Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Computing an indefinite integral(2) $$\int \sqrt{1- \sin(2x)}dx =?$$
My attempt:
$$\int \sqrt{1- \sin(2x)}dx = \int \sqrt{(\cos x - \sin x)^{2}} dx = \int|\cos x- \sin x| dx = ??$$
| Notice, $$\cos x-\sin x=\sqrt 2\left(\frac{1}{\sqrt 2}\cos x-\frac{1}{\sqrt 2}\sin x\right)=\sqrt 2\sin\left(\frac{\pi}{4}-x\right)$$
but $\sin \theta\ge 0\iff 2k\pi\le \theta\le 2k\pi+\pi$
hence, solving the inequality
*
*$$2k\pi\le \left(\frac{\pi}{4}-x\right)\le 2k\pi+\pi$$
$$ 2k\pi\le \left(\frac{\pi}{4}-x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Examples for $ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$ I saw the answers here about how to prove:
$$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrat... | If $x>1$ then $0< \frac 1x < 1$ so $\lfloor \frac 1x \rfloor = 0$ and the equation reduces to
$$x + \frac{1}{x} = 1 + \lfloor x \rfloor$$
Any number $x>1$ can be written on the form $x = n + r$ where $n$ is an integer and $0<r<1$. Take this form for $x$ in the equation above. Since $\lfloor n + r\rfloor = n$ the equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $1234^{1234} \pmod{5379}$? Note: $5379 = 3 \times 11 \times 163$.
I tried Chinese Remainder Theorem and Fermat's Little Theorem, got as far as:
$$
1234^{1234} = 1 \pmod{3} \\
1234^{1234} = 5 \pmod{11}
$$
With a bit more work:
$$1234^{1234} = 93^{100} \pmod{163}$$
But $93^{100}$ doesn't really help?
Wolf... | Well this is far from perfect,but it works if you have enough time or a calculator.
$$93\equiv -70\pmod{163}$$
$$\begin{align}
93^{100}&\equiv(-70)^{100}\\
&= 490^{50}\cdot10^{50}\\
&\equiv 10^{50}\\
&= 2^{50}\cdot 5^{50}\\
&= 1024^5\cdot 3125^{10}\\
&\equiv 46^5\cdot 28^{10}\\
&=2^{25}\cdot 23^5\cdot 7^{10}\\
&= 2^{25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$ How to solve this? Any advice?
$$1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$$
Next step I do this
$\sum\limits_{n=0}^\mathbb{\infty}(-1)^n \tan^nx = \frac{\tan 2x}{1+\tan2x} $
But I don't know next step. I am cule... | Notice, for the sum of infinite series on LHS, $|\tan x|<1$
$$1-\tan x+\tan^2 x-\tan^3 x+\ldots =\frac{\tan 2x}{1+\tan 2x}$$
$$\frac{1}{1-(-\tan x)}=\frac{\frac{2\tan x}{1-\tan^ 2x}}{1+\frac{2\tan x}{1-\tan^ 2x}}$$
$$\frac{1}{1+\tan x}=\frac{2\tan x}{1-\tan^2 x+2\tan x}$$
$$1-\tan^2 x+2\tan x=2\tan x+2\tan^2 x$$
$$3\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step:
Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$
Now, this is what I have done:
$$
(\frac{a+x}{a-x})^{\frac{3}{2}} \\
\implies \frac{\delta}{\delta y}\frac{f}{g} \\
\implies gf' = (a-x)^{\frac{3}{2}} \t... | i think the right answer is this here
$$\frac{3}{2} \sqrt{\frac{a+x}{a-x}}
\left(\frac{a+x}{(a-x)^2}+\frac{1}{a-x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the probl... | the case $x+2>0$, you do not need to solve $-1/2\le 1/(x+2)\leftarrow x+2>0$.
the case $x+2<0$, the inequality you solve it's not the original one.
