Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Volume of a tetrahedron whose 4 faces are congruent. Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume?
PS. The reason for me tagging calculus and linear algebra is that I figur... | The volume of a tetrahedron satisfies
$$36V^2 = a^2 b^2 c^2\left(\;1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma\;\right) \tag{1}$$
where $a$, $b$, $c$ are lengths of edges coinciding at a vertex, and $\alpha$, $\beta$, $\gamma$ are the angles between those edges ($\alpha$ between $b$ and $c$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambd... | Doing long division:
$$A^5-27A^3+65A^2=(A^2+5A-8)(\underbrace{A^3-5A^2+6A-5}_{=0})+73A-40.$$
Hence it is:
$$A^5-27A^3+65A^2=73A-40I=\begin{pmatrix}-40&0&73\\ 219&33&0\\-146&73&252\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and... | $27(m^3+n^3) +27(m^2-n^2) + 9(m+n) = 3^z$
$(m+n)(3m^2 +3n^2-3mn+3m-3n+1) = 3^{z-2}$
since $z \ne 2$, we have $3(m^2 +n^2-mn+m-n)+1$ divides a power of $3$ or is equal to $1$, both of which is not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Elementary proof that $24^n$ is the sum of 3 perfect squares Proof that $24^n$ is the sum of 3 perfect squares for every $n >= 1$ . I found this problem in a math contest for 5th graders http://www.mategl.com/atasamente%20pentru%20site/OLM%202014%20Prahova%20(5-12).rar , so obviously a very very elementary proof is re... | \begin{align}
24^{2n + 2} &= 24^2 * 24^{2n} = (2^6 + 2^8 + 2^8)* 24^{2n} = (8* 24^n)^2 + (16*24^n)^2 + (16*24^n)^2 \\
24^{2n+1} &= 24 * 24^{2n} = ( 2^2 + 2^2 + 2^4 ) * 24^{2n} = (2*24^n)^2 + (2*24^n)^2 + (4*24^n)^2\\
\end{align}
P.S: Thus the OP was the right track, if they would have computed it for $24^3$ and $24^4$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability of trials without replacement using conditional probability
A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?
Attempt:1
$$
E:\text{first ma... | I use the elementary events for the answer. Firstly I use your definition of events:
$E:\text{first marble is red}\\
F:\text{at least one of the marbles drawn be black}\\$
You have to find the probability that you draw at least 3 black marbles given the first marble is red. This is $P(F|E)$
In total we have these 8 cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why can't $\frac{1}{\sin(x)\cos(x)}$ be expressed in the form $\frac{A}{\sin x}+\frac{B}{\cos x}$ I've tried expressing $\frac{1}{\sin(x)\cos(x)}$ as partial fractions:
$$\frac{1}{\sin(x)\cos(x)} = \frac{A}{\sin x}+\frac{B}{\cos x}
\implies 1=A\cos x + B\sin x$$
I let $x=\frac{\pi}{2}$, getting $A=-1$. Then I let $x=\p... | \begin{align}
\sin \sin x + \cos x \cos x &= 1 \\
\dfrac{\sin x \sin x}{\sin x \cos x} + \dfrac{\cos x \cos x}{\sin x \cos x}
&= \dfrac{1}{\sin x \cos x} \\
\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}
&= \dfrac{1}{\sin x \cos x} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$ (induction) Problem
Prove $$\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$$
Attempt to solve
I would try to prove this with induction. We have sum and the sum as function of $p(n)$. Now i try to prove that the sum equals $p(n)$ with induction. $p(n)=\frac{1}{4}n^2(n+1)^2$
$... | As @packsciences said, your mistake is after "so we have". The correct way to do it is:
$$\frac{1}{4}n^2(n+1)^2 + (n+1)^3 = (n+1)^2 (\frac{1}{4}n^2 + n+1) = \frac{1}{4}(n+1)^2 (n^2 + 4n + 4) = \frac{1}{4} (n+1)^2 (n+2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... |
Could $\frac{1}{0}$ be a valid limit?
You should be asking instead :"Is the following true?" $$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 0$$
where $\lim_{n \rightarrow \infty} f(n) = K $, where $K$ is finite and $\lim_{n \rightarrow \infty} g(n) = +\infty$. The answer is yes. This is one of the limit rules.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
} |
Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is ... | Denote $$S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}.\tag1$$Then $$\frac{1}{2}S_n=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots+\frac{n}{2^{n+1}}.\tag2$$Thus by $(1)-(2)$, we obtain $$\frac{1}{2}S_n=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^n}\right)-\frac{n}{2^{n+1}}< 1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Fourier series of $f(x)=\sin^2x\cos^2x$ at $(-\pi,\pi)$ Find Fourier series of $f(x)=\sin^2 x \cos^2 x$ at $(-\pi,\pi)$
$f(x)$ is even so we only have to evaluate $a_0,a_n$
$$
a_0 = \frac{1}{\pi}
\int_{-\pi}^{\pi} \sin^2 x \cos^2 x dx
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\sin^2(2x)
= \frac{1}{4\pi}
... | Simply write
$$\sin^2x\cos^2x=\dfrac14\sin^22x=\dfrac18-\dfrac18\cos4x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of ways to pay (generating functions) I've just started learning generating functions.
*
*Let $a_n$ be the number of ways in which you can pay $n$ dollars using 1 and 2 dollar bills. Find the generating function for $(a_0, a_1, a_2, \ldots)$ and general term $a_n$.
*Prove that the number of ways to pay $n$ d... | we have with $n_1$ 1 dollar bills and $n_2$ 2 dollar bills we have: $n_1+2n_2 = n$. which has a solution for every $n_1$ such that $n-n_1$ is even.
So if $n$ is odd then there are $(n+1)/2$ possible ways of paying.
So if $n$ is even then there are $(n/2)+1$ possible ways of paying.
Hence problem 1 is done.
Let the solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\}\cap\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$
Find $A \cap B$, where
$A=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\},\quad B=\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$
I know that $(\sqrt{3},0)$ and $(-\sqrt{3},0)\in A \cap B$ so $A\cap B\neq\emptyset$, but how can I find... | An.option:
Ellipse $2x^2+3y^2=6$, or
$3(x^2+y^2) =6+x^2$.
With $d^2$ the (distance)$^2$ from the origin:
$d^2:= x^2+y^2= 2 +(1/3)x^2 \le 3$,
since $|x| \le √3$ for points on the ellipse.
$d^2 = 3$ , for $x = ^{+}_{-}√3$, else $d^2<3$, i.e
only $2$ points on the circle with $r^2=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
In $\triangle ABC$, $\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$ In $\Delta ABC$ , prove $$\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$$
I tried to ta... | Hint: Note that $$\begin{align}\frac{\cot(A)+\cot(B)}{\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)}&=\frac{\sin(B)\cos(A)+\sin(A)\cos(B)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\Bigg(\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)+\cos\left(\frac{B}{2}\right)\sin\left(\frac{A}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
| Well, you know that $$\frac{(n+4)!}{(n+2)!} = (n+4)\cdot(n+3) = n^2+7n+12=90$$$$n^2+7n-78=0$$The solutions to this quadratic are $n=6$ and $n=-13$, so we take $n=6$ to be our answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.
