Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the s... | Hint:
There's no need to split the integral that way (IMO), we can instead write, $$\begin{align}
\int \frac{3x+2}{x^2+x+1}\,\mathrm dx&=\frac{3}{2}\int\frac{2x+\tfrac{4}{3}}{x^2+x+1}\,\mathrm dx\\
&=\frac32\int\dfrac{2x+1+\tfrac13}{x^2+x+1}\,\mathrm dx\\
&=\dfrac32\left(\int\dfrac{2x+1}{x^2+x+1}\,\mathrm dx+\int\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $16\cos^5A-20\cos^3A+5\cos A=\cos5A$
Prove that
$$16\cos^5A-20\cos^3A+5\cos A=\cos5A$$
My solution begins here;
$$
\begin{align}
\text{RHS} & =\cos5A \\
& =\cos(A+4A) \\
& =\cos A\cos4A-\sin A\sin4A \\
& =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\
& =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\
& =... | Use
$$
2\cos x=e^{ix}+e^{-ix}
$$
Then
$$
32\cos^5x=e^{5ix}+5e^{3ix}+10e^{ix}+10e^{-ix}+5e^{-3ix}+e^{-5ix}=
2\cos5x+10\cos3x+20\cos x
$$
Moreover
$$
8\cos^3x=e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}=
2\cos3x+6\cos x
$$
Therefore
\begin{align}
16\cos^5x-20\cos^3x+5\cos x
&=(\cos5x+5\cos3x+10\cos x)-5(\cos3x+3\cos x)+5\cos x\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the value of :Solving $\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$ I have trouble finding the value of the following limit:
$$\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$$
For now I have rewritten the term into:
$$ \lim_{n \to \infty} \dfrac{\sin\left({\sqrt{n+3... | $\sqrt{n + 3} - \sqrt{n - 2} = \frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}$
This yields the following term:
$$\sqrt{n}\sin(\sqrt{n + 3} - \sqrt{n - 2}) = \frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} \frac{\sin\left(\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}\right)}{\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}}$$
Now use the fact that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding the minimum of $x^2+y^2$ for $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$. If $x,y \in \mathbb {R}$, find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$.
This problem was inspired by a problem which asked if $x,y \in \mathbb {R}$ and $xy \neq 0$, find the minimum of $x^2+y^2$ when $xy(x... | EDIT: Just realized $xy≠0$. Here is the sadly tough solution.
Instead of writing out the minutiae, I will just sketch out the solution method.
To begin, rewrite the initial expression as follows:
$xy(x-y)^2(x^2+xy+y^2)=(x+y)(x^2+y^2-xy)$
We did this by factoring the sum and the difference of cubes. (For which there a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find the limit $\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)$ $$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)\;=\;?\quad(n\in I) \\ \text{where $\lfloor\cdot\rfloor$ is the greatest integer function.}$$
This is what I did:
Since $[x] = x - ... | For every natural number, we have
$$\lfloor \sqrt{n^2+n+1} \rfloor =n$$
because $n^2\leq n^2+n+1< n^2+2n+1=(n+1)^2$. So we get
$$\begin{align}\lim_{n\to\infty} (\sqrt{n^2+n+1}- \lfloor \sqrt{n^2+n+1} \rfloor)&=\lim_{n\to\infty}(\sqrt{n^2+n+1}-n)\\
&=\lim_{n\to\infty}\frac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\\
&=\lim_{n\to\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$$
Now by inductive hypothesis:
$$\frac{1\cdot2^2+2\cdot3^2+\c... | This is also a proof by induction and the result is more general.
The identity
$$\sum_{m=0}^{n}\binom{m}{k}=\binom{n+1}{k+1} \tag{1}$$
can be easily proofed by induction using the well known identity of the Pascal Triangle
$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1} \tag{2}$$
$(1)$ says that the sum of this $k$-th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i... | Let $\cos\theta=\frac2{\sqrt5}$. Then
$$\cos3\theta=4\cos^3\theta-3\cos\theta=\frac2{5\sqrt5}$$
and so
$$\tan3\theta=\frac{\sqrt{(5\sqrt5)^2-2^2}}2=\frac{11}2\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Why is ln(x) bigger than log(x)? I couldn't find any results. I would like to know why $\ln x$ stays higher than $\log x$. Does it have to do with ase $10$ of $\log$? base $e$ with $\ln x$?
| Take two different bases: $2$ and $4$. Now let's look at what exponent is required to equal some numbers (this is what the logarithm computes):
\begin{align}
2^? = 16 \rightarrow 2^4 = 16 && 4^? = 16 \rightarrow 4^2 = 16\\
2^? = 64 \rightarrow 2^6 = 64 && 4^? = 64 \rightarrow 4^3 = 64 \\
2^? = 1024 \rightarrow 2^{10} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:
*
*$0^2+1^2=1^2$
*$3^2+4^2=5^2$
*$20^2+21^2=29^2$
*$119^2+120^2=169^2$
*$696^2+697^2=985^2$
*$4059^2+4060... | The values $\; 1,5,29,169,\cdots\;$ are subset of pell number as shown in the OEIS sequence A000129. You are in fact using the odd-value numbers in the sequence.
If we define Euclid's formula for generating Pythagorean triples as:
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
the $(m,k)$ values needed to gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
} |
Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus.
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2... | $$y_1 = x_1^2, \qquad y_2 = -x_2^2 - 16x_2 -65$$
$$d = \sqrt{(x_2-x_1)^2 + (-x_2^2 - 16x_2 -65-x_1^2)^2}$$
Thus we want to minimize the following: (Note the change of variables, simply to make the notation easier to follow)
$$f(u,v) = (u-v)^2 + (-u^2 - 16u -65-v^2)^2$$
$$f_u(u,v) = 4u^3 + 96u^2 + u(4v^2 + 774) + 32v^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Number of terms in the expansion of $\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$
Number of terms in the expansion of $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$$
$\bf{My\; Try::}$ We can write $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n=\frac{1}{x^{2n}}\left(1+x+x^2\right)^n$$
Now we have to calculate number of t... | When you expand $(1 + x + x^2)^n$, the highest degree is $2n$, and the lowest is 0. None of them is vanished, and it can be seen easily that there is one term of any degree between 0 and $2n$. So there should be $2n + 1$ terms.
Cheers,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Can you multiply a natural log term by a constant to take power inside log term? For Calculus 2 homework I must prove that $\displaystyle\int \frac{du}{u^2-a^2} = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C$
The instructor wants us to use trigonometric substitution to solve.
My question is, may I take the term (acqu... | The substitution you can use in this case is
$$
u=\frac{a}{\cos2t}
$$
so
$$
\frac{1}{u^2-a^2}=\frac{\cos^22t}{a^2\sin^22t},
\qquad
du=\frac{2a\sin2t}{\cos^22t}
$$
and the integral becomes
$$
\frac{2}{a}\int\frac{1}{\sin2t}\,dt=
\frac{1}{a}\int\frac{\cos^2t+\sin^2t}{\sin t\cos t}\,dt=
\frac{1}{a}\int
\left(\frac{\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
integrate $\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$$
$(2-x)^2(4+x^2)=x^4-4x^3+8x^2-16x+16$
$$\int \frac{x^4+4x+6}{x^4-4x^3+8x^2-16x+16}$$
Is there is anything to do else than long division?
| I know that you asked without long division, but
$$\int \frac{x^4+4x+6}{(x-2)^2(x^2+4)}\stackrel{\text{L.div}}{=}\frac 1 8 \int\frac{11x-8}{x^2+4}dx+\frac{21}{8}\int\frac{dx}{x-2}dx+\frac{15}{4}\int \frac{dx}{(x-2)^2}+\int1 dx=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
For all $x$ which are real numbers, prove that
$$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$
I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
| I will try to show
$\lfloor nx \rfloor = \sum_{k=0}^{n-1} \lfloor x+\frac{k}{n} \rfloor$.