It should be $-1/2 \le 1/(x+2) \le 3/2$ and in case $x+2<0\rightarrow 1/(x+2)<3/2$, so it's just $-1/2 \le 1/(x+2)$.
Think it simply.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to compute this finite sum $\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$? I do not know how to find the value of this sum:
$$\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$$
(Yes, the last term is added twice).
Of course I've already plugged it to wolfram online, and the answer is $$2-\frac{1}{2^{n-1}}$$
But I do not kn... | Let $$S=\sum_{k=1}^{n}\frac{k}{2^k}$$
or, $$S=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{n}{2^n}$$
and $$\frac{S}{2}= \frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{n-1}{2^n}+\frac{n}{2^{n+1}}$$
Subtracting we get,
$$\frac{S}{2}= \frac{1}{2^1}+\frac{1}{2^2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
prove polynomial division for any natural number Show that for any natural numbers $a$, $b$, $c~$ we have $~x^2 + x + 1|x^{3a+2} + x^{3b+1} + x^{3c}$.
Any hints on what to use?
| We have
$$x^3-1=(x-1)\left(x^2+x+1\right)$$
This means that
$$\begin{align}x^3-1&\equiv0\pmod{x^2+x+1}\\x^3&\equiv1\pmod{x^2+x+1}\end{align}$$
Now, substitute it
$$\begin{align}x^{3a+2}+x^{3b+1}+x^{3c}&\equiv x^2\left(x^3\right)^a+x\left(x^3\right)^b+\left(x^3\right)^a\\&\equiv x^2\cdot1^a+x\cdot1^b+1^c\\&\equiv x^2+x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !
| There are some errors in your calculation, e.g. a missing factor 2 in
$$
(\sqrt{x-1}+\sqrt{y-1})^2 = x - 1 + y - 1 + 2\sqrt{x-1}\sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x \ge 1$, $y \ge 1$ you can square the inequality (since both
sides are non-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
how to prove that the sum of the power two of three reals is less than 5? $a, b$ and $c$ are three real number such that: $(\forall x \in [-1,1]): |ax^2+bx+c|\leq 1$ .
Prove that:
1) $|c|\leq 1$ .
2) $-1\leq a+c\leq 1$ .
3)Deduce that : $a^2+b^2+c^2\leq 5$.
The first and second questions are easy to prove ( just take... | $x=0$ gives us $|c| \le 1$. Similarly set $x = \pm1$ to get $|a \pm b+c| \le 1$. So we have the inequalities:
$$-1 \le c \le 1 \iff -1 \le -c \le 1\tag{1}$$
$$-1 \le a+b+c \le 1 \iff -1\le -a-b-c \le 1 \tag{2}$$
$$-1 \le a-b+c \le 1 \iff -1 \le -a+b-c \le 1\tag{3}$$
Note the right side of the above three inequalities... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | By the mean value theorem - MVT, for all $x >0$, it exists $\zeta_x \in (x, x+1)$ such that $$\frac{1}{2\sqrt{\zeta_x}}((x+1)-x)=\frac{1}{2\sqrt{\zeta_x}}=\sqrt{x+1}-\sqrt{x}$$ As $y \to \frac{1}{2\sqrt{y}}$ is stricly decreasing, you have $$\frac{1}{2\sqrt{\zeta_{x+1}}}=\sqrt{x+2} -\sqrt{x+1} < \sqrt {x+1}-\sqrt{x}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$? $$
\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx
$$
A bit confused with how to integrate this question. I though it was partial fractions but was unsure about the what to do after that.
| Partial Solution
Well, assuming you did the partial fraction decomposition already (as you said you did), you should get the following integral
$$\int\left(\frac{9-x}{2(x^2+1)}+\frac{1}{x-1}+\frac{1}{2 (x+1)}\right) dx$$
$$=\frac{9}{2}\int\frac {dx}{x^2+1} - \frac{1}{2}\int \frac {x}{x^2+1}dx + \int \frac{dx}{x-1}+ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Linear Transformations, Linear Algebra Let T:P3→P3 be the linear transformation such that $T(−2x^2)= −2x^2 − 2x$, $T(0.5x + 2)= 3x^2 + 4x−2$, and $T(2x^2 − 1)= 2x + 1$.