I tried the following concept, how to sum 9 using 1-9 only without repetition.
1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,
6:1+2+6 ,7:1+3+5, 8:2+3+4
The answer is... | Yes we can count directly the cases that is
*
*$x^9$
*$x\cdot x^8$
*$x\cdot x^2\cdot x^6$
*$x\cdot x^3\cdot x^5$
*$x^2\cdot x^7$
*$x^2\cdot x^3\cdot x^4$
*$x^3\cdot x^6$
*$x^4\cdot x^5$
to obtain $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Problem with simple inverse Laplace transform I have the function:
$$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)}$$
and I have to make inverse Laplace. I tried to collect $s^3$ from the first and second element of the numerator in order to obtain:
$$\frac{2s^2+4s+4}{(s+3)(s^2+1)} + \frac{s^3}{s^2+1}$$
for the first, ... | Because you have a mistake when computing the Residues, you should get
$$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)} = \frac{1}{s+3} + \frac{1}{s^2+1} + s \tag{1} $$
which answers your question on why you have an extra $\cos(t)H(t)$ appearing. To see why you have done a calculation mistake, take a look here:
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=... | Hint: you need $f^\prime$ to be non-negative for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac... | hint
At $-\infty$, $x$ becomes negative, thus
$$\sqrt{x^2}=|x|=-x$$
$$\frac{\sqrt{x^2-5x+1}}{x}=\frac{\sqrt{x^2-5x+1}}{-(-x)}=$$
$$-\sqrt{\frac{x^2-5x+1}{x^2}}.$$
the denominator goes to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Cubic Diophantine with two variables The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.
I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$... | Based on your argument, we can split into two cases:
*
*$y + x + 4$ and $y - x - 4$ are relatively prime.
In this case, as you said, we can assume that $y + x + 4 = b^3$ and $y - x - 4 = a^3$. Then we have $x = ab$. Hence, $b^3 - a^3 = 2x + 8 = 2ab + 8$. This means that $b > a$ and have the same parity, so $b - a \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$
I am trying to find the Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ when $0<|z-1|<1$.
I thought that if
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-2}+\frac{1}{(z-2)^2} \\
&=\frac{1}{z-1}+\frac{1}{2-z}+\frac{d}{dz}\left(\frac{1}{2-z}\right) \\
&=\frac{1}{z-... | Use this one: let $z-1=w$ to make it easier then with $0<|w|<1$
$$\dfrac{1}{1-w}=\sum_{n\geq0}^\infty w^n$$
differentiating shows
$$\dfrac{1}{(1-w)^2}=\sum_{n\geq1}^\infty nw^{n-1}$$
therefore
$$\dfrac{1}{w(w-1)^2}=\sum_{n\geq1}^\infty nw^{n-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\sum\limits_{\text{cyc}}x=\frac\pi2$, then $2\sqrt{\sum\limits_{\text{cyc}}\tan x}\le\sum\limits_{\text{cyc}}\frac{\sqrt{\tan x}}{\cos x}$
Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=\frac{\pi}{2}$. Then,
$$2\sqrt{\tan x+\tan y+\tan z}\leq \frac{\sqrt{\tan x}}{\cos x}+\frac{\sqrt{\tan y}}{\cos y}... | Let $\tan{x}=a$, $\tan{y}=b$ and $\tan{z}=c$.
Thus, $a$, $b$ and $c$ are positives, $ab+ac+bc=1$ and we need to prove that
$$\sum_{cyc}\sqrt{a(1+a^2)}\geq2\sqrt{a+b+c}.$$
Indeed, by C-S and Schur we obtain:
$$\sum_{cyc}\sqrt{a(1+a^2)}=\sum_{cyc}\sqrt{a(a^2+ab+ac+bc)}=\sum_{cyc}\sqrt{a(a+b)(a+c)}=$$
$$=\sum_{cyc}\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Sinc... | We have
$$2^{6k+1}\equiv 8^{2k}\cdot 2\equiv 2\pmod{9} $$
from which
$$2^{6k+2}\equiv 4\pmod {18} $$
hence by Fermat's little theorem
$$2^{2^{6k+2}}\equiv 2^4\equiv -3\pmod {19} $$
For the second $2^4\equiv 3\pmod {13} $ and $2^{12}\equiv 1\pmod {13}$ by Fermat little theoren hence
$$2^{70}+3^{70}\equiv 2^{70}+2^{280... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's Find limit of
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$
I started by defining $\ x = y + \frac{\pi}{4} $
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$
T... | $$\frac{\tan\left(y+\dfrac\pi4\right)-1}y=\frac{\dfrac{\tan y+\tan\dfrac \pi4}{1-\tan y\tan \dfrac \pi4}-1}y=2\frac{\tan y}{y}\frac1{1-\tan y}\to 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving a convexity inequality
Given $f: \mathbb{R} \to \mathbb{R}$ convex, show that:
$$ \frac{2}{3}\left(f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right)\right) \leq f\left(\frac{x+y+z}{3}\right) + \frac{f(x) + f(y) + f(z)}{3}.$$
I have tried some ideas, such as transforming... | Let $x\geq y\geq z$.
Consider two cases.
*
*$x\geq y\geq\frac{x+y+z}{3}\geq z$.
From here we see that $2y\geq x+z$ and easy to check that
$$\left(\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3},z\right)\succ\left(\frac{x+z}{2},\frac{x+z}{2},\frac{y+z}{2},\frac{y+z}{2}\right),$$ which by Karamata gives
$$3f\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find $n$ such that there are $11$ non-negative integral solutions to $12x+13y =n$
What should be the value of $n$, so that $12x+13y = n$ has 11 non-negative integer solutions?
As it is a Diophantine equation, so we check whether the solution exists? it exists if $gcd(12,13)|n$ that is $1|n$ and hence integer solution... | Note that from the following hint
$$-1\cdot 12+1\cdot 13=1 \implies (-1+k \cdot 13)\cdot 12+(1-k\cdot 12)\cdot 13=1$$
we have
$$-n\cdot 12+n\cdot 13=n \implies (-n+k \cdot 13)\cdot 12+(n-k\cdot 12)\cdot 13=1$$
and we need
*
*$-n+k \cdot 13\ge 0$
*$n-k\cdot 12\ge 0$
that is
$$\frac n{13}\le k\le\frac n{12}$$
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$, then what's the value of $4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$? I was trying to solve this problem:
If
$$\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$$
then what is the value of
$$S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin... | I don't know if this helps, but this is what I've discovered so far
The first equation can be written as
$$\cos x \sin y + \sin x \cos y = -\cos y \sin y$$
or using the double angle formula
$$\sin (x+y) = -\cos y \sin y$$
By the triple angle formulas for sine and cosine, we have that
$$\sin^3(y)=\frac{3\sin(y)-\sin(3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Differential equation $y''+y=\tan(x)$ Given the following differential equation:
\begin{align}
y''+y=\tan(x)
\end{align}
I tried this way:
\begin{align}
y_H(x)=A\cos(x)+ B\sin(x)\\ A,B\in \Re \\
y_p(x)=A(x)\cos(x)+B(x)\sin(x)
\end{align}
I continue with the Wronski method
\begin{align}
W(F)= \begin{bmatrix}
\cos(x) & \... | Your solution is correct just note that $(\sin(x/2)\pm \cos+x/2))^2=1\pm \sin(x)$. Now use the log property that $\log(a^b)=b\log(a)$ and you are done. And if you are confused with the discrepancy of plus and minus sign before the log expression we have $\log(\frac{1}{x})=-\log(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\i... | Try this-
$$\int\sqrt{\frac{1+x}{1-x}}dx=\int\frac{1+x}{\sqrt{1-x^2}}dx=\int\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}dx$$
The first integral is $\arcsin x$ and the second is evaluated by substituting $u=x^2\implies du=2xdx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos... | Your integral is equal to $$2\int_{0}^{\pi}\frac{\sin^2x}{a-b\cos x} \, dx$$ and we can express the integrand as $$\frac{\cos x} {b} +\frac{a}{b^2}+\frac{b^2-a^2}{b^2(a-b\cos x)} $$ and hence the integral in question is $$2\cdot 0+\frac{2\pi a} {b^2}+\frac{2(b^2-a^2)}{b^2}\int_{0}^{\pi}\frac{dx}{a-b\cos x} $$ The last ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Approximation of square roots Recently, I've seen a YouTube video where they approximate square roots real quick.