Let
$m = \lfloor x \rfloor$
and
$d = x - m$,
so
$0 \le d < 1$.
Let
$j
=\lfloor nd \rfloor
$,
so
$0 \le j \le n-1
$
and
$\frac{j}{n}
\le d
< \frac{j+1}{n}
$.
If
$0 \le k \le n-j-1$,
$\begin{array}\\
m
&\le m+d+\frac{k}{n}\\
&< m+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove by induction that: $\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i}$ Basis step:
For $n=1$, equation holds.
Inductive step: Now,
$$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i} \tag{i. h.}$$
Now we want to show that
$$\sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^{n+1}_{i=1}\frac{1... | Change the indices in the sum: $$\sum_{i=1}^n\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{i=0}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\
=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{i=1}^{n+1}\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots
$$
up to $n$ terms. I need help in solving this sum. I trie... | HINT:
The $r(\ge1)$th term
$$=\dfrac1{r(r+1)(r+2)(r+3)} =\dfrac{r+3-r}{3r(r+1)(r+2)(r+3)}=v_r-v_{r+1}$$
where $v_m=\dfrac1{3m(m+1)(m+2)}$
$$\sum_{r=1}^n u_r=\sum_{r=1}^n(v_r-v_{r+1})=v_1-v_{n+1}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Induction Example Here's a (possibly sloppily written) Induction proof that has me stumped. It's the first example from here and supposedly the statement holds?
(*) For n > 1, 2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^n+1 – 2
Let n = 1. Then:
2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^1 = 2
...and:
2^n+1 – 2 = 2^1+1 – 2 = 2^2 – 2... | The goal of the induction step is to show that (*) holds for $n = k+1$, in particular, we want to show $$2 + 2^2 + \cdots + 2^{k+1} = 2^{(k+1)+1}-2$$
The induction hypothesis says $2 + 2^2 + \cdots + 2^{k} = 2^{k+1}-2$. So we start with the LHS of what we want to show, and transform it to the RHS, using things we know ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there a mathematical way to solve this problem? The question goes something like this:
How many different ways can you add up 2, 3, and 5 to get a sum of 12. Numbers can be repeated, and all three numbers do not have to be used for each solution. For instance, 5 + 2 + 2 is an answer, and so is 2 + 5 + 2.
The answe... | Here's a way to enumerate all sums of $5$, $3$, and $2$ that add up to $n$,
if we do not consider the order of the terms in the sums to matter.
Let's consider just sums of $2$ and $3$ at first.
If $n > 1$, there is at least one such sum that adds up to $n$.
To find a specific sum, take the remainder of $n$ after dividi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove Lucas numbers and Fibonacci numbers relation $F_n = \frac{L_{n - 1} + L_{n + 1}}{5}$ $F_n$, is the $n$th term of the Fibonacci sequence.
$L_n$ is the $n$th Lucas number.
I want to prove that $F_n = \dfrac{L_{n-1}+L_{n+1}}5 $.
Things I know:
*
*$L_n$+$L_{n+1}=L_{n+2}$
*$F_{n-1} + F_{n+1}=L_n$.
The base case... | The Fibonacci numbers are defined recursively by
$$
F_n =
\begin{cases}
1 & \text{if $n = 1$}\\
1 & \text{if $n = 2$}\\
F_{n - 1} + F_{n - 2} & \text{if $n > 2$}
\end{cases}
$$
and the Lucas numbers are defined recursively by
$$
L_n =
\begin{cases}
2 & \text{if $n = 0$}\\
1 & \text{if $n = 1$}\\
L_{n - 1} + L_{n - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding formula for sum $S_n = \sum_{i=1}^n i(i-1)$ How do I find the formula for this sum
$S_n = \sum_{i=1}^n i(i-1)$
Wolfram alpha gives me the correct formula which is $ \frac{1}{3} (n-1)n(n+1)$, but I'm interested in how I get to that and how I can approach this kind of problem in general.
| This is much simpler. Let's use the notation $x^{\underline{k}} = x (x - 1) \dotsm (x - k + 1)$ for arbitrary $x$ and $k \in \mathbb{N}_0$.
Prove by induction on $n$ that:
$\begin{align}
\sum_{0 \le k < n} k^{\underline{m}} = \frac{n^{\underline{m + 1}}}{m + 1}
\end{align}$
Base: For $n = 0$, this is true, the left ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How do I solve the system of congruences $x \equiv 1 \pmod{3}$, $x \equiv 2 \pmod{4}$, $x \equiv 4 \pmod{5}$, $x \equiv 2 \pmod{7}$? Sorry for the weird title, but I don't know how else to phrase this.
Say I have a set of numbers, and the remainders each get by dividing them by a certain number. For instance:
$x \equi... | Since $x\equiv2\pmod{4}$ and $x\equiv2\pmod{7}$, we must have that $x\equiv2\pmod{28}$.
We can use the Extended Euclidean Algorithm as implemented in this answer to solve the following equivalences.
Since $140=5\cdot28$, solve
$$
\begin{align}
x&\equiv1\pmod{3}\\
x&\equiv0\pmod{140}
\end{align}
$$
getting $x\equiv280\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \... | As the terms grow linearly, their average is also the average of the extreme values. Hence the sum for the $n$ first terms
$$S_n=n\cdot\frac{0+8(n-1)}2+1=4n^2-4n+1=(2n-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
A hemisphere is inscribed in a cube Finding the largest cube inscribed in a hemisphere has been considered here previously. So let's consider the reverse relationship:
A hemisphere is inscribed in a cube with an edge of $1m$. What is the maximum radius of the hemisphere ?
Obviously a whole sphere of radius $\frac{1}... | I forward and edit this proof from a middle school math teacher
Note that a sphere inscribe to a cubiod is tangent to all the six faces of the cuboid, then a hemisphere inscribe to a cuboid should be tangent to at least three cuboid-faces adjacent to each other.
This means, the (sphere) center of the hemisphere locates... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$. I found this problem in a book, I can't solve it unfortunately.
Prove that for all integer values $n$, $n^9 - 6n^7 + 9n^5 - 4n^3$ is divisible by $8640.$
So far I've noticed that $8460 = 6! \times 12$, also I've tried to simplify that expression and I've found that ... | We can further note that $-1$ is a root of $n^3-3n-2$, and that $1$ is a root of $n^3-3n+2$.
You can find these roots by the Rational Root theorem:
Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.
Other techniq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
An alternative way to compute $a^n+b^n+c^n+d^n$ I was doing the following problem:
given $a, b, c, d \in \mathbb{R}$ and
$a+b+c+d=1$, $a^2+b^2+c^2+d^2=2$, $a^3+b^3+c^3+d^3=3$ and $a^4+b^4+c^4+d^4=4$.