Find $T(1), T(x), T(x^2)$, and $T(ax^2 + bx + c)$, where a, b, and c are arbitrary real numbers.
I understand how to find $T(x^2)$ where you just divi... | We can write $1=-(-2x^2)-(2x^2-1), x=2(0.5x+2)-4$, so
$$T(1)=T(-(-2x^2)-(2x^2-1))=-T(-2x^2)-T(2x^2-1)=2x^2-1$$
$$T(x)=T(2(0.5x+2)-4)=2T(0.5x+2)-4T(1)=-2x^2+8x$$
Then we get
$$T(ax^2+bx+c)=a(x^2+x)+b(-2x^2+8x)+c(2x^2-1)$$
$$=(a-2b+2c)x^2+(a+8b)x-c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$.
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
$0<u<\sqrt{2}$ and $v>0$.
I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\fr... | You can simply think $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ is the square of distance between two points $a$ and $b$ in the plane, where
$$a=(u,\sqrt{2-u^2}),b=(v,\frac{9}{v}).$$
Clearly, $a$ satisfies $x^2+y^2=2,(0<x<\sqrt{2})$ and $b$ satisfies $y=\frac{9}{x},(x>0)$. Then from the geometry graph, we easily know the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What are the residues of $\frac{z^2 e^z}{1+e^{2z}}$? I hope to know the singularity and pole of $\frac{z^2 e^z}{1+e^{2z}}$.
I try $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{e^{-z}+e^{z}}$ and observe that the denominator seems like cosine function. So I think the singularites are i(2n+1)$\pi$/2. But when I try to evaluate ... | Since $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{2\cosh z}$, let's start with
$$f(z)=\frac{z^2}{2\cos iz}$$
Now both $z^2$ and $\cos iz$ are entire. The function can be singular only where $z_{0_n} = iz = \frac{\pi +2\pi n}{2}$ and since the zeroes of the denominators are first order, we have
$$\begin{align*}
Res f(z_{0_n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
What is the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$ I want to find the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$
from $n(n-1)(n+1)$ 2 and 3 should divide this expression for all positive n. how can I find the rest? which python says $(2, 3, 6, 7, 9, 42, 14, 18, 21, 63,126)$
| If $$n\equiv 0 \mod 3,$$ then the factor $$n^2+3\equiv 0 \mod 3.$$ Otherwise
$$n^2\equiv1 \mod 3$$ and thus $$n^2+5\equiv 0\mod3.$$
Therefore $$(n^2+3)(n^2+5)\equiv0\mod3.$$
Let us now look modulo $7$ and compute the number for all elements of $\mathbb Z_7$ (which means to check if the number is a multiple of $7$ for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a... | Hint: as we can have only cubic terms in the symmetric polynomial sums, the only terms which can be used are of form $(\sum x)^3, \sum x \sum xy$ and $\sum xyz$. Then it is a matter of testing $3$ coefficients...
$$\sum_{cyc} x_1^3 = \left(\sum_{cyc} x_1 \right)^3-3\left(\sum_{cyc} x_1 \right)\left(\sum_{cyc} x_1 x_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$
show that
$$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$
I have one methods to solve this problem:
Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$
... | Let $A (x) = (1-x+x^2-x^3+\cdots-x^{99}+x^{100})$ and $B (x) = 1+x+x^2+\cdots+x^{100}$. Then we have $g(x) = A (x) B (x)$. We have $(x + 1) A (x) = x^{101} + 1$ and $(x - 1) B (x) = x ^ {101} - 1$. Then $$g (x) = \frac {x^ {202} - 1} {x^2 - 1} = x^{200} + x^{198} + \cdots + 1,$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Solve $x-\lfloor x\rfloor= \frac{2}{\frac{1}{x} + \frac{1}{\lfloor x\rfloor}}$ Could anyone advise me how to solve the following problem:
Find all $x \in \mathbb{R}$ such that $x-\lfloor x\rfloor= \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{\lfloor x\rfloor}},$ where $\lfloor *\rfloor$ denotes the greatest integer function.