They use this approximation :
$$\sqrt{x} \approx \lfloor \sqrt{x} \rfloor+\dfrac{x-(\lfloor \sqrt{x} \rfloor)^2}{2\lfloor \sqrt{x} \rfloor}$$
I want to know the math behind this approximation. Can someone h... | Let, $n \in N $
Suppose, we want to evaluate $\sqrt{n^2+c}$
$$\implies \sqrt{n^2(1+\dfrac{c}{n^2})}$$
$$\implies n \sqrt{1+\dfrac{c}{n^2}}$$
Now, we shall use the binomial expansion for $\sqrt{1+\dfrac{c}{n^2}}$
We know, $\sqrt{1+\dfrac{c}{n^2}}=1+\dfrac{c}{2n^2}-\dfrac{c^2}{8n^4}+\cdots$
So, $n \sqrt{1+\dfrac{c}{n^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How evaluate $ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2} \;dx$ How evaluate (without using Complex analysis)
$$ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2}\; dx\quad (a\gt0)$$
My Attempt:
I used the expansion of th... | Using OP's series expansion, we have
\begin{align*}
I
&= 2 \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \int_{0}^{\infty} \frac{\sin[(2m+1)ax]\sin[(2n+1)ax]}{1+x^2} \, dx \\
&= \pi \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \left( e^{-a|2m-2n|} - e^{-a(2m+2n+2)} \right).
\end{align*}
We will split the sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
limit of $\sqrt{(x^2+1)/(x^3+1)}$ as $x$ approaches negative infinity $\lim_{x \rightarrow - \infty } \sqrt{ \frac{x^2+1}{x^3+1} } $
My teacher says that no limit exists, but Wolfram Alpha says the limit is 0. I'm confused.
Any helps are welcome.
| The answer depends on the underlying domain of the expression. To see this, you may write
$$\sqrt{ \frac{x^2+1}{x^3+1} } \stackrel{x<-1}{=} \sqrt{-1} \cdot \frac{1}{\sqrt{|x|}} \cdot \sqrt{ \frac{1+\frac{1}{x^2}}{1+\frac{1}{x^3}} } = \begin{cases} \mbox{not defined} & \mbox{ in } \mathbb{R} \\ \pm i \cdot \frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the values of $p$ and $q$
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(\text{mod} \, q)$ and $q = -1(\text{mod} \, p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic e... | Clearly $p\neq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $p\mid q+1$ and $q\mid p^2+1$. Write
$$q+1=ap\qquad\text{ and }\qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1\equiv0\pmod{p}$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Probability of getting a particular sum from n m-sided dice c# I am trying to generate a probability of getting a specific number x from n dice, with no guarantee of them having the same number of sides. (eg, 1d6 + 2d10)
Does there exist a mathematical formula for this?
Does there exist a formula for getting above x?
| The real trick here is to use generating functions.
The way they work is by assigning the probabilities to a polynomial, with each term having coefficient equal to its probability of being rolled, and power the value of the roll.
Then, one simply looks at the coefficient of the term with power equivalent to the desired... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Probability of at least once out of the three tosses Suppose a pair of fair dice is tossed 3 times. Let $X$ be the sum of the outcomes at each toss. Find the probability of getting a sum of 7 (i.e., $X=7$) at least once out of the three tosses.
My attempt:
The sample space is $6^3=216.$ There are 15 combinations that g... | let X = getting total of 7 then sample space S={(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)}=6 times
Total number of outcomes =
{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1).....................(2,6)
(3,1).....................(3,6)
(4,1)
(5,1)
(6,1),,,,,,,,,,,,,,,,,,,,,(6,6)}=6x6=36 outcomes
P(getting total 7)=p(x)=6/36=1/6
q=5/6
get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using Vieta's formula to find the sum of the roots for a given cubic equation. Vieta's formula states that, if a cubic equation has three different roots, the following is true:
$$\begin{eqnarray*}
x_1 + x_2 + x_3 &=& -b/a\\
x_1x_2 + x_1x_3 + x_2x_3 &=& c/a \\
x_1x_2x_3 &=& -d/a
\end{eqnarray*}$$
Then, how is the foll... | Use that $$(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$$
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $(a-b+5)x^2 + (2b+1)y^2 = b^2-2b-11$ is the equation of a unit circle, then find the sum of the possible values of $ab$. If
$$(a-b+5)x^2 + (2b+1)y^2 = b^2-2b-11$$
is the equation of the unit circle, then what is the sum of the values that $ab$ can take?
| Here's an extended hint ...
The standard equation for an origin-centered unit circle is $x^2+y^2=1$. More generally, such a circle has the equation
$$p x^2 + p y^2 = p$$
for any non-zero $p$. Therefore, for $$(a-b+5)x^2+(2b+1)y^2=b^2-2b-11 \tag{1}$$
to be the equation of a unit circle, the coefficients of $x^2$ and $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$
If $z_{1}\;,z_{2}$ are two complex number $(|z_{1}|\neq |z_{2}|)$ satisfying
$\bigg||z_{1}|-4\bigg|+\bigg||z_{2}|-4\bigg|=|z_{1}|+|z_{2}|$
$=\bigg||z_{1}|-3\bigg|+\bigg||z_{2}|-3\bigg|.$Then minimum of $\bigg||z_{1}|-|z_{2}|\bigg|$
Try: Let $|z_{1}|=a$ and $|z_{2}|=b$ an... | From the equality
$$|a-4|+|b-4|=a+b = |a-3|+|b-3|$$
we have that $a,b\ge 3$ is not possible.