Find $a^n+b^n+c^n+d^n$ (I am looking for a closed kind of formula)
Now, I can already do it in one way. We can find the... | If we take $a,b,c,d$ as the roots of a monic polynomial
$$ p(x)=x^4-e_1 x^3+e_2 x^2-e_3 x+e_4 $$
Newton's identities give:
$$ \left\{\begin{array}{rcl}e_1 &=& 1\\ 2e_2 &=& e_1-2 \\ 3e_3 &=& e_2 - 2e_1 +3 \\ 4e_4 &=& e_3-2e_2+3e_1-4 \end{array}\right. $$
hence it follows that:
$$ e_1=1,\quad e_2 = -\frac{1}{2},\quad e_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ax^2 + (b+c)x + d+ e=0$ has solutions on $[1,\infty)$, so that $ax^4+bx^3+cx^2+dx+e=0$ has solutions on $[1,\infty)$
Let $ax^2 + (b+c)x + d+ e=0, a, b, c, d, e \in \mathbb R$ has solutions on the interval $[1, +\infty)$. Prove that the polynomial $f(x) = ax^4 + bx^3 + cx^2+ dx + e =0$ has solutions on $[1, +\infty... | Nobody can prove that, since it is false. Suppose that $a=b=1$, that $c=-5$, that $d=4$, and that $e=0$. Then\begin{align}ax^2+(b+c)x+d+e=0&\iff x^2-4x+4=0\\&\iff(x-2)^2=0\\&\iff x=2.\end{align}So, yes, $ax^2+(b+c)x+d+e=0$ has solutions on $[1,+\infty)$ (actually, all of its solutions belong to that interval). However,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Good idea to find below sum this is the question.
Purpose is to find sum of $$f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\left(\frac{3}{14}\right)+...+f\left(\frac{13}{14}\right)$$for function $$f(x)=\frac{4^x}{4^x+2}$$Is there a nice method to find $f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\l... | Yup, a nice thing to notice is the value of
$$
\begin{align}
f(x)+f(1-x)
&=\frac{4^x}{4^x+2}+\frac{4}{4+2\cdot4^x}\\
&=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}\\[3pt]
&=\cdots\,?
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Vectors that form basis in $\mathbb{R}^3$ and transition matrix
Consider the following set of vectors in $\mathbb{R}^3:$
$u_0 = (1,2,0),~ u_1 = (1,2,1), ~u_2 = (2,3,0), ~u_3 = (4,6,1)$
Explain why each of the two subsets $B_0 = \left\{u_0, u_2,
u_3\right\}$ and $B_1 = \left\{u_1, u_2, u_3\right\}$ forms a basis of
... | The given vector $x$
$$
x = 4 e_1+ 4 e_3
$$
can be represented according the first or the second basis too.
\begin{align}
x
&= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)}
= (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\
&= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} b_3^{(1)}
= (x_1^{(1)},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!
| Hint. You may use an integration by parts
$$
\begin{align}
\int(x+2)\ln(x-3)\:dx&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+2)^2}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+7)(x-3)+25}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\left(x+7+\frac{25}{x-3}\right)\:dx.
\end{align}
$$ Then conclude e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 0
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Solving linear recursive equation $a_n = a_{n-1} + 2 a_{n-2} + 2^n$. I wish to solve the linear recursive equation:
$a_n = a_{n-1} + 2a_{n-2} + 2^n$, where $a_0 = 2$, $a_1 = 1$.
I have tried using the Ansatz method and the generating function method in the following way:
Ansatz method
First, for the homogenous part, ... | The $1-2z$ in your implicit equation for $A(z)$ is not correct, it should be $1-3z$:
$$\begin{eqnarray*}A(z)&=& a_0 + a_1 z + \sum_{n \geq 2} a_n z^n \\ &=& 2+ z + \sum_{n \geq 2} a_{n-1} z^n + \sum_{n \geq 2} 2 a_{n-2} z^n + \sum_{n \geq 2} 2^n z^n \\
&=&2+ z + \sum_{n \geq 1} a_n z^{z+1} + \sum_{n \geq 0} 2 a_n z^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Matrix representation of Heisenberg group Equiped with the law $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab'),$ the matrix representation of the Heisenberg group $H^3$ is given by
$$
\begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}; \quad
a,b,c\in \mathbb{R}.
$$
My question, what the... | I think the answer is as follows, we have:
*
*$$ \begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
1 & a' & c'\\
0 & 1 & b'\\
0 & 0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
1 & a+a' & c+c'+ab'\\
0 & 1 & b+b'\\
0 & 0 & 1\\
\end{pmatrix} $$
is corresponds to $(a,b,c)\circ (a',b',c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find sum of $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$ Find the sum of the following series :
$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots$$ and
$$1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\fra... | The first is $\sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{2(2i+1) + 2}]= \sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{4(i+1)}]$
which is $\sum_{i=0}^{\infty}[\frac {4(i+1)}{4(2i+1)(i+1)} - \frac {2(i+1)}{4(2i+1)(i+1)} - \frac {2i+1}{4(2i+1)(i+1)}]=\sum_{i=0}^{\infty}\frac {1}{4(2i+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2... | Hint: $f(x,y) = f(x) = \dfrac{4}{4-x^2} +\dfrac{9x^2}{9x^2-1}$. Let $u = x^2$, then $0 \leq u < 4$, and you get a relatively nice rational function in $u$ that you can use derivative to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Problem with indefinite integration I introduce shift $t^5=2x -3$,but it doesn't help $$\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx$$
| Starting from
$$I=\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\text dx$$
One substitution approach is $u=\sqrt[5]{(2x-3)^4},\,\text du=\frac 8{5\sqrt[5]{2x-3}}\text dx,$ which yields
$$I=\frac 58\int\frac{2u^\frac 14-1}{u^\frac 54+1}\text du$$
If $v^4 = u,4v^3\text dv=\text du$, then we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$A, B$ and $C$ can do a piece of work
$A, B$ and $C$ can do a piece of work in $18$, $24$ and $30$ days respectively. $A$ and $B$ work together for a certain number of days. Then $B$ leaves and $C$ joins. They work together for half the number of days for which $A$ and $B$ had worked together. After that $A$ goes away... | Let:
$x$ the number of days that $A$ and $B$ worked together,
then $A$ and $C$ worked together $x/2$ days,
and $B$ and $C$ worked $x/3$ days together.
So, we have
\begin{align}
x\left(\frac{1}{18}+\frac{1}{24}\right)+\frac{x}{2}\left(\frac{1}{18}+\frac{1}{30}\right)+\frac{x}{3}\left(\frac{1}{24}+\frac{1}{30}\right)&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Inequalities Using the AM-GM inequality, the Cauchy-Schwarz Inequality, and brute force Prove that if $x,y,z$ are positive real numbers then the following inequality holds
$$ \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . $$
I have tried everything to listed in ... | We have $a^2+b^2\ge 2ab$ for all $a,b\in\mathbb R$, because this is equivalent to $(a-b)^2\ge 0$, which is true.
Since $x,y,z>0$, we get:
$$\frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2}$$
$$\leq \frac{x+y}{2xy}+\frac{y+z}{2yz}+\frac{z+x}{2zx}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
Equality holds if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.
Here,
*
*$\angle DOE=45°$
*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.
*$\triangle DEF$ i... | The hardest step is the first one: guessing the correct answer. By symmetry, it's plausible that $OF$ bisecting $\angle DOE$ will give either a maximum or minimum length. Another likely guess is when $O$, $D$, and $E$ are collinear, but for the given angles of $45^\circ$ and $60^\circ$ that would make $F$ lie outside t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
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Simple epsilon-delta question Is it true that for every $ε > 0$, there is $δ > 0$, such that $0 < |x−2| < δ ⇒ |(x^2 −x)−2| < ε$?
Now I know that $|(x^2 −x)−2|$ is same as $|(x-2)(x+1)|$, but I am not sure how to link that with the first bit of info given. In general epsilon-delta proofs confuse me.
So I start by sayin... | Let the $\epsilon = \epsilon_0$ satisfying $|x^2-x-2| < \epsilon_0$. Initially choose $\delta$ to be $1$. We will refine this delta.
$-\epsilon_0 < x^2-x-2 < \epsilon_0$
$\implies -\epsilon_0+\frac{9}{4} < x^2-x+\frac{1}{4} < \epsilon_0+\frac{9}{4}$
$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 < \epsilon_0+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integral of $\int\frac{dx}{(x^4-1)^{3/2}}$ Can $$\int\frac{dx}{(x^4-1)^{3/2}}$$ be represented in terms of elementary integrals?