He... | Completing the square gives
$$
x^2-2x\lfloor x\rfloor+\lfloor x\rfloor^2=2\lfloor x\rfloor^2
$$
so
$$
(x-\lfloor x\rfloor)^2=2\lfloor x\rfloor^2
$$
Since $0\le x-\lfloor x\rfloor<1$, we conclude $\lfloor x\rfloor=0$ that's disallowed by the starting equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_\limits{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1}$ For all $n,k \in N a,b > 0$
$$\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{(1+ \frac{ax}{n})(1+ \frac{bx}{k})- 1}= \lim_{x \to 0}\frac{x}{x(\frac{a}{n} + \frac{b}{k}) + \frac{ab}{nk}x^2} = \lim_{x... | The limit is fine. Perhaps with a view to formalizing the procedure is appropriate to introduce an infinite series , and then Landau notation.
$$O(g(x)) = \left\{\begin{matrix} f(x) : \forall x\ge x_0 >0 , 0\le |f(x)|\le c|g(x)| \end{matrix}\right\}$$
We will also use:
$$f_1=O(g_1)\wedge f_2=O(g_2)\implies f_1f_2=O(g_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What does it mean to represent elements of an ideal? Say I have the polynomial $x^9 + 1$
Then: $x^9 + 1 = (x+1)(x^2 + x + 1)(x^6 + x^3 + 1)$
is a complete factorization over $GF(2)$ of $x^9 + 1$
The dimension of each ideal is: length $n - deg(ideal)$
So for $n=9$, dimension of $(x+1)$ = $9-1=8$
of $(x^2 + x + 1) = 9 -... | Let’s call $f=x^6+x^3+1$. You want three linearly independent elements of the ideal $(f)$ of the ring $R=\Bbb F_2[x]/(x^9+1)$. Since $(f)$ is just the set of multiples of $f$, you certainly have $1\cdot f$, $xf$, and $x^2f$. Notice that $x^3f=x^9+x^6+x^3=1+x^6+x^3=1\cdot f$, already counted. I’ll leave it to you to sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \... | $$
\lim_{x \to 0} \left( 1 - \frac {x^2}6\right)^{\frac 1{x^2}} = \lim_{x \to 0} \left [ \left( 1 - \frac {x^2}6\right)^{\frac 6{x^2}} \right ]^{\frac 16} = \left [ \left (e^{-1} \right ) \right ]^{\frac 16} = e^{-\frac 16}
$$
Here, I used somewhat modified limit regarding Euler's number
$$
\lim_{t \to 0} \left( 1 - t\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Positive equilibria for a system of eqautions I have the following system of equations
\begin{align}
\frac{dx}{d \tau} &= x \left(1-x-\frac{y}{x+b} \right) \\
\frac{dy}{d \tau} &= cy \left(-1+a\frac{x}{x+b} \right)
\end{align}
I am asked to show that if $a<1$, the only nonnegative equilibria are $(0,0), (1,0)$.
So fir... | If $x \ne 0$, We have
$$ 1 - a\frac{x}{x+b} = 0 $$
$$ a\frac{x}{x+b} = 1 $$
$$ ax = x + b$$
$$ (a-1)x = b$$
$$ x = \frac{b}{a-1} $$
If $a < 1$ and $b > 0$, $x < 0$ and therefore is not a non-negative solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
calculate $\lim\limits_{x \to 1}(1 - x)\tan \frac{\pi x}{2}$ I need to calculate $$\lim_{x \to 1}\left((1 - x)\tan \frac{\pi x}{2}\right)$$.