So, assume $a\ge 3, b\le 3$. Then
$$a+b = a-3+3-b=a-b$$ which is only possible if $b=0.$ In such a case
$$|a-4|+4=a =a-3+3=a.$$ This is only possible if $a\ge 4.$
Now assume $a,b\le 3.$ In such a case
$$4-a+4-b=a+b=3-a+3-b.$$ B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$ Where is my mistake in the below:
$$\frac{1}{c} \lim_{x \to 0} \frac{x^3 \arctan x}{x^{4+n} + x^n \cos x + 3x^n} = \frac{1}{c} \lim_{x \to 0} \frac{x^4}{x^{4+n}+x^n \cos x + 3x^n} \\ =\frac{1}{c} \lim_{x \to 0} \frac{1}{x^n + x^{n... | It should be obvious that $c\neq 0$ otherwise the definition of $\sim$ is in trouble. Next we can note that the given condition implies $$\lim_{x\to 0}\frac{cx^n(3+\cos x+x^4)}{x^3\arctan x} =1$$ Using the limit $\lim_{x\to 0} (1/x)\arctan x=1$ we can see that the above condition is equivalent to $$4c\lim_{x\to 0}x^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1}... | Show that
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}\leq 2-\frac{An+B}{2^n}$$
for some $A,B\geq 0$. Then the base case is satisfied if
$$\frac{1}{2}\leq 2-\frac{A+B}{2}$$
that is $A+B\leq 3$.
The induction step works if for all $n\geq 1$,
$$2-\frac{An+B}{2^n}+\frac{n+1}{2^{n+1}}\leq2-\frac{A(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left... | This is a
major revision
of the last few lines.
The conclusion is that
$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}
=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}
$
where
$\psi^{(1)}$
is a polygamma function
(reference below).
$\begin{array}\\
\sum_{n=1}^\infty \frac{\sin\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Finding the centroid of a tetrahedron I have four points to form a tetahedron
$$A=(0,-\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\B=(0,\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\C=(-\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})
\\D=(\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})$$
it looks like this :
I'm asked to ... | Few days ago, using vectog calculus, I have proved an analogous formula for centroid of a triangle. To appply this method here, two things are needed to check.
*
*The centroid is a common point of four lines, each of them joins a vertex with the centroid of the opposite face.
*The centroid divides this face in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how is that expression is generated
I have no idea that how is red arrowed mammoth term is generated from the yellow arrowed term
Please explain
| The author starts by substituting $t$ for $e^x - 1$, hence
$$ \sin (e^x - 1) = \sin t $$
Now the power series of the sine function is plugged in (yellow arrowed line on the right)
$$ \sin (e^x - 1) = \sin t = t -\frac{t^3}{3!} + \frac{t^5}{5!} \mp \cdots $$
Now we resubstitute $e^x -1$ for $t$, at the same time we us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
how to find subdifferential of a function $x^2+ |x-1|+|x-2|$ Given a function
$f(x) = x^2+ |x-1|+|x-2| $
find it's subdifferential.
My approach to solving this was to divide the answer into 5 parts:
*
*For |x-1|>1 and |x-2|>2
$f(x) = x^2+ x-1+x-2$ and $f'(x) = 2x+2$
*For |x-1|<1 and |x-2|<2
$f(x) = x^2-(x-1)-(x-2)... | First, I have to admit I am not familiar with a "subdifferential". It look to me like you are just taking the derivative. Second, dividing with things like "|x- 1|> 1 and |x- 2|> 2" is confusing and probably not what you want. Instead divide the real line into three intervals: $x\le 1$, $1< x\le 2$, and $x> 2$.
For ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivati... | By the chain rule,
$$\left(\arctan\sqrt{\frac{x+1}{x-1}}\right)'=-\frac2{(x-1)^22\sqrt{\dfrac{x+1}{x-1}}\left(1+\dfrac{x+1}{x-1}\right)}=-\frac1{2x(x-1)\sqrt{\dfrac{x+1}{x-1}}}
\\=-\frac1{2x\text{ sgn}(x-1)\sqrt{x^2-1}}.$$
The claim follows from
$$\text{ sgn}(x-1)=\text{ sgn}(x)$$ (when $|x|>1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R... | If $p=2$ or $p\equiv 1\pmod{4}$, then $R$ is obviously $0$ because $-1$ is a quadratic residue modulo $p$. We assume from now on that $p\equiv 3\pmod{4}$.
Let $f(x)$ denote the polynomial $x^p-x$ over the Galois field $K=\operatorname{GF}(p)$. Note that $f$ factorizes into
$$f(x)=\prod_{t\in K}(x-t).$$
Let $E$ be the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= ... |
When terms are in A.P in denominator, we use difference
of last and first terms in product to distribute the terms
of denominator over numerator, to change into
telescoping series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Divisibility 1,2,3,4,5,6,7,8,9,&10 Tried:
Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could end)
$8325971640,$
$8365971240,$
$8317956240,$
$8291357640,$
$8325971640,$
$8235971640,$
$1357689240,$
$1283579640,$
$1783659240,$
$1563729840,$
$1763529840,$
$165372... | The number which are divisible by $8$ is also divisible by $2$ and $4$.
The number which are divisible by $9$ is also divisible by $3$.
The number of which is divisible by $6$ is also divisible by $2$ and $3$.
The number which are divisible by $10$ is also divisible by $2$ and $5$.
Also also the number we expect is div... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Trigonometric Identities: Given $\tan(2a)=2$ and $\frac{3\pi}{2}Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)}
\implies\frac{5}{2\sqrt 5}\left(1-\fra... | Just use the double-angle identity.
$$\tan(2a) = \frac{2\tan{a}}{1-\tan^2(a)}$$
$$\implies \frac{2\tan{a}}{1-\tan^2(a)} = 2$$
$$\implies 2\tan{a} = 2-2\tan^2{a} \implies \tan{a} = 1-\tan^2{a} \implies \tan^2{a}+\tan{a}-1 = 0$$
Set $t = \tan{a}$ and solve for $t$.
$$t^2+t-1 = 0$$
$$t = \frac{-1\pm\sqrt{5}}{2}$$
Plug in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$ without induction I want to prove that $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$, where $F(0) = 0$, $F(1) = 1$ and for all $n \geq 2$, $F(n) = F(n-1) + F(n-2)$ without using induction.
I want to prove it without induction because I want to implement the equation within another induct... | Use the general term formula: $$F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}, \ \ \text{where} \ \ \phi=\frac{1+\sqrt{5}}{2}, \ \ \psi=\frac{1-\sqrt{5}}{2}.$$
Use the sum of geometric progression:
$$\begin{align}\sum^{n}_{i=0}{F(i)} + 1 &= \frac{\phi^0-\psi^0}{\sqrt{5}}+\frac{\phi^1-\psi^1}{\sqrt{5}}+\cdots + \frac{\phi^n-\psi^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$ How would I find $a$, $b$, $c$, and $d$ in :
$$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$
I have already worked out that $a=2$, but I may be wrong. I am not sure how to find the values of the other letters.
Thanks in advance!
| Hint: $$(ax+b)^2(x+c)={a}^{2}c{x}^{2}+{a}^{2}{x}^{3}+2\,abcx+2\,ab{x}^{2}+{b}^{2}c+x{b}^{2}=a^2x^3+x^2(a^2c+2ab)+x(2abc+b^2)+b^2c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know... | $$2x^3+x^2-7x-6$$ and $$2x^3+19x^2+53x+36$$
have a common root, which is such that
$$x^2-7x-6=19x^2+53x+36.$$
By solving the quadratic, this root is one of $-1$ or $-\dfrac73$. Then $-1$ fits and by long division you reduce to
$$2x^2-x-6=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many $4$-element subsets of $S = \{a, b, c, d, e, f, g, h, i\}$ contain at least one of $a$ and $b$? Consider the set $S = \{a,b,c,d,e,f,g,h,i,j\}$. How many $4$-element subsets of $S$ contain at least one of $a$ and $b$?