I have tried multiple substitutions, such as $x^2=\sec(t)$ or $\sqrt{x^4-1}=t$, but nothing seems to be working. Please provide some insight.
| Case $1$: $|x^4|\leq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{i}{(1-x^4)^\frac{3}{2}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$
Case $2$: $|x^4|\geq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
| For $x\ne n\pi$ (it's true, because $ \cos{x}\cdot\cos{2x}\cdot\cos{4x}\,\Big|_{x=n\pi} \ne\frac{1}{8}$)
$$ \cos{x}\cdot\cos{2x}\cdot\cos{4x}=\frac{\sin{x}\cdot\cos{x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}} = \\
=\frac{1}{2}\cdot \frac{\sin{2x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{4}\cdot \frac{\sin{4x}\cdot\cos{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3]... | This is an elementary but very good question. We have $$k = 2\sqrt{3} + 3\sqrt[3]{2} - 1$$ so that $$\sqrt[3]{2} = \frac{k + 1 - 2\sqrt{3}}{3}$$ and if $k$ is rational then it means that that $\sqrt[3]{2}$ is a quadratic irrationality i.e. it is a root of a quadratic equation with rational coefficients.
Hardy proves in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Inequality with Square roots and Condition Given $a+b+c=1$, prove that $\sqrt{a+\frac{(b-c)^2}4}+\sqrt{b}+\sqrt{c}\le \sqrt{3}$.
So far, I have tried to apply cauchy schwarz somehow because this works well with square roots and the inequality signs match up. However, this nonhomogeneity is tripping me up, so I would li... | By Cauchy-Schwarz inequality
$$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2
\le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2)
$$
$$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$
since $1-a=b+c$
$$\Longleftrightarrow \dfrac{(b-c)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$ I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.
Find $x$, $y$ integers such that
$$ (x+1)^2 + (y+1)^2 + xy(x+y+3)=2 $$
Some basic algebra will lead to:
$$
(x+y)^2 + x(xy+2) + y(xy+2) +... | The origin $(0,0)$, $(0,-2)$ and $(-2,0)$ are the only diophantine (integer) solutions. Simplifying the equation and taking note of the symmetry about $y=x$ ($x$ & $y$ are interchangeable, so we will eventually only see symmetric terms such as xy and x+y), we see that
$$\begin{align}
0 &= (x+1)^2 + (y+1)^2 + xy(x+y+3) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find Nth formula of recursive formula $a_n=a_{n-1}+n(n-1)a_{n-2}$ $$a_n=a_{n-1}+n(n-1)a_{n-2}$$
$$a_0=1, a_1=-\frac{1}{2}$$
Is it possible to find explicit formula for $a_n$ just by using $a_0$ and $a_1$?
I know how to solve this problem if $a_n=Aa_{n-1}+Ba_{n-2}$ where $A$ and $B$ is some real number(constant), but i... | For the second question, note that
$$\begin{align*}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}&=\begin{bmatrix}Aa_{n-1}+Ba_{n-2}\\ a_{n-1}\end{bmatrix}=\begin{bmatrix}A &B\\ 1 & 0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}A&B\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\ a_{n-3}\end{bmatrix}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would ... | Raising to the $6^{th}$ power we get
$$
\eqalign{
64i^6\sin^6 x &= \bigl(z-{1\over z}\bigr)^6 \cr
&= z^6 - 6z^5{1\over z} + 15z^4{1\over z^2} -20z^3{1\over z^3} + 15z^2{1\over z^4} - 6z{1\over z^5} + {1\over z^6} \cr
&= z^6 - 6z^4 + 15z^2 -20 + 15{1\over z^2} - 6{1\over z^4} + {1\over z^6} \cr
&= z^6 + {1\over z^6} - 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Integrate the expression using any method I have an expression to integrate, which I don't know how to integrate, tried to factorize and solve, but got really ugly expression at the end. Would appreciate any help:
$$\int \frac{dx}{(x-1)\sqrt{x^2-x+1}}$$
| That edit makes it easier. Let $x-\frac12=\frac{\sqrt3}2\tan\theta$. Then
$$\begin{align}\int\frac{dx}{(x-1)\sqrt{x^2-x+1}} & =\int\frac{d\theta}{\frac{\sqrt3}2\sin\theta-\frac12\cos\theta}=\int\frac{d\theta}{\sin(\theta-\pi/6)} \\
& =-\ln(\csc(\theta-\pi/6)+\cot(\theta-\pi/6))+C_1 \\
& =\ln\left(\frac{\sin(\theta-\pi/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerat... | $\sin(1+x)=x+\mathcal O(x^3)\;$, and likewise for the logarithmic function, so:
$$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\cong$$
$$\frac{\left(\sin^3x\cos^2x\right)\cot\left(x^3\right)\left(x^4\right)}{\left(\sqrt{x^2+2}-\sqrt2\right)\left(x^2\right)}=\cos x^3\;\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the two powers are equal
Prove that: $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1} = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}$$ where: $$a = \dfrac{1+\sqrt{3}}{\sqrt{2}}.$$
This seemed like a very challenging question but the fac... | HINT:
Observe that $a^2=2+\sqrt3$
$\implies1+\dfrac{\sqrt3}2=\dfrac{a^2}2$
and $1-\dfrac{\sqrt3}2=\dfrac1{2a^2}$
Use Componendo and Dividendo to find $\dfrac{a^2-1}{a^2+1}=?$
Now
$$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}=\left[\left(\dfrac{a^2}2\right)^{180}-\left(\dfrac1{2a^2}\right)^{180}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let... | Assume $a\in\mathbb{R}^+$
$$\int_{0}^{a}\frac{\sqrt{a+x}}{\sqrt{a-x}}\space\text{d}x=$$
Substitute $u=\sqrt{a+x}$ and $\text{d}u=\frac{1}{2\sqrt{a+x}}\space\text{d}x$.
THis gives a new lower bound $u=\sqrt{a+0}=\sqrt{a}$ and upper bound $u=\sqrt{a+a}=\sqrt{2a}$:
$$\int_{\sqrt{a}}^{\sqrt{2a}}\frac{2u^2}{\sqrt{2a-u^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding dy/dx with two elements? I don't understand the problem. How can I find $\frac{dy}{dx}$, given that $y = \frac{u^2 - 1}{u^2 + 1}$ and $u = \sqrt[3]{x^2 + 2}$?
For the next step, the book says:
$\frac{dy}{du} = \frac{4u}{(u^2+1)^2}$.
Am I supposed to place the value of $u$ into the $y$ fraction?
If I put it into... | You have two functions:
$$
y=f(u)=\frac{u^2-1}{u^2+1} \qquad u=g(x)=\sqrt[3]{x^2+1}
$$
so $y$ is a composite function:
$$
y=f(g(x))=\frac{\left(\sqrt[3]{x^2+1}\right)^2-1}{\left(\sqrt[3]{x^2+1}\right)^2+1}
$$
Maybe that, in this form, you know how find the derivative using the chain rules:
$$
\frac{dy}{dx}=f'(g(x))g'(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the integral $\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$ $$\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}$$
$$\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c$$
$$\int \frac{1}{2\sqr... | Hint. Let $u^{1/5}=v$ so $$\int\frac{du}{u^{6/5}+u^{1/5}}=5\int\frac{v^4}{v^6+v}dv=5\int\frac{v^3}{v^5+1}dv$$
and factorize $v^5+1$ as $$(1 + v)(v^4 - v^3 + v^2 - v + 1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and vari... | $$\int\frac{2x^3+3x^2}{\left(2x^2+x-3\right)\sqrt{x^2+2x-3}}\space\text{d}x=$$
Using the Euler subsitution:
$x=\frac{u^2+3}{2u+2}$ and $\text{d}x=\left(\frac{2u}{2u+2}-\frac{2\left(u^2+3\right)}{\left(2u+2\right)^2}\right)\space\text{d}u$.