I used MacLaurin for $\tan$ and got $\frac{\pi x} {2} + o(x)$. Then the full expression comes to $$\lim_{x \to 1}\left(\frac {\pi x} {2} - \frac {\pi x^2} {2} + o(x)\right) = 0$$B... | $L=\lim\limits_{x \to 1}((1 - x)\tan \frac{\pi x}{2})$
$L=\lim\limits_{(x-1) \to 0}((1 - x)\cot(\frac{\pi }{2}- \frac{\pi x}{2}))$
$L=\frac{2 }{\pi}\lim\limits_{\frac{\pi }{2}(x-1) \to 0}(\frac{\pi }{2}(1 - x)(-)\frac{cos\frac{\pi }{2}(x- 1)}{sin\frac{\pi }{2}(x- 1)})$
$L=\frac{2 }{\pi}\lim\limits_{\frac{\pi }{2}(x-1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Limit of $\lim\limits_{n\to\infty}\frac{\sum_{m=0}^n (2m+1)^k}{n^{k+1}}$ I wanted to find the limit of: ($k \in N)$
$$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}}.$$
Stolz–Cesàro theorem could help but $\frac{a_n-a_{n-1}}{b_n-b_{n-1}}$ makes big mess here:
$$\lim_{n \to \infty}{\frac{-0^k+1^k-2^k+3^... | Using Faulhaber's formula,
$\lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}} = \frac{1}{k+1}$.
Then,
$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}} = \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+(2n+1)^k}{n^{k+1}}} - 2^k \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}}$
$ = \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to manually calculate the sine? I started studying trigonometry and I'm confused.
How can I manually calculate the sine? For example: $\sin(\frac{1}{8}\pi)$?
I was told to start getting the sum of two values which will result the sine's value. For $\sin(\frac{7}{12}\pi)$, it would be $\sin(\frac{1}{4}\pi + \frac{1}... | For $\frac{\pi}{8}$ you need to use double angle formula.
Recall that $\sin[2\theta)=2\sin\theta\cos\theta=2\sin\theta\sqrt{1-\sin^2\theta}$
Then let $\theta=\frac{\pi}{8}$:
$$\sin\left(\frac{\pi}{4}\right)=2\sin\frac{\pi}{8}\sqrt{1-\sin^2\frac{\pi}{8}}$$
$$\frac{1}{2}=4\sin^2\frac{\pi}{8}\left(1-\sin^2\frac{\pi}{8}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Converting a sum of trig functions into a product Given,
$$\cos{\frac{x}{2}} +\sin{(3x)} + \sqrt{3}\left(\sin\frac{x}{2} + \cos{(3x)}\right)$$
How can we write this as a product?