So my thought is :
1) number of $4$-element subsets in the set $S = 210$
2) $4$-element subse... | You have correctly calculated that the number of four-element subsets of the ten-element set $S$ is
$$\binom{10}{4} = 210$$
that the number of subsets of $S$ that contain exactly one of $a$ or $b$ is
$$\binom{2}{1}\binom{8}{3} = 112$$
and that the number of subsets of $S$ that contain both $a$ and $b$ is
$$\binom{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refe... | Your answer is correct. For all $n \in \mathbb{Z}$:
$$\tan(A-B) = 1 \implies A-B = \frac{\pi}{4}+\pi n$$
Recall that secant, like cosine, is an even function. (Which you've apparently referred to as well, after the format editing.)
$$\sec(-\theta) = \sec(\theta)$$
$$\sec(A+B) = \frac{2}{\sqrt 3} \implies \cos(A+B) = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find diagonalizable decomposition of $A$.
Let $A=\begin{pmatrix}1&2\\-2&-3\end{pmatrix}.$ Find $P$ and $D$, such that: $$A=PDP^{-1}$$
First let's calculate the characteristic polynomial for $A$: $$P_a(x)=\det(A-xI_2)=(x+1)^2.$$
So the eigenvalues of $A$ are $\lambda=-1$ with $a(\lambda)=2$ where $a(\lambda)$ is the a... | In order to solve for the eigenvectors, you simply put them back in the equation. Suppose that $A$ is
$$ A = \begin{pmatrix} 1& 2 \\ -2 & -3 \end{pmatrix} \tag{1}$$
then if solved for our eigenvalues by the equation
$$ \det(A- \lambda I) =0 \tag{2} $$
it gave us $\lambda_{1}, \lambda_{2}$ which you below $\lambda_{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few c... | Notice that the LHS is:
$$\frac{3n \times (3n-1) \times (3n-2)}{3!} = \frac{n \times (3n-1) \times (3n-2)}{2}$$
$$\to \frac{9n^3-9n^2+2n}{2}$$
For the RHS:
$$3 \times \frac{n\times(n-1)\times(n-2)}{3!} + 6n \times \frac{n\times(n-1)}{2!} + \frac{2n^3}{2}$$
$$\Longrightarrow \frac{n\times(n-1)\times(n-2)}{2} + \frac{6n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is... | $$\alpha+\beta+\gamma=-2,$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=3$$ and
$$\alpha\beta\gamma=-3.$$
Thus, $$\sum_{cyc}\frac{\alpha}{\alpha+1}=\frac{\sum\limits_{cyc}\alpha(\beta+1)(\gamma+1)}{\prod\limits_{cyc}(\alpha+1)}=\frac{\sum\limits_{cyc}(\alpha\beta\gamma+2\alpha\beta+\alpha)}{\alpha\beta\gamma+\alpha\beta+\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) Fo... | Without any derivative, just high-school theory of quadratic equations:
A quadratic polynomial (with real coefficients) has a global extremum at the arithmetic mean of its roots. Further, this extremum is a maximum if its leading coefficient is negative, a minimum if the leading coefficient is positive.
So here, the ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What is the biggest possible sum $|X_1-X_2|+|X_2-X_3|+\cdots+|X_{n-1}-X_n|$ where $X_1,X_2,\cdots,X_n$ are first $n$ positive integers? What is the biggest possible sum $|X_{1}-X_{2}|+|X_{2}-X_{3}|+\cdots+|X_{n-1}-X_{n}|$ where $X_{1},X_{2},\cdots,X_{n}$ are first $n$ positive integers?
| Be greedy. $n$ should go next to $1,2$, then $n-1$ next to $1$ and so on. There are other arrangements that give the same sum, but none better. For $n=12$ it gives $7,5,9,3,11,1,12,2,10,4,8,6$ for $2+4+6+8+10+11+10+8+6+4+2$ and for $n=13$ it gives $8,5,10,3,12,1,13,2,11,4,9,6,7$ for $3+5+7+9+11+12+11+9+7+5+3+1$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Determine that the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$ is continous with the $\varepsilon$-$\delta$-definition of limit/criterion
[Proof-verification] Determining whether the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$; $x_0\in \mathbb{R}$ is continous $\color{red}{\text{ in }x_0}$ or not with the
... | First of all, the function $f$ is not defined on $\mathbb{R}$ but for
$$
x\in(-\infty,-2]\cup[3,+\infty).
$$
So one can only talk about its continuity of $f$ at $x=3$ from the right.
For $x>3$, you are right to get
$$
|f(x)-f(3)|=\sqrt{x-3}\sqrt{x+2}.
$$
Note that you don't need the absolute value for the square roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$. Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$.
Answer:
To do this we have to make partial fractions as follows:
$ \frac{2}{(s-1)^3(s-2)^2}=\frac{A}{S-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)}+\frac{J}{(s-1)^3}+\frac{K}{(s-2)^2}+\frac{L}{s-2}$
Am I righ... | You have one too many $\frac {1}{(s-1)}$ terms.
$\frac {2}{(s-1)^3(s-2)^2} = \frac {A}{(s-1)} + \frac {B}{(s-1)^2} + \frac {C}{(s-1)^3} + \frac {D}{(s-2)} + \frac {E}{(s-2)^2}$
Traditionally, you would then multiply through to clear out the denominators.
$2 = A(s-1)^2(s-2)^2 + B(s-1)(s-2)^2 + C(s-2)^2 + D(s-1)^3(s-2) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & ... | Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Let ${x_n}=2^{n}a_n$, and $a_{n+1}=\sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}, a_0=1$, how to prove ${x_n}$ converges? How to show $x_n=2^na_n$ converges, where $a_{n+1}=\sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}$
The question originated from Professor David McKinnon.
Attempt:
I did prove it is increasing but failed to show it is bounde... | Since $$\frac{a_{n+1}^2}{a_n^2} = \frac{1 - \sqrt{1-a_n^2}}{2a_n^2} = \frac{1}{2(1+\sqrt{1-a_n^2})} \le \frac{1}{2}$$
We have
$$a_n^2 = a_0^2 \prod_{k=1}^n \frac{a_k^2}{a_{k-1}^2} \le 2^{-n}
\quad\implies\quad
\sum_{k=0}^\infty a_k^2 \le \sum_{k=0}^\infty 2^{-k} = 2$$
Since
$$\frac{x_{n+1}^2}{x_n^2} = \frac{4a_{n+1}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2989388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$ This question is I am working on is the extension of the domain of zeta function.$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+ \cdots$$\eta(x)$=$1-\frac{1}{2^x}+\frac{1}{3^x}-\frac{1}{4^x}+\frac{1}{5^x}-\frac{1}{6... | An idea for you to develop:
$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}=\sum_{n=1}^\infty\frac1{(2n)^s}+\sum_{n=1}^\infty\frac1{(2n-1)^s}=\frac1{2^s}\zeta(s)+\sum_{n=1}^\infty\frac1{(2n-1)^s}\implies\ldots$$
Observe you're going to get lots of "poles"...which in fact are removable singularities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using the Well Ordering Principle to Show that n^3 - n is divisible by 6. A Guide On Well-Ordering Principle
MIT OCW Guide on WOP
I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the following problem:
*... | Define $f(n) = n^3-n$.