And when we substitute back we get $u=x+\sqrt{x^2+2x-3}$:
$$\int\frac{\left(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integral of product of two normal distribution densities I want to compute the integral:
$\displaystyle \int^{\infty} _{-\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-x)^2}{2}} \frac{1}{\sqrt{2\pi}ab} e^{-\frac{x^2}{2(ab)^2}} dx$
Maybe we can use that for a normal distribution with mean $\mu$ and variance $\sigma^2$ we ... | Actually, you're on the right track. Just keep on going with
$$\begin{align}(ab)^2(y^2-2xy+x^2)+x^2 & =(a^2b^2+1)x^2-2a^2b^2yx+a^2b^2y^2 \\
& =(a^2b^2+1)\left[x^2-2\frac{a^2b^2y}{a^2b^2+1}x\right]+a^2b^2y^2 \\
& =(a^2b^2+1)\left[\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2-\frac{a^4b^4y^2}{(a^2b^2+1)^2}\right]+a^2b^2y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Elliptic integrals Knowing that
$$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}t+b^2\sin^{2}t}dt＝\int_{0}^{2\pi}\sqrt[]{1+\cos^{2}t}dt$$
how to find the value of $a$ and $b$?
Elliptic integral? Thank you!
| $$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}(t)+b^2\sin^{2}(t)}dt = |a|\int^{2\pi}_{0}\sqrt{1-\left(1-\frac{b^2}{a^2}\right)\sin^{2}(t)}dt =4|a|\text{E}\left(1-\frac{b^2}{a^2}\right)$$
E$(X)$ is the complete elliptic integral of second kind.
$$\int^{2\pi}_{0}\sqrt{1+\cos^{2}(t)}dt =\sqrt{2}\int^{2\pi}_{0}\sqrt{1-\frac{1}{2}\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral through $u$-substitution v. multiplying out I have the integral:
$ \int (x^2)(x^3-1)^2 \, dx $ Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $ which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$.
However, if rather than using $u$-substitution, I multiply wit... | Ahhh, the question is what is the function at the point (1,0). And the constant will change if I use the first equation v. the second. So the first equation will have a constant of 0, which has a solution function of $ \frac {1}{3} * \frac {(x^3-1)^3}{3}$ with its constant = 0.
And the second way $ \frac{x^9}{9} - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
partial fractions
$$\frac{1}{1-x^2}$$
$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
|
Notice:
*
*$$\frac{1}{1-x^2}=\frac{1}{-(x-1)(x+1)}=-\frac{1}{(x-1)(x+1)}$$
So, we get:
$$-\frac{1}{(x-1)(x+1)}=-\frac{a}{x-1}-\frac{b}{x+1}$$
And, now we can see that:
$$a(x+1)+b(x-1)=1\Longleftrightarrow a-b+(a+b)x=1$$
So, we get:
*
*$$a-b=1\Longleftrightarrow a=b+1\Longleftrightarrow a=-\frac{1}{2}+1=\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3 Prove that if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3
| $$
(3k)^2 = 3k' \; , (3k+1)^2 = 3k'+1 \; and \; (3k+2)^2 = 3k'+4=3k''+1
$$
(so there's no square of the form $3k + 2$)
If $a = 3k$ or $b=3k$ we are done, assume the opposite:
$a\mod3 \neq b \mod 3 \neq 0$
$$
(3m+1)^2+(3n+2)^2 = 3p+1+4=3p'+2 = c^2
$$
(not possible)
$a\mod3 = b \mod 3 \neq 0 $
$$
(3m+1)^2+(3n+1)^2 = 3p+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral $\int_0^{2\pi} \frac{d \theta}{5-3 \cos \theta}$ $$\int_0^{2\pi} \frac{d \theta}{5-3 \cos \theta}$$
My attempt:
Let $z=e^{i\theta}$ which gives $d\theta = \frac{dz}{iz}$
Thus,
$$\oint_C \frac{1}{5-3(\frac{z+z^{-1}}{2})}\frac{dz}{iz}$$
$$=\frac{1}{i}\oint_C \frac{dz}{(3z-1)(z-3)}$$
We can ignore th... | It is true that in order to find the roots, you can multiply both sides of
$$
5z-\frac{3z^2} 2 - \frac 3 2 = 0
$$
by $-2$, getting
$$
-10z + 3z^2 + 3 = 0,
$$
and then using your favorite method of solving quadratic equations, concluding that the roots are $3$ and $1/3$.
But that does not mean that $-10z+3z^2 + 3z$ is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Jordan form exercise What am I doing wrong?
I've been learning how to put matrices into Jordan canonical form and it was going fine until I encountered this $4 \times 4$ matrix:
$A=\begin{bmatrix}
2 & 2 & 0 & -1 \\
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4 \\
\end{bmatrix}
$
Which has as only eigenvalue $\lam... | Take a look at
$$
(A-2I)^2 = \begin{bmatrix}
0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2 \\
\end{bmatrix}^2
$$
The $a_{1,2}$ entry is non-zero. What does that tell you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Sum of the n nth roots of unity 1) If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$
2) Then apparently: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}=\frac{0}{1-z}=0$
Could someone please explain why: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}$?
| Let $z^0+z^1+z^2...z^{n-1}= p(z)$.
Then $z \ p(z)=z^1+z^2+z^3...z^{n}$.
Hence $p(z)-zp(z)=1-z^n$.
$p(z)(1-z)=1-z^n$.
$p(z)=\frac{1-z^n}{1-z}$
If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. Then there is a formula (that is easy to prove) that states for any $i \in \mathbb{R}$, $z^i=\cos\frac{2i\pi}{n}+i\sin\frac{2i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
| If $\sin4x=\sin3x,$
$4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$ $$4\cos x\cos2x=3-4\sin^2x\iff4\cos x(2\cos^2x-1)=3-4(1-\cos^2x)$$
$$\iff8\cos^3x-4\cos^2x-4\cos x+1=0$$
Again if $\sin4x=\sin3x,$
$4x=n\pi+(-1)^n3x$ where $n$ is any integer
For even $n=2m$(say), $x=2m\pi\implies \sin x=?$ where $m$ is any in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Contour integral of $f(z) = \frac{1}{z^2+iz+6} $ Need help evaluating a certain contour integral.
$f(z) = \frac{1}{z^2+iz+6} $
Steps so far:
Poles: $ z^2+iz+6 \rightarrow \frac{-i \pm \sqrt{-1-24}}{2}=0 \rightarrow z_0 = +2i, -3i $
Residues: $ a_{-1} = \lim{z \to z_{i}} [ \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^n f(z)] ... | Let $f(z)$ be defined as
$$\begin{align}
f(z)&=\frac{1}{z^2+iz+6}\\\\
&=\frac{1}{(z+i3)(z-i2)}\\\\
&=\frac{1}{i5}\left(\frac{1}{z-i2}-\frac{1}{z+i3}\right)
\end{align}$$
Therefore, $f(z)$ has poles at $z=-i3$ and $z=i2$. Integration of $f(z)$ around a closed rectifiable contour $C$ yields
$$\oint_C f(z)\,dz=\begin{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Irreducible factorisation of polynomial over quotient field Let $F=\mathbb{Z}_3[x]/<x^2+1>$.
Factor $x^4+2$ into irreducibles in $F[x]$.
I know that $F$ is a field since $x^2+1$ is irreducible.