Some things I have tried:
*
*Grouping like arguments with each other. Wolfram Alpha gives $$\cos{\frac{x}{2}} + \sqrt{3}\sin{\frac{x}{2}... | First of all
$$
A\cos\alpha+B\sin\alpha=\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos\alpha+\frac{B}{\sqrt{A^2+B^2}}\sin\alpha\right)\\
=\sqrt{A^2+B^2}(\sin\beta\cos\alpha+\cos\beta\sin\alpha)=\sqrt{A^2+B^2}\sin(\beta+\alpha)
$$
you only have to find $\beta$ such that
$$
\left\{
\begin{align}
\sin\beta&=\frac{A}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why I am getting wrong answer to this definite integral? $$\int_0^{\sqrt3} \sin^{-1}\frac{2x}{1+x^2}dx $$
Obviously the substitution must be $x=tany$
$$2\int_0^{\frac{\pi}{3}}y\sec^2y \ dy $$
Taking $u=y $, $du=dy;dv=sec^2y \ dy, v=\tan y $
$$2\Big(y\tan y+\ln(\cos y)\Big)^{\frac{\pi}{3}}_{0} $$
Hence $$2\frac{\pi}{\s... | $$I=\int_{0}^{\sqrt{3}} \sin^{-1} \frac{2x}{1+x^2} dx$$
Notice that $$\frac{d}{dx} \sin^{1} \frac{2x}{1+x^2}= \frac{2}{1+x^2}~~ \frac{|1-x^2|}{(1-x^2)}= \frac{2}{1+x^2}, ~if~~ x^2<1,~~\frac{-2}{1+x^2}~if~ x^2>1.$$
So let us integrate by parts taking 1 as the second function. Then
$$I=\left ( x \sin^{-1} \frac{2x}{1+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| HINT: Note that $$\color{Green}{x^2=-x+1}.$$
By multiplying the given expression by $x,$ we can obtained $$\color{Green}{x^3=2x-1}.$$ Again by multiplying $x$ we have $$\color{Green}{x^4=-3x+2}$$ and so on..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question
| By mathematical induction, say $P(n):1+4+9+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$
Now $P(1)$ is true as L.H.S. $= 1$ and R.H.S. $= 1$
Say $P(k)$ is true for some $k \in \mathbb{N}$ and $k>1$.
Therefore $1+4+9+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$
Now for $P(k+1)$,
\begin{align}
& 1+4+9+\cdots+k^2+(k+1)^2 \\[10pt]
= {} & \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the characteristic polynomial of $(M^{-1})^3$ Given that M is a square matrix with characteristic polynomial
$p_{m}(x) = -x^3 +6x^2+9x-14$
Find the characteristic polynomial of $(M^{-1})^3$
My attempt:
x of $(M^{-1})^3$ is $1^3$, $(-2)^3$ , $7^3$ = $1$ , $-8$ , $343$
$p_{(m^-1)^3} = (x-1)(x+8)(x-343)$
or $-(x-1)(x... | Note that $-x^3+6x^2+9x-14=-(x-1)(x+2)(x-7)$, so we may assume that $M$ is the diagonal matrix with entries $1,-2,7$ on the diagonal. Then it's easy to see that the characteristic polynomial of $M^{-3}$ is given by $(x-1)(x+(1/2)^3)(x-(1/7)^3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to eliminate $\theta$? While doing a sum I was stuck in a particular step:
$$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$
How to eliminate $\theta$ ?
| $$\begin{cases}
\displaystyle
r_1=\frac{4a\cos\theta}{\sin^2\theta}\\
\displaystyle
r_2=\frac{4a\sin\theta}{\cos^2\theta}\\
\end{cases}$$
Try to find $\sin\theta$ and $\cos\theta$ as following:
$$r_1\sin^2\theta=4a\cos\theta\tag{1}$$
$$r_2\cos^2\theta=4a\sin\theta\tag{2}$$
To equation $(1)$ plug computed $\sin\theta$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:
Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations
What I've Tried
*
*$ \frac{dx}{dt} = -5\sin(t) $
*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of... | You computed the slopes as $\frac{dy}{dx}=\frac{3}{5}$ and $\frac{dy}{dx}=-\frac{3}{5}$, but for some reason when you wrote the equations of the tangent lines you took the reciprocal of these slopes and wrote $y=\frac{5}{3}x$ and $y=-\frac{5}{3}x$. They should be $y=\frac{3}{5}x$ and $y=-\frac{3}{5}x$. Other than that,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Complex equation: $z^8 = (1+z^2)^4$ What's up with this complex equation?
