum\text{Res}
\end{align}
where
$$
Y(s) = \frac{2}{(s^2 + 2s + 5)^2} + \frac{2s + 3}{s^2 + 2s + 5}
$$
Then the poles... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Complex number calculation I'm supposed to show $ z^{10} $, when z = $ \frac{1+ \sqrt{3i} }{1- \sqrt{3i} } $
I can work it out to $ \frac{(1+\sqrt{3}\sqrt{i})^{10}}{(1-\sqrt{3}\sqrt{i})^{10}} $
However this is inconclusive because I need to show $ z^{10} $ in the form x+yi, and I can't figure out the real and imaginary... | It seems to me that @MichaelAlbanese is right. It should be
$ \frac{1+\sqrt{3} i}{1-\sqrt{3} i} $. After removing any complex part from denominator we get
$$
\frac{1+\sqrt{3} i}{1-\sqrt{3} i} = \frac{1+\sqrt{3} i}{1-\sqrt{3} i} \frac{1+\sqrt{3} i}{1+\sqrt{3} i} = \frac{-2 + 2 \sqrt{3} i}{4} = -\frac{1 - \sqrt{3} i}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sqrt{n^2 + 2}$ is irrational
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.
From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a p... | $$\begin{array}{l}
\sqrt {2 + n^2 } = \frac{p}{q};\quad and\quad p \wedge q = 1 \\
\sqrt {2 + n^2 } = \frac{p}{q} \Leftrightarrow 2 + n^2 = \left( {\frac{p}{q}} \right)^2 \\
\Leftrightarrow 2 = \left( {\frac{p}{q}} \right)^2 - n^2 \\
\Leftrightarrow 2 = \left( {\frac{p}{q} - n} \right)\left( {\frac{p}{q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Laurent Series, region of convergence I want to find the laurent series for
$$
f(z) = \frac{z}{z^2 - (1+i)z +i}
$$ in powers of $z-1$ and find the region of convergence. I am not quite sure how to do this.
I know that
$$
f(z) = \frac{z}{(z-1)(z-i) }
$$
but I do not know where to go from here. Any help would be great... | First rewrite $f$:
$$
f(z) = \frac{z}{z^2 - (1+i)z +i} = \frac{z}{(z-1)(z-i)} = \left(\frac{A}{z-1} + \frac{B}{z-i}\right) = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-i}\right).
$$
Since you want the expansion around $1$, let's write everything in terms of $(z-1)$:
$$
\begin{align}
f(z) &=\frac{1}{1-i}\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find cosine of acute angles in a right triangle.
If sides of a right triangle are in Geometric Progression, then find the cosines of acute angles of the triangle.
[Answer] $\frac{\sqrt{5}-1}{2}$,$\sqrt\frac{\sqrt{5}-1}{2}$
My work:
Using Pythagoras Theorem, $a^{2}+a^{2}r^{2}=a^{2}r^{4}$
$1+r^{2}=r^{4}$
$r^{4}-r^{2}-1=... | You actually do have the right answers; they just merely look different. A quick check on wolfram alpha will tell you that they are equivalent.
First answer: http://cl.ly/image/1C3F0O3U1A16
Second answer: http://cl.ly/image/0E0X3z3q0M0B
Both of these calculations can be very quickly justified by multiplying both side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Existence of Holomorphic function (Application of Schwarz-Lemma) Let, $D=\{z\in \mathbb C:|z|<1\}$. Which are correct?
*
*there exists a holomorphic function $f:D \to D$ with $f(0)=0$ & $f'(0)=2$.
*there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=\dfrac{3}{4}$ & $f'\left(\dfrac{2}{3}... | We want to determine whether for given $a,b,c,d$, there exists a holomorphic $f\colon D \to D$ with
*
*$f(a) = b$, and
*$f'(c) = d$.