The usual way to find out that a polynomial is irreducible is that it has no roots. But how to do it if the element of F is no... | we find that $F = \mathbb{Z}_3[x] / (x^2+1)$ (which is the field with $9$ elements) has a cyclic multiplicative group with $8$ elements, whose generator is $x+1$
(proof : $(x+1)^2 = 2x$, $(2x)^2 = 2$, $2^2 = 1$, hence $(x+1)^8 = 1$ but $(x+1)^4 \ne 1$)
hence in $F$ the equation $y^4 = 1$ has $4$ solutions for $y$ : $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For large $n$, show that $\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $ nearly equals $\frac{1}{2}$. For large $n$, show that $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $$ nearly equals $\frac{1}{2}$.
Integrating by parts we get $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2}=\Bigg(\frac{x^n}{1+x^2}\Bigg)^{1}_{0} - \... | As commented:
$$\left.\left(\frac{x^n}{1+x^2}\right)\right|_0^1=\frac1{1+1}-\frac0{1+0}=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloo... | One way to do it is to write $x=kn+r+\epsilon$, where $0\le r\le n-1$ is an integer and $0\le\epsilon\lt1$, so that $0\le r+\epsilon\lt n$. Then
$$\left\lfloor{\lfloor x\rfloor\over n}\right\rfloor=\left\lfloor{kn+r\over n}\right\rfloor=\left\lfloor k+{r\over n}\right\rfloor=k=\left\lfloor k+{r+\epsilon\over n}\right\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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What is the indefinite integral of $\sqrt[4] {\tan \left( x\right) }$ $$\int\sqrt[4] {\tan \left( x\right) } dx$$
I'm really stuck right now with this integral, so any kind of advice would be appreciated.
Perhaps a pretty nifty substitution that can save me from partial fraction decomposition.
| $$ \small \left( x^2 + \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 + \sqrt{2 - \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 - \sqrt 2} \; x + 1 \right) = x^8 + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| Let $$I = \int\frac{x^2+20}{(x\sin x+5 \cos x)^2}dx$$
Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $x^{2\times 5-2} = x^8$
So $$I = \int\frac{x^8(x^2+20)}{x^8(x\sin x+5\cos x)^2}dx = \int\frac{x^8(x^2+20)}{(x^5\sin x+5x^4\cos x)^2}dx$$
Now put $x^5\sin x+5x^4 \cos x=t\;,$ Then $x^3\cos x(x^2+20)dx=dt$
So $$I = \int x^5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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What is the infinite sum $$S = {\frac{2}{5} \\+ \frac{2}{5}\cdot\frac{3}{7} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{... | If we group the series in units of pair, the first few pairs are
$$\require{cancel}
\begin{array}{rll}
p_0 :& \frac25 \left(1+\frac37\right) &= \frac25 \frac{10}{7} = \frac47\\
p_1 :& \frac{\color{red}{2}}{\cancel{5}}\frac37\frac{\cancel{5}}{\color{red}{11}}\left(1+\frac{9}{13}\right)
&= \color{red}{\frac{2}{11}}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$
Find the minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$
$\alpha=\sqrt{3-2\sqrt{2}} \implies -\alpha^2-3=2\sqrt{2} \implies (\frac{-\alpha^2-3}{2})^2-2=0$
$\implies (\frac{-\alpha^2-3}{2})^2-2=\frac{\alpha^4+6\alpha^2+9}{4}-2=\a... | Hint: Note that $(1-\sqrt 2)^2=3-2\sqrt{2}$.
Therefore, $\alpha = \pm(1-\sqrt 2)$. Since $\alpha>0$, we must have $\alpha = -1+\sqrt2$.
Then $\alpha^2 = 3-2\sqrt{2} = -2\alpha +1$ and so the minimal polynomial of $\alpha$ is $x^2+2x-1$.
Clearly, $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2)$, because $\alpha = -1+\sqrt2 \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $ Show that if $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $
$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?
| By dividing $d$ on both sides and using $\gcd(ac,bc)=c\gcd(a,b)$, we may assume that $\gcd(a,b)=d=1$, so you only need to show that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$. Suppose that $p$ is a prime that divides $a^2$ and $b^2$, then $p$ divides $a$ and $b$, this contradict to the assumption $\gcd(a,b)=1$, so there ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\f... | See $\sec^{-1}(\sqrt{5})+\csc^{-1}(\frac{\sqrt{10}}{3})=135$ so $x=45$ if we take $\theta=\pi/4$ then sum would be $\pi/2$ and not $\pi$ thus book has taken $\theta=3\pi/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| Hint. One may observe that
$$
6+2\sqrt{5}=(\sqrt{5})^2+2\sqrt{5}+1=(\sqrt{5}+1)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of $\int \frac{1-\sin x}{(1+\sin x)\cos x}dx$ Evaluate $$I=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}$$
I tried in the following way:
$$1-\sin x=1-\cos\left(\frac{\pi}{2}-x\right)=2 \sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
Similarly $$1+\sin x=2 \cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
So
$$I=... | Multiply both numerator and denominator by $\;1+\sin x\;$ , so you get ( observe that $\;\cos x=(1+\sin x)'\;$):
$$\int\frac{\cos x}{(1+\sin x)^2}dx=-\frac1{1+\sin x}+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Let $a_{n}=(1-\frac{1}{\sqrt{2}})\ldots(1-\frac{1}{\sqrt{n+1}}), n \geq 1 $. Then $\lim_\limits{n \rightarrow \infty} a_{n} $ Options are (A) equals 1 (B) does not exist (C) equals $\frac{1}{\sqrt{\pi}}$ (D) 0.
I have multiplied each numerator with $\sqrt{n+1}+1$, $\forall n = 1$ to $\infty$ and then got $a_{n}=\frac{... | We can write
$$a_n = \frac{\sqrt{2}-1}{\sqrt{2}}\frac{\sqrt{3}-1}{\sqrt{3}}\cdots \frac{\sqrt{n+1}-1}{\sqrt{n+1}} = \frac{1}{\sqrt{2}(\sqrt{2}+1)}\frac{2}{\sqrt{3}(\sqrt{3}+1)}\cdots \frac{n}{\sqrt{n+1}(\sqrt{n+1}+1)}$$
For every positive integer $k$, $\sqrt{k}(\sqrt{k}+1) > \sqrt{k}\cdot\sqrt{k} = k$; therefore
$$a_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Catalan numbers formula derivation I'm trying to follow a proof of the Catalan numbers being equal to $\frac{1}{n+1} {2n \choose n}$ from the recurrence relation $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-2}C_{1}+C_{n-1}C_0$
Now it's seen that the generating function satisfies $xf^2-f+1=0$ so $f=\frac{1-\sqrt{1-4x}}{2x}$ si... | Hint: By binomial theorem, we have $$(1 + z)^\alpha = \sum\limits_{k \geq 0} \binom{\alpha}{k} z^k.$$
Use this for $\alpha = \frac{1}{2}$ on the square root term, and equate coefficients.
Edit: Here, we define $$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)\cdots(\alpha - (k-1))}{k!}.$$
Note that this agrees with our u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the remainder when $169\times144^{25}$ is divided by $13^{4}$
Find the remainder when $169\times144^{25}$ is divided by $13^{4}$
Meanwhile I reduced it to
$\dfrac{13^{2}\times 144^{25}}{13^{4}} \\
=\dfrac{144^{25}}{13^{2}} \\
=\dfrac{12^{50}}{13^{2}} \\
$
and Euler function of $13^{2}$ is coming to be $156$
I ... | The problem boils down to computing $144^{25}\pmod{13^2}$, or $25^{25}\pmod{13^2}$, or $25^{26}\pmod{13^2}$, or $25^{13}\pmod{13^2}$. By the binomial theorem:
$$ 25^{13} = (26-1)^{13} = \sum_{k=0}^{13}\binom{13}{k}(-1)^{13-k}(26)^k $$
but for every $k\geq 2$ we have $26^k\equiv 0\pmod{13^2}$ and $\binom{13}{1}(26)^1\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think:
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. N... | The point is that since the exponent $3$ is coprime to the order $5$ of $a$, the orders of $a$ and $a^{3}$ are the same, and thus so are the subgroups generated by each one. Therefore the centralizer of $b$, which contains $a^{3}$, also contains $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that $U_{14}\cong U_{18}$? Because $|U_{14}|=|U_{18}|$ and both of them is cyclic and commutative, so i just need define a function $f : U_{14}\to U_{14}$ that f is homomorphism and bijective.