Since $f(-n) = -f(n)$, we only have to prove that $f(n)$ is divisible by $6$ for $n \ge 0$. So our domain is the well-ordered set of natural numbers and we always have $f(n) \ge 0$.
We note that both $f(0)$ and $f(1)$ are divisible by $6$.
Let $k \gt 1$ be the least number such that $f(k)$ is not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Parametrization of a rotated ellipse I need to parametrize an ellipse centered at the origian in cartesian coordinates given by
$a x^2 + b xy + c y^2 = 1$
Now I dug up from old notes that this can be brought into normal form through a rotation $\tan 2 \theta = \frac {b}{a-c}$, which eliminates the $xy$ term in rotated ... | Rewriting equation is quicker : $4a^2x^2 + 4abxy + 4acy^2 = 4a$ (multiplying by $4a$) ; $(2ax + by)^2 - b^2y^2 + 4acy^2 = 4a$ (it is a trick removing the $xy$ term) ; $(2ax + by)^2 + (4ac - b^2)y^2 = 4a$, then you can compare with equation $X^2/p + Y^2/q = 1$ of a conic (ellipse or hyperbola) for which you know a para... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I find a polynomial that fits a given table?
The first difference is 1, 4, 7, 10, 13, 16
The second difference is 3, 3, 3, 3, 3, 3
Since the second difference is constant this would be quadratic and I would have
$\frac{3}{2}n^{2}$
So now I will take the differences between the original sequence and the values... | Let $\Delta_k$ be the $k^{\text{th}}$ finite diffetence. then
$$P(x)=P(x_0)+$$
$$\sum_{k=1}^n\frac{\Delta_k}{k!}(x-x_0)(x-x_1)...(x-x_{k-1})=$$
$$3+1.(x-x_0)+\frac 32(x-x_0)(x-x_1)=$$
$$3+(x-0)+\frac 32(x-0)(x-1)=$$
$$3+x+\frac 32x^2-\frac 32x=$$
$$\frac 32x^2-\frac 12x+3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Explanation for binomial sums $\sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1}$ I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$\sum_{n=0}^{\infty} \binom{n+2}{3} x^n = \sum_{n=0}^{\infty} \binom{n+2}{n-1} x^n= \sum... | Index transformation (replace $n$ by $n+1$) gives
$$ \sum_{n=0}^\infty \binom{-4}{n-1}(-1)^{n-1} x^n
= \sum_{n=-1}^\infty \binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ \binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $\binom{-4}{-1} = 0$. Hence
$$ \sum_{n=0}^\infty \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For any $2$ x $2$ matrix $A$, does there always exist a $2$ x $2$ matrix $B$ such that det($A+B$) = det($A$) + det($B$)? For each invertible $2$ x $2$ matrix $A$, does there exist an invertible $2$ x $2$ matrix $B$ such that the following conditions hold?
(1) $A + B$ is invertible
(2) det($A+B$) = det($A$) + det($B$)
... | Since we aretalking of $2\times2$ matrices, its slightly easier to write down explicitly.
So $\det(A+B)=\det(A)+\det(B)$ happens when $(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})=(a_{11}a_{22}-a_{12}a_{21})+(b_{11}b_{22}-b_{12}b_{21})$
$\implies a_{11}b_{22}+b_{11}a_{22}-a_{12}b_{21}-b_{12}a_{21}=0=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), o... | Another approach is this. Rewrite the DE as
$$
\frac{yy'}{y^2-1}=\frac 1x
$$
Which is the same as
$$
\frac 12 \frac{d}{dx}\ln(y^2-1)=\frac 1x
$$
Upon integration, one gets
$$
\ln\dfrac{y^2(x)-1}{y^2(x_0)-1}=2\ln\frac{x}{x_0}
$$
Hence, one obtains
$$
y^2(x)=1+[y^2(x_0)-1]\left(\frac{x}{x_0}\right)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$1+\frac {1}{4}(1+\frac {1}{4}) +\frac {1}{9}(1+\frac {1}{4} +\frac {1}{9})+....$ Show that $$1+\frac {1}{4} \bigg(1+\frac {1}{4}\bigg) +\frac {1}{9} \bigg(1+\frac {1}{4} +\frac {1}{9}\bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$\bigg(\s... | $$
2S = \sum_{i\leq j} \frac{1}{i^{2}j^{2}} + \sum_{i\geq j} \frac{1}{i^{2}j^{2}} = \left(\sum_{n\geq 1}\frac{1}{n^{2}}\right)^{2} + \sum_{n\geq 1}\frac{1}{n^{4}} = \frac{\pi^{4}}{36} + \frac{\pi^{4}}{90}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
| The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=\sum_{i+j+k=n}\binom{n}{i,j,k}a^ib^jc^k
$$
where $\dbinom{n}{i,j,k} = \dfrac{n!}{i! \, j! \, k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,\qquad i+k\le 7
$$
Hence $k=2i-8$ and $3i-8\le 7$, so $i\ge4$ and $i\le 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$
How do you find the Laurent series for $f(z) = \dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?
I know that $f(z) = \dfrac{z-12}{z^2 + z - 6} = \dfrac{-2}{z-2} + \dfrac{3}{z+3}$
It is easy for me to extract a series for $\dfrac{3}{z+3}$, but have no idea ho... | Hint. Starting from your partial fraction decomposition, we have that
$$\frac{z-12}{z^2 + z - 6} = -\frac{2}{u-1} + \frac{3}{u+4}=-\frac{2/u}{1-1/u} + \frac{3/u}{1+4/u}$$
where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}=\frac{n!}{x(x+1)\cdots(x+n)}$. Given the following formula
$$
\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,.
$$
How can I show that this is equal to
$$
\frac{n!}{x(x+1)\cdots(x+n)}\,?
$$
| Induction step:
$$\begin{align}
\sum_{k=0}^{n+1}&\frac{(-1)^k}{x+k}\binom{n+1}k=\frac1x+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\left[\binom nk+\binom n{k-1}\right]
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\binom{n}{k-1}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
how to determine the bounds of this integral?
Let $ G:= \{ (x,y) \in \mathbb{R}^2 \mid x^2 + 4y^2 >1, x^2+y^2 <4 \}$
I want to determine $ \int_G x^2+y^2 d(x,y)$.
So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $\frac{1}{2}$ in $y$-direction, right?