A typical way to attack such a problem is the Schwarz-Pick lemma, resp. its differential version
$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can w... | If $y=x-3$ and $y^3-42y=105$, then take $y=t+r\implies t^3+r^3+(3tr-42)y=105$, then:
$$3tr-42=0$$$$t^3+r^3=105$$
$$3tr-42=0\implies tr=14\implies t^3r^3=2744$$
Then $t^3$, and $r^3$ are roots of:
$$Z^2-105Z+2744=0$$
Then $Z_1=49$ and $Z_2=56$, $\implies t^3=49$ and $r^3=56$, $\implies t=\sqrt[3]{7^2}$ and $r=2\sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Formula for tangent derivatives, how to prove? How to prove?
$$(\tan x)^{(s-1)}=\pi^{-s}\Gamma(s)\left(\zeta\left(s, \frac12-\frac x\pi\right)+(-1)^s\zeta\left(s, \frac12+\frac x\pi\right)\right) $$
| We start with the following claim (I hope it has been proven on this web site, but I couldn't find a link):
$$\sum_{k=-\infty}^\infty\frac{1}{(k-\frac{x}{\pi})^2}=\frac{\pi^2}{\sin^2(x)}$$
Because
$$\frac{d}{dx}\tan(x)=\frac{1}{\cos^2(x)}=\frac{1}{\sin^2(x-\frac{\pi}{2})}$$
it follows that
$$\frac{d}{dx}\tan(x)=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
(Non?)-uniqueness of sums of squares (I've had almost no exposure to number theory, so please keep answers as elementary as possible.)
Write $\mathbb{N} = \{0,1,2,3,\ldots\}$ for the natural numbers. Then every element of $\mathbb{N}$ can be expressed as a sum of squares. For example:
$$6 = 1+1+1+1+1+1$$
Usually, we ca... | There are many counterexamples, of which the smallest is $$5^2 + 5^2 = 7^2 + 1^2.$$
(I earlier stated that $25$ was the smallest counterexample, since $3^2+4^2 = 0^2+5^2$, but in your terminology, it has a potency of 1, not 2, so it is not a counterexample.)
Brahmagupta's identity shows that if $x$ and $y$ are each exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Number of Sets of Partitions I looked at the partitions of numbers, like let's say $n=5$. You get
$$
\begin{eqnarray}
5&=&5\\
\hline
&=&4+1\\
&=&3+2\\
\hline
&=&3+1+1\\
&=&1+2+2\\
\hline
&=&2+1+1+1\\
\hline
&=&1+1+1+1+1\\
\end{eqnarray}
$$
where I grouped the partitions according to their distribution (i.e. appearance)... | As pointed out, there is no easy closed form for the partition function without any restrictions. However, given some restrictions, there are some nicer closed forms. For example, if we let a composition of $n$ be a partition of $n$ in which order matters, then let $P(n)$ be the number of compositions of $n$ which only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was abl... | We have by the Riemann sum
$$c_n=\frac1n\sum_{k=2}^n\frac1{\sqrt k}=\frac1{n^{3/2}}\sum_{k=2}^n\frac1{\sqrt{\frac kn}}\sim\frac1{\sqrt n}\int_0^1\frac{dx}{\sqrt x}=\frac2{\sqrt n}$$
so clearly $c_n$ tends to $0$ but we found also that
$$\lim_{n\to\infty}c_n\sqrt n=2$$
which we can't find it by applying the Cesàro's the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Double integral with two parameters $\int_{x_1=1}^{x_2=2}\int_{y_1=0}^{y_2=x}\arctan\left(\frac{y}{x}\right)\,dx\,dy$ Given the following integral:
$$\int_{y_1=0}^{y_2=x}\,dy\int_{x_1=1}^{x_2=2}\arctan\left(\frac{y}{x}\right)\,dx$$
I thought of using $u$-substitution:
$$\begin{align}
u &= \frac{y}{x} \\
w &= \arctan(u)... | $$
\int \arctan(\frac{y}{x})\;dx, \;\text{ Let } u=\arctan\frac{y}{x}\; \Rightarrow \;
du = \frac{1}{(\frac{y}{x})^2+1}\cdot \frac{-y}{x^2}dx \; \Rightarrow \; du = \frac{-y}{x^2+y^2}dx \\ \text{and let } dv=dx \; \Rightarrow \; v=x \\ \text{Therefore} \\ \int u \;dv = uv \; - \; \int v\;du \;\; \Rightarrow \; \; \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If a,b,c are in AP and $a^2,b^2,c^2$ are in HP, then prove either $a=b=c$ or $a,b,- \frac c2 $ are in GP As the title says.
Although first part of the proof is obvious, I'm still able to prove it.
And for the second part, I'm essentially trying to prove $b^2=-c/a$ (which is possible only when c<0 Xor a<0).