$f(1)=1, f(3)=5, f(5)=7, f(9)=11, f(11)=13, f(13)=17$
Is that function homomorphism and bijective? If not, what is the fun... | Since
\begin{align}
\varphi(14)
&=14\left(1-\frac{1}{2}\right)\left(1-\frac{1}{7}\right)=6 \\[6px]
\varphi(18)
&=18\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=6
\end{align}
we know the two groups can be isomorphic.
If we consider $3\in U_{14}$ and its powers, we get
$$
1=3^0,\quad
3=3^1,\quad
9=3^2,\quad
13=-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove if $A$ is positive definite, then $A_{11}$,$A_{22}$ are positive definite. Here ${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\
{{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}}
\end{array}} \right)$.
I can prove $\mathbf{A}_{11}$ is positive definite but need help ... | Positive definite implies that the scalar product of $x^TAx\geq 0$ for all $x$, then you just have to take $x=e_2$ and you get your result
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove $\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$ Today I found the identity : $$\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$$.
How to prove or disprove this?
Thank you.
| Now I can solve it.
Let $a=e^{\frac{\pi}{2n+1}i}$. We will get $$\sum_{k=1}^n(-1)^k\cos^{2n+1}\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^{2n+1}}\sum_{k=1}^n(-1)^k(a^k+a^{-k})^{2n+1}$$
Using the binomial theorem and switching the sums yield
$$\frac{1}{2^{2n+1}}\sum_{k=1}^n(-1)^k(a^k+a^{-k})^{2n+1}=\frac{1}{2^{2n+1}}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Limit of recursive sequence $n^2q_n=1+(n-1)^2q_{n-1}+2(n-2)q_{n-2}$ When looking at this riddle, I came across the following sequence for the frequency of sampled integers between 1 and $n$ in a without replacement/without neighbour sampling:
$$q_1=1,\quad q_2=1/2,\quad q_n=\frac{1}{n^2}+\frac{(n-1)^2}{n^2}q_{n-1}+\fra... | Let $a_n = (n+1)q_{n+1}$. We have $a_0 = a_1 = 1$ and
$$(n+1)a_n = 1 + na_{n-1} + 2a_{n-2}\quad\text{ for }\quad n > 1\tag{*1}$$
Let $f(z) = \sum\limits_{n=0}^\infty a_n z^n$, multiply $(*1)$ by $z^n$ and start to sum from $n = 2$, we get
$$\begin{align}
& \left(z\frac{d}{dz} + 1 \right)(f(z) - 1 - z) = \frac{z^2}{1-z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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For each pair of events A and B, find P(A|B) and P(B|A). I've got a simple problem here, but I just want to ensure that I'm not losing a simple concept.
Relevant equations:
$$P(A|B) = \frac{P(A \cap B)}{P(B)},$$
$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
Let $(S,P)$ be a sample space with $S = \{1, 2, 3, 4, 5\}$, and,... | Your approach is correct, as are your answers. However, your solutions in part (d) are incorrect.
Since $A = \{1, 2, 3, 4, 5\}$ is the entire sample space, $P(A) = 1$. Since $P(4) = 0.2$ and $P(5) = 0.4$ and $B = \{4, 5\}$, $P(B) = P(4) + P(5) = 0.2 + 0.4 = 0.6$. Since $B \subseteq A$, $P(A \cap B) = P(B)$. Hence, ... | {
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"url": "https://math.stackexchange.com/questions/1759064",
"timestamp": "2023-03-29T00:00:00",
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| $$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(a+c)$$
$$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(2-c)+bc(2-a)+ca(2-b)$$
$$12=8+2(ab+bc+ca)-3abc$$
Now,
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$2(ab+bc+ca)=4-6=-2$$
Thus,
$$12=8-2-3abc$$
$$abc=-2$$
Let
$$f(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
$$f(x)=x^3-2x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How to show that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is strictly increasing for $x \geq 1$? I am trying to prove that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is a strictly increasing function for $x \geq 1$.
I try to do this by showing that $f'(x)>0$ for all ... | I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a):
$$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without calculating them determine whether $36^2+1$ and $154^2+1$ are prime and find the prime factors if not prime I know that $36^2 + 1$ is prime, $154^2 + 1$ is not, both are equal to $1 \bmod 4$. The prime divisors of $154^2 + 1$ should also be of the form $1 \bmod 4$. Tried showing this by Wilson's theorem however... | The square roots of $-1 \pmod 5$ are $2,3 \pmod 5$ but $154 \equiv 4 \pmod 5.$
The square roots of $-1 \pmod {13}$ are $5,8 \pmod {13}$ but $154 \equiv 11 \pmod {13}.$
The square roots of $-1 \pmod {17}$ are $4,13 \pmod {17}$ but $154 \equiv 1 \pmod {17}.$
The square roots of $-1 \pmod {29}$ are $12,17 \pmod {29}$ but ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$? I was looking up riddles for my math classes to work on for the end of the year and found the following riddle. http://mathriddles.williams.edu/?p=129
I followed the advice and started working with examples of small numbers and stumbled upon a... | If you factor it as $\frac{1}{n}(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{n-1})$ it telescopes nicely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? The solution is supposed to be derived using Inclusion-Exclusion.
Here is my attempt at a solution:
Let $A_0... | The structure of the analysis is the same as yours. We count the bad strings, where some digit appears exactly twice.
We first count the strings where say the digit $0$ appears exactly twice. Where the $0$'s are can be chosen in $\binom{5}{2}$ ways. For each of these ways, the remaining $3$ places can be filled in $9^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Hard summation involving binomial and quadratic What is $$\sum \frac{2r^2-98r+1}{(100-r)({100\choose r})}$$ Where $r\in [1,99]$I have reduced it to $$\frac{(2r^2-98r+1)}{(100){99\choose r}}$$ what to do further? Partial fractions don't seem to help.
| Just to note that, in general $\sum_{r=0}^n \frac{P(r)}{\binom nr}$ can be expressed in terms of $\sum_{r=0}^n \frac1{\binom nr}$ and $n$ only.
Let's denote $c_{np} = \sum_{r=0}^n \frac{r^p}{\binom nr}$. Note that
$$
c_{n1} = \sum_{r=0}^n\frac r{\binom nr}
= \sum_{r=0}^n\frac{n-r}{\binom n{n-r}}
= \sum_{r=0}^n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$ Is it possible to prove that
$$\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$$
without evaluating the integral?
| Without evaluating this integral:
Note $1\le 2-\sin x\le 2$ on $\;[0,\frac\pi2]$, hence $\;\dfrac{x^2}2\le \dfrac{x^2}{2-\sin x}\le x^2 $, and by the positivity of the integral:
$$\frac{\pi^3}{48}=\int_0^\tfrac\pi2\dfrac{x^2}2\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2\dfrac{x^2}{2-\sin x}\,\mathrm d\mkern1mu x\le \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$1+x^4\leq 2(y-z)^2$ and switching of $x,y,z$ Find all triples of real numbers $x,y,z$ such that $1+x^4\leq 2(y-z)^2$, $1+y^4\leq 2(z-x)^2$, and $1+z^4\leq 2(x-y)^2$.