How c... | Let
$$ x = r \cos \theta, \quad y = r \sin \theta. $$
Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
$$ r^2\cos^2\theta+4r^2\sin^2\theta = r^2+3r^2\sin^2\theta > 1 \quad \iff \quad r > \frac{1}{\sqrt{1+3\sin^2\theta}}. $$
Thus
\begin{align}
\int_G x^2+y^2 \, \mathrm d(x,y) &= \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Simplfying expression efficiently What would be the best way to simplify such an expression? Simply foiling out the expression? could I take out $\frac{b-a}{4}$ and simplifying?
$$\frac{3}{4}(b-a)\left( \frac{2a}{3} +\frac
{b}{3} \right)^2 + \left( \frac{b-a}{4} \right)b^2$$ I know that this is equivalent to $\frac{b^3... | Well, first we notice that there's a common factor of $\frac{b - a}{4}$ in there, so we'll pull that out, leaving
$$\left(\frac{2a+b}{3}\right)^2 + b^2.$$
Expanding that bracket out, we obtain
$$\frac{4a^2 + 4ab + b^2}{3} + b^2$$
Combining those fractions, that's
$$\frac{4a^2+4ab+4b^2}{3}$$
Pulling out the $\frac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\... | Yes because $H_n-\log n$ is bounded by $1$ and $\log_2n$ is a multiple of $\log n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $a$
Find $a$ for which $f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \;$ decreases for all $x$ with $a\neq 1$ and $a\geq -4$.
My try:
For $f(x)$ to decrease
$$5x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) -3 <0\implies x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) < 3/5 $$
How can I proceed... | HINT
We have
$$f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \implies f'(x) = 5\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4-3\le 0$$
then we need
$$\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4- \frac35 \le 0 $$
that is true for all $x$ when
$$\frac{\sqrt{a+4}}{1-a} -1\le0 \iff \frac{\sqrt{a+4}+a-1}{1-a}\le0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Closed form for $K(n)=[0;\overline{1,2,3,...,n}]$ I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $n\in\Bbb N$,
$$K(n)=[0;\overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=... | Following up on Daniel Schepler's comment. Let $$P_n(x) = \frac{1}{1 + \frac{1}{2 + \ddots \frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
\begin{align*}
P_1(x) &= \frac{1}{x+1} \\
P_2(x) &= \frac{x+2}{x+3} \\
P_3(x) &= \frac{2x+7}{3x+10} \\
P_4(x) &= \frac{7x+30}{10x+43} \\
P_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate limit of \Gamma function Show that
$$\lim _{x \to \infty} \log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right) = \frac{1}{2} \log(2),$$ where $\Gamma$ is the Gamma function.
I reduced this problem to calculate the limit:
$$\lim_{x \to \infty} \frac{B\left(... | Old topic, but there is a much easier way to prove it.
First of all, the following relationship exists between Beta and Gamma functions:
$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
So we have:
$$\Gamma \left( \frac{x+1}{2}\right) = \Gamma \left( \frac{x}{2}+\frac{1}{2}\right) = \frac{\Gamma(\frac{x}{2})\Gamma(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
System of 3 equations problem
If
$$x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1$$
Then how much is
$$x^4+y^4+z^4$$
So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.
I did this:
$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+... | Alternativley, working with trinomials is cumbersome, so make them binomials:
$$\begin{cases}x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1\end{cases} \Rightarrow \begin{cases}x+y=1-z\\x^2+y^2={3\over2}-z^2\\x^3+y^3=1-z^3\end{cases}.$$
$(1)^2-(2)$:
$$2xy=2z^2-2z-\frac12$$
$(1)^3-(1)$:
$$3xy(x+y)=3z(z-1) \Rightarrow 3xy(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$
How can we evaluate $$\liminf_{n\to\infty}n\{n\sqrt2\},$$where $\{\cdot\}$ denotes the fractional part of $\cdot$?
The first thing came to my mind is Pell's equation $x^2-2y^2=1$.
Knowing that $\sqrt2$ has a continued fraction $[1;2,2,2\cdots]$, I tried to estimate the li... | For $n > 0$, let $m = \lfloor n\sqrt{2}\rfloor$. Since $\sqrt{2} \not\in \mathbb{Q}$,
$$n\sqrt{2} > m \implies 2n^2 > m^2 \implies 2n^2 \ge m^2 + 1 \implies \sqrt{2}n \ge \sqrt{m^2+1}$$
This implies
$$n\{n\sqrt{2}\} \ge n(\sqrt{m^2+1} - m) \ge \frac{1}{\sqrt{2}}\sqrt{m^2+1}(\sqrt{m^2+1}-m)\\ = \frac{1}{\sqrt{2}}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Trying to find a modulo for $2^{24}$ I was trying to figure out which modulo $n$ would make $2^{24}$ congruent to $1 \bmod n$. One answer is $241$, and I wonder if there is a good way to find it.
I tried to use Fermat's little theorem or Chinese Remainder theorem, but $24$ is not a prime and $241$ is not composite.
| We want to find factors of
\begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\\
&= (2^6-1)(2^6+1)(2^{12}+1)
\end{align}
Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.
Also note that $$2^{24}=(2^3)^8=(7+1)^8$$
Hence we can pick $n=7$.
Also $$2^{24}=(2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find all natural numbers $n$, such that polynomial $n^7+n^6+n^5+1$ would have exactly 3 divisors.
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. ... | You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
\begin{eqnarray} (n^7+n^5)+(n^6+1) &=& n^5(n^2+1) + (n^2+1)(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$ I am in the middle of proving that
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving ... | Another Approach is to employ Feynman's Trick:
Let
$$I(x) = \int_{0}^{1} \frac{\ln\left| x^2\left(t^2 - t\right) + 1\right|}{t^2 - t}\:dt$$
Note $I = I(1)$ and $I(0) = 0$
Thus
\begin{align}
I'(x) &= \int_{0}^{1} \frac{2x\left(t^2 - t\right)}{\left(x^2\left(t^2 - t\right) + 1\right)\left( t^2 - t\right)}\:dt = \frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Factorising $X^{16}- X$ over $\mathbb F_4$. I need to factorise $X^{16}- X$ over $\mathbb F_4$. How might I go about this? I have factorised over $\mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?