$ z^8 = (1+z^2)^4 $
To start with, there seems to be a problem when we try to apply root of four to both sides of the equation:
$ z^8 = (1+z^2)^4 $
$ z^2 = 1 + z^2 $
which very clearly doesn't have any solutions, but we know there are solutions: ... | The solutions of $z^8 = (1+z^2)^4$, and the problem you see, are clearly seen from a factorization process:
$$\begin{align}
z^8 - (1+z^2)^4 &= 0 \,, \\
[z^4 + (1+z^2)^2] [z^4 - (1+z^2)^2] &= 0 \,, \\
[z^2 + i(1+z^2)] [z^2 - i (1+z^2)] [z^2 - (1+z^2)] [z^2 + (1+z^2)] &= 0 \,.
\end{align}$$
From the last line, it is cle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Is This Matrix Diagonalizable? Consider the matrix A below:
$$A = \begin{pmatrix} 3 & 2 & 2 \\ -1 & 0 & -2 \\ 1 & 2 & 4\end{pmatrix}$$
Is the matrix A diagonalizable? If so, then find a diagonal matrix D and an invertible matrix P such that $P^{-1}AP=D$.
I know it is supposed to be diagonalizable but I have tried to so... | Assuming $A$ is an $n\times n$ matrix, $A$ diagonalizable if $P_A(\lambda)$ has $n$ distinct real roots or if, for each eigenvalue, the algebraic multiplicity equals the geometric multiplicity. So let's find out:
$$P_A(\lambda) = det(A - \lambda I) = det \begin{pmatrix} 3-\lambda & 2 & 2 \\ -1 & -\lambda & -2 \\ 1 & 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of... | Here's a solution that exploits a comment of Oscar Lanzi under my other answer (using an observation that I learned from a note of Noam Elkies [pdf]). In particular, it avoids both the identity $\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6}$ and using integration.
Let $\Bbb P$ denote the set of prime numbers and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Recurrence relation for the determinant of a tridiagonal matrix Let
$$f_n := \begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & c_{n-1} & a_n
\end{vmatrix}$$
Apparently, the determinant of the tridiagional matrix above is given by the recurrence relation
$... | For $n \ge 2$ using Laplace expansion on the last row gives
\begin{align}
f_n &=
\begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-3} \\
& & & c_{n-3} & a_{n-2} & b_{n-2} \\
& & & & c_{n-2} & a_{n-1} & b_{n-1} \\
& & & & & c_{n-1} & a_n
\end{vmatrix}
\\
&=
(-1)^{2n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more c... | We don't need to use any integration by parts.
We start by completing the square
$$\int \frac{x+1}{(x^2-x+8)^3} dx = \int \frac{x-\frac{1}{2}+\frac{3}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx =$$
$$\underbrace{ \int \frac{x-\frac{1}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx}_{=:I_1}+ \frac{3}{2}\underbrace{\int\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | 0/0 is indeterminate. Which means it has infinite number of values.
(0)(1) = 0 => 0/0 = 1
(0)(2) = 0 => 0/0 = 2
(0)(4) = 0 => 0/0 = 3 .....
From above it can be shown that 0/0 has infinite number of values which make it meaningless. Hence the term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 15
} |
What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| Notice, for $(Bx+C)$ part, you should use separation as follows $$\int \frac{x-1}{(x+3)(x^2+1)}\ dx=\int \frac{-2}{5(x+3)}+\frac{2x-1}{5(x^2+1)}\ dx$$
$$=\frac{1}{5}\int \left(-\frac{2}{x+3}+\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\ dx$$
$$=\frac{1}{5} \left(-2\int \frac{1}{x+3}\ dx+\int \frac{d(x^2)}{x^2+1}-\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Linear Algebra - seemingly incorrect result when looking for a basis
For the following matrix $$ A = \begin{pmatrix}
3 & 1 & 0 & 0 \\
-2 & 0 & 0 & 0 \\
-2 & -2 & 1 & 0 \\
-9 & -9 & 0 & -3
\end{pmatrix} $$
Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value.