Beside $(1,0,-1)$ and permutations, I can't find any others. We cannot have two equal numbers. Maybe the inequality $1+a^4\geq 2a^2$ helps? It yields $|x... | The only solutions are $(-1,0,1)$ and permutations.
Subtracting $2x^2,$ $2y^2,$ and $2z^2$ respectively from each inequality gives
\begin{align*}
(1-x^2)^2&=1+x^4-2x^2\leq 2(y-z)^2-2x^2&=&-2(x-y+z)(x+y-z)\\
(1-y^2)^2&=1+y^4-2y^2\leq 2(z-x)^2-2y^2&=&-2(x+y-z)(-x+y+z)\\
(1-z^2)^2&=1+z^4-2z^2\leq 2(x-y)^2-2z^2&=&-2(-x+y+z... | {
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"timestamp": "2023-03-29T00:00:00",
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Can anyone show that $\int_0^1\left[\frac{B-1}{1-x}+\frac{B(x^B-x)}{(1-x)(1-x^B)}\right]\sum _{n=1}^{\infty}x^{B^n-1}dx=\gamma$ Euler's constant
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{k}-\ln(n)=\gamma$$
It numerical value is $\gamma=0.5772166...$
Let B be valid for all integers, where $B\ge 2$
Show that,
$$\int_0^... | Let we set $f_B(x)=\sum_{n\geq 1}x^{B^n}$. We have to compute:
$$ I_B=\int_{0}^{1}\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right)\,f_B(x)\,\frac{dx}{x} $$
but since $f_B(x^B) = x^B+f_B(x)$, by setting $x=z^B$ it follows that:
$$ I_B = \int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)\cdot(z^B+f_B(x))\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of three consecutive cubes equals a perfect square I have found this problem in an old German textbook:
Find all sets of three consecutive integers such that the sum of their cubes is a perfect square.
We can write
$$S = (x-1)^3 + x^3 + (x+1)^3 = (x-1+x+1)((x-1)^2 - (x-1)(x-1) + (x+1)^2) + x^3$$
which reduces to
$... | Here is an approachable (but not exhaustive) algebraic route to obtain the solution $23,24,25$.
OP obtains $S=3x(x^2+2)$. Plainly, $x^2+2$ cannot itself be a square (there are no two integer squares which differ by $2$), so OP makes the (limiting) assumption that to obtain a square, you must set $3x=x^2+2$. It is also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Evaluate a double integral.
To find:
$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$
R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$
$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$
$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$
i.e.
$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$
$$I=\int_0^2\int_{x^2}... | \begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
I&=& \int_0^4\left(\int_0^{\sqrt{y}}\frac{x}{\sqrt{1+y^2}}dx\right)dy\\
&=& \int_0^4\frac{1}{\sqrt{1+y^2}}\left(\int_0^{\sqrt{y}}xdx\right)dy \\
&=& \frac{1}{2} \int_0^4\frac{y}{\sqrt{1+y^2}} dy\\
&=&\frac{1}{2}\left[\sqrt{1+y^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$ So, my teacher gave us this to compute yesterday, and I'm completly confused on how should I proceed :
$$\frac{1^4 + 2012^4 +2013^4}{1^2 + 2012^2 + 2013^2}$$
I've tried several ways, but most of them are very long, for example I've simplified bot... | Hint The given triple $(1, 2012, 2013)$ has the special feature that $1 + 2012 = 2013$. This motivates writing the quotient as
$$\frac{a^4 + b^4 + (a + b)^4}{a^2 + b^2 + (a + b)^2}$$
for $a = 1, b = 2012$.
Additional hint Expanding the numerator, we can see that it factors as $2(a^2+ab+b^2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Inequalities involving the Roots of unity Let $\epsilon \ne 1$ be a nth root of unity. Then prove that,
$\bullet \ |1-\epsilon|\ge\frac{2}{n-1}$
$\bullet |\sin\frac{k}{n}| \ge \frac{1}{n-1}$
Here are my solutions:
$\bullet$ To prove $\implies \ |1-\epsilon| \ge\frac{2}{n-1}$
We have $$\sum_{i=1}^{n-1} {\epsilon}^i = ... | HINT:
Let $f(x)=\sin\left(\frac{\pi}{x}\right)-\frac{1}{x-1}$ for $x\ge 2$. Note that $f(0)=0$ and $\lim_{x\to \infty}f(x)=0$. The second derivative is given by
$$f''(x)=-\frac{2}{(x-1)^3}+\frac{2\pi\cos\left(\frac{\pi}{x}\right)}{x^3}-\frac{\pi^2\sin\left(\frac{\pi}{x}\right)}{x^4}$$
Show that $f''(x)>0$ for $x\ge 2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt... | \begin{align*}
&\left(10+\sqrt{108}\right)^{1/3}-\left(-10+\sqrt{108}\right)^{1/3}\\
&=\left(10+6\sqrt{3}\right)^{1/3}-\left(-10+6\sqrt{3}\right)^{1/3}\\
&=\left((1+\sqrt{3})^3\right)^{1/3}-\left((\sqrt{3}-1)^3\right)^{1/3}\\
&=(1+\sqrt{3})-(\sqrt{3}-1)\\
&=2
\end{align*}
The formula:
$$(A+B\sqrt{3})^3=A^3+3\sqrt{3}A^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Since
$$
\begin{align}
\frac{x^2+3}{x^6(x^2+1)}
&=\frac1{x^6}+\frac2{x^6(x^2+1)}\\
&=\frac1{x^6}+\frac{2(x^6+1)}{x^6(x^2+1)}-\frac2{x^2+1}\\
&=\frac1{x^6}+\frac{2(x^4-x^2+1)}{x^6}-\frac2{x^2+1}\\
&=\frac3{x^6}-\frac2{x^4}+\frac2{x^2}-\frac2{x^2+1}
\end{align}
$$
we have
$$
\int\frac{x^2+3}{x^6(x^2+1)}\,\mathrm{d}x
=C-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 20... | I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5a^2}\geq \frac{a+b}{18}$$
Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^3}{13x^2+5}+\frac{1}{13+5x^2}\geq \frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)\geq 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 4
} |
How to solve $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } }$ for $\cos(\alpha)$ using half-angle formula. I have $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
How can I get $\cos(\alpha ) $ from this?
I know this identitiy.
$\cos(\frac{\alpha }{2} )=\sqrt{\frac{1+\cos(\alpha ) }{2} } $
But j... | If we have that $a,b$ are given then we can use the above answers to say that
$$\sqrt{\frac{1+\cos(\alpha)}{2}}=\frac{a}{\sqrt{a^2+b^2}}$$
Then
$$\cos(\alpha) = \frac{2a^2}{a^2+b^2}-1$$
If instad you have $\sin(\alpha/2) = \frac{a}{\sqrt{a^2+b^2}}$ you can use that $\sin(\alpha/2) = \sqrt{1 - \cos^2(\alpha/2)} = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
I want to know if my attempt is correct.
First $a^{13} \equiv (a^3)^4 \cdot a \equiv a^4 \cdot a \equiv a^3 \cdot a^2 \equiv a \cdot a^2 \equiv a^3 \equiv a \pmod 3$.
Second, $(a^7)^2 \cdot a^{-1} \equiv a^2 ... | By the Chinese remainder theorem, we have that $ R = \mathbb{Z}/(3 \cdot 7 \cdot 13)\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z} $, so the equation $ X^{13} - X = 0 $ holds for $ R $ if it holds for each factor ring on the right. On the other hand, we know by Ferm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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.