Here ... | Let $q:=4$. You are factorizing $X^{q^2}-X$ over $\mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X)\,f_1(X)\,f_2(X)\,\cdots\,f_k(X)$$
where $f_1(X),f_2(X),\ldots,f_k(X)\in\mathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $\mathbb{F}_q$ (whence $k=\dfrac{q^2-q}{2}$, which equals $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I al... | For positive variables we need to prove that
$$\ln\left((1+a)(1+b)(1+c)\right)\geq\ln64$$ or
$$\sum_{cyc}(\ln(1+a)-2\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\right)\geq0,$$
which is true because
$$f(a)=\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Different answers with $\sec(x) = 2\csc(x)$ My son and I were solving this last night and we get different answers depending on which identities we use. The question also did specify $0 \leqslant x < 2\pi$
Here's our work:
$$\sec x = 2 \csc x$$
$$\frac 1 {\cos x} = \frac 2 {\sin x}$$
cross multiply:
$$2 \cos x = \sin ... | 1) $a^2 = b^2$ => $a = b$ or $a = - b$
2) $a = b$ => (square both sides) $a^2 = b^2$
The idea here is that in the first case you have that $a = -b$, but in the second case (which is your case), your $a^2 = b^2$ inevitably adds the $a = -b$ solutions to your total, which obviously are wrong since your original equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^... | Use the trick twice for the numerator to obtain
$$\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}=\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot \dfrac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}=$$
$$=\dfrac{x^4}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}... | By Taylor's expansion we have
$$\left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}=\left(1-\frac{n}{n^{2} +n +1}\right)^{n^2}=e^{n^2 \log\left(1-\frac{n}{n^{2} +n +1}\right)}=e^{n^2 \left(\frac{-n}{n^{2} +n +1}+O(1/n^2)\right)}\sim \frac c{e^n}$$
and then refer to limit comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof verification for $\lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty$
Show that:
$$
\lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty
$$
I've tried the following way. Consider the following sum:
$$
\sqrt n + \sqrt{n-1} + \dots + \sqrt{n-\frac{n}{2}} + \dots + \sqrt{2} + 1
$$
No... | Just to give an alternative, note first that the sequence is increasing:
$$\begin{align}
{1\over n+1}(1+\sqrt2+\cdots+\sqrt n+\sqrt{n+1})-{1\over n}(1+\sqrt2+\cdots\sqrt n)
&={\sqrt{n+1}\over n+1}-{1+\sqrt2+\cdots+\sqrt n\over n(n+1)}\\
&\gt{\sqrt{n+1}\over n+1}-{n\sqrt n\over n(n+1)}\\
&={\sqrt{n+1}-\sqrt n\over n(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Ideals generated by two elements in $\mathbb{Z}[x]$ Consider the following ideal $(2+x,x^2+5)$ in $\mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $... | Note that $\langle 1-4x,x^2+5\rangle=\langle x+20,81\rangle$. This is because
$$x^2+5=(x-20)\cdot(x+20)+5\cdot 81\,,$$
$$1-4x=(-4)\cdot (x+20)+1\cdot 81\,,$$
$$x+20=4\cdot(x^2+5)+x\cdot(1-4x)\,,$$
and
$$81=16\cdot(x^2+5)+(1+4x)\cdot(1-4x)\,.$$
Similarly, $\langle 1-2x,x^2+5\rangle=\langle x+10,21\rangle$. This is bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the following integrals? How the calculate the following integrals? Therein $D$ is a constant.
$$(1)\;\;\int_{0}^{2\pi}\frac{1}{1-D\cdot\cos\theta} d\theta$$
and
$$(2)\;\;\int_{0}^{2\pi}\frac{1}{1+D\cdot\cos\theta} d\theta$$
| $$I_2=\int_0^{2\pi}\frac{d\theta}{1+D\cos(\theta)}=\int_0^{\pi}\frac{d\theta}{1+D\cos(\theta)}+\int_{\pi}^{2\pi}\frac{d\theta}{1+D\cos(\theta)}$$ Let $\theta=\tan(\frac{x}{2})$
$$I_2=\int_0^{\infty}\frac{1}{1+D\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}+\int_{-\infty}^0\frac{1}{1+D\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}=2\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding matrix $A^n,$ when $\lim_{n \to \infty}$
Finding $\lim_{n\rightarrow \infty}\begin{pmatrix}
1 & \frac{x}{n}\\ \\
-\frac{x}{n} & 1
\end{pmatrix}^n$ for all $x\in \mathbb{R}$
Try:
Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$
Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}... | You seem to be arguing that the terms proportional to $x$, $x^2$, $x^3$, etc. vanish in the limit $n \to \infty$. But the problem with this argument is that the numerical coefficients in front of these terms also grow as $n \to \infty$, and diverge without bound. To see this, define
$$
A_n = \begin{bmatrix}1 & x/n\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
maximum value of $x_{1}+x_{2}+x_{3}+\cdots+x_{10}$
Consider the following quantities $x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}$
and $-1\leq x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}\leq 1.$ and
$x^3_{1}+x^3_{2}+\cdots+x^{3}_{10}=0.$
Then maximum of $x_{1}+x_{2}+x_{3}+\cdots\cdots+x_{10}$
Try : Let $x_{i} = \sin(a_{i})$ ... | Let us transform the variables using $f(t)=\sqrt[3]t$, to equivalently maximise $ \sum f(x_k)$ subject to $\sum x_k = 0$ and $x_k \in [-1, 1]$, where index $k\in \mathbb N, 1\leqslant k\leqslant 10$. WLOG using symmetry we may assume the $x_k$ are in non-descending order.
Noticing $f(x)$ is convex for $x< 0$, if the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of piecewise function with $\sin\frac{1}{x}$ term I was going through my calculus book, and I am not sure I understand this part
$f(x) = \begin{cases} \frac{x^2}{4}+x^4\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f'(x) = \begin{cases} \frac{x}{2}-x^2\cos(\frac{1}{x})+4x^3\sin(... | If $x\neq0$, then\begin{align}\frac{f'(x)-f'(0)}x&=\frac{\frac x2-x^2\cos\left(\frac1x\right)+4x^3\sin\left(\frac1x\right)}x\\&=\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\end{align}and therefore\begin{align}f''(0)&=\lim_{x\to0}\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\\&=\frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$
Find the geometric locus of the set of $z \in \mathbb C$ so that
$$\frac{z+2}{z(z+1)}\in \mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution pr... | A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$\dfrac{z+2}{z(z+1)}=\dfrac1{z+1}+\dfrac{2(z+1-z)}{z(z+1)}=\dfrac2z-\dfrac1{z+1}$$
Now the imaginary part of $\dfrac2z$ is $-\dfrac{2y}{x^2+y^2}$
and that of $\dfrac1{z+1}$ is $-\dfrac y{(x+1)^2+y^2}$
Finally, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Floor equation $\lfloor 3x-x^2 \rfloor = \lfloor x^2 + 1/2 \rfloor$
Solve the equation:
$$\left \lfloor 3x-x^2 \right \rfloor = \left \lfloor x^2 + 1/2 \right \rfloor$$
In the solution it writes
We notice that $x^{2}+\frac{1}{2}> 0$ therfore $\left \lfloor x^2 - 1/2 \right \rfloor \geq 0$ . From there $\left \lfloo... | If we put $-x^2+3x$ into translated form we get:
$$-(x^2-3x)=-\left[\left(x-\frac 3 2\right)^2-\left(\frac32\right)^2\right]=-\left(x-\frac 3 2\right)^2+\frac94$$
We can now see that this is $x\mapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $\frac 3 2$ uni... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | Here's a solution using the dual problem. Assume that the solution of the problem occurs at $z = re^{i\theta}.$ Consider
$$
F_r(t) = |re^{it}+r^{-1}e^{-it}|^2 = r^2 + r^{-2} + e^{2it}+e^{-2it} = r^2+r^{-2}+2\cos 2t.
$$ If $z= re^{i\theta}$ is the maximizer, we claim that $F_r'(\theta)=0$. That is, given $|z|=r$, the an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | $$\displaystyle x+\frac{3x-y}{x^2+y^2}=3\cdots (1)$$
$$\displaystyle y-\frac{x+3y}{x^2+y^2}=0\cdots\cdots(2)\times i $$
Now adding these two equations
and substituting $z=x+iy$ and $\bar{z}=x-iy$
and $$|z|^2=x^2+y^2$$
So we have $$z+\frac{3-i}{z}=3\Rightarrow z^2-3z+(3-i)=0$$
On solving that equation we have
$$z=2+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
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.