The relati... | The AP condition gives $2b = a+c$ and the HP condition gives
$$\frac2{b^2}= \frac1{a^2}+\frac1{c^2} \iff \frac8{(a+c)^2}=\frac{a^2+c^2}{a^2c^2} \iff (a-c)^2(a^2+4ac+c^2)=0$$
Now either $(a-c)^2=0 \implies a=b=c$
or $a^2+4ac+c^2 = 0 \implies (a+c)^2+2ac=0 \implies 2b^2+ac=0 \implies a, b, -\dfrac{c}2$ are in GP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Give the explicit form of the following parametrized surface Let $\boldsymbol{X}:\boldsymbol{R}^2\to \boldsymbol{R}^3$ be the paramtrized surface given by$$\boldsymbol{X}(s,t)=(s^2-t^2,s+t,s^2+3t)$$
I'm trying to describe the parametrized surface by an equation of the form $z=f(x,y)$
| You have three equations: $x=s^2-t^2$, $y=s+t$ and $z=s^2+3t$. You need to eliminate $s$ and $t$.
Rearranging $y=s+t$ gives $t=y-s$. Substituting this into the other two equations gives
$$\begin{eqnarray*}
x &=& 2ys-y^2 \\
z&=& s^2-3s+3y
\end{eqnarray*}$$
Now we can rearrange $x = 2ys-y^2$ to solve for $s$. We get $\di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating a limit of integral Computing the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$
I made the substiution $t=nx$ then, we have: $$I=\frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan t}dt$$ where $I$ is the inside integral. How would you continue?
| Asymptotics of the integral:
When $x$ is large,
\begin{align}
\frac{1}{\arctan{x}}
=&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\arctan\left(\frac{1}{x}\right)}\tag1\\
=&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)}\tag2\\
=&\frac{2}{\pi}\left(1+\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Integral with contour integration I want to evaluate the integral:
$$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
using contour integration.
I re-wrote it as: $\displaystyle \int_{0}^{\infty}\frac{2x^2-1}{x^4+1}\,dx$. I am considering of integrating on a semicircle contour with center at the origin. I considered the fu... | Let's try and hope I have not made any mistakes here.
We will calculate the integral:
$$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
Since the integrand function is even we can rewritte it as:
$$\int_{-\infty}^{\infty}\frac{2x^2-1}{x^4+1}\,dx=2\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
Considering the function $\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
| Have a look at the series representation of $e^x$ which is
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$
Taking derivative of this gives
$$\left(e^x\right)'=\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right)'=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 4
} |
Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$ I am working on $\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx$
First: I use integrating by part then get
$$ x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx $$
then I have $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$ because $x\arctan(\sin x)... | Let be
$$\begin{align} I=\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\operatorname{d}x&=-\int_0^\pi \arctan(\sin x)\operatorname{d}x \\
&=-2\int_0^{\pi/2} \arctan(\sin x)\operatorname{d}x\\
&=-2\int_0^1\frac{\arctan t}{\sqrt{1-t^2}}\operatorname{d}t
\end{align}$$
Let $$ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan at}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving $\frac{\sin x}{x}$ Can anybody explain me, why the following inequality is true?
$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|... | This uses the fact that $$\int_a^b |f(x)|\ge (b-a)\inf_{x\in(a,b)} |f(x)|$$and that $\left|\frac{\sin x}x\right|$ is minimised on $(k\pi +\frac\pi4, (k+1)\pi -\frac\pi4)$ when $x=(k+1)\pi -\frac\pi4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there a "counting groups/committees" proof for the identity $\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$? This is exercise number $57$ in Hugh Gordon's Discrete Probability.
For $n \in \mathbb{N}$, show that
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$
My algebraic solution:
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$... | $\binom{\binom n2}2$ counts pairs of (distinct) 2-element subsets of $n$-element set. Union of such pair is either 4-element set (and each 4-element set is counted 3 times: there are 3 ways to divide 4-set into 2 pairs) or 3-element set (and each 3-element set is also counted 3 times). That gives $3\binom n4+3\binom n3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ without the use of a calculator. It is clear that $2003^4+1$ has a $082$ at the end of its number so $2003^4+1$ only has one factor of 2, while $2003^3+1$ has a $028$ at the end of its num... | Let $d=\text{ GCD }(2003^3+1,2003^4+1)$, then $d$ divides $2003(2003^3+1)-(2003^4+1)=2003-1=2002=2\cdot 7\cdot 11 \cdot 13$, and since both of $2003^3+1$ and $2003^4+1$ are even we have $d\geq 2$.
Since $2003=2002+1=2\cdot 7\cdot 11 \cdot 13 + 1$ it follows $2003^4+1$ gives residue $2$ when is divided by $7, 11$ or $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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.