First step is to find the ... | You're missing an eigenvalue! I think you've made a typo in your characteristic polynomial, writing $-\lambda^4$ instead of $\lambda^4$. The characteristic polynomial factorizes to $(x-2)(x+3)(x-1)^2$, hence the eigenvalues are $-3$, $2$, and $1$. The vector you have found is indeed a basis for the 2-eigenspace, and y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Adding and multiplying piecewise functions How do I add and multiply two piecewise functions?
$$
f(x)=
\begin{cases}
x+3 &\text{if }x<2\\
\dfrac{x+13}{3} &\text{if }x>2
\end{cases}
$$
$$
g(x)=
\begin{cases}
x-3 &\text{if }x<3\\
x-5 &\text{if }x>3
\end{cases}
$$
| It helps to think about the fact that you can only add and multiply stuff that actually exists (i.e. is defined):
We can only add and multiply these functions in places where they both exist at the same time, namely:
$x<2\\
2<x<3\\
x>3$
The final step is to check what the functions equal at each of these segments, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\g... | Using the information you supplied, you could also just square both sides of the original inequality to obtain $a^2+2ab+b^2 \geq c^2$. Divide the inequality by 2 and subtract $ab$ to the other side. Given the AM-GM, we know that $\frac{a^2+b^2}{2} \geq ab $. Adding this inequality to the previous one we obtain $2\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playi... | Notice, the area bounded by the curves from $x=0$ to $x=2$, is given as $$\int_{0}^{2}\left(\frac{2x}{x-3}-(x^2-4x)\right)\ dx$$
$$=\int_{0}^{2}\left(\frac{2(x-3)+6}{x-3}-x^2+4x\right)\ dx$$
$$=\int_{0}^{2}\left(2+\frac{6}{x-3}-x^2+4x\right)\ dx$$
$$=\left(2x+6\ln|x-3|-\frac{x^3}{3}+2x^2\right)_{0}^{2}$$
$$=4-\frac{8}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \inf... | The series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges if $|x|<\frac{1}{3}$
Indeed $$(3^n + n)\cdot x^n\sim_{\infty} 3^n \cdot x^n$$ but this is a geometric series that converges only if $|x|<\frac {1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Sum of all values of $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$ Find sum of all possible values of the parameter $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$.... | Since $f(x) = (x-b)^2 + (1 - b^2)$, the vertex will be at $(b, 1-b^2)$.
Also $f(1) = 2-2b$ and $f(0) = 1$.
When the vertex is at or to the left of $x=0$, then $f(x)$ is increasing on $[0, 1]$. So
$$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(1) - f(0) = 1-2b$$
and $1-2b=4$ when $b = -\frac 32$
When the vertex ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd ... | The result is $a_{n + 1} = 3 a_n - a_{n - 1}$.
HINT: Let $a_{k + 1} = a_k + b_k$. Then, by the original equation, we get $$a_{n - 1} + b_{n - 1} + b_n = \frac {(a_{n - 1} + b_{n - 1})^2 + 5} {a_{n - 1}} = a_{n - 1} + 2 b_{n - 1} + \frac {b_{n - 1}^2 + 5} {a_{n - 1}}.$$ Hence, $$b_n = b_{n - 1} + \frac {b_{n - 1}^2 + 5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Trig : With point $(X,Y)$ and a end point $(x_2,y_2)$, work out the 2 angles So imagine your arm, you have a shoulder point be $X,Y (0,0)$ you have a angles off this, then you have a elbow point $(x_1,y_1)$ and another angles which ends at your hand.
If you know the arm length (4 and 4), you know your starting point $(... | Draw a line from the origin to the hand, called the distance line, then you have (based on your image, I assumed $y = -2$):
$$
\begin{eqnarray}
l &=& 4 & \textrm{length of an arm segment} \\
d &=& \sqrt{x^2 + y^2} = 2\sqrt{10} & \textrm{distance from the origin to the hand} \\
\tan \theta_1 &=& \frac{y}{x} = -\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$ I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$
I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:
$$ \frac{... | $$
\begin{align}
\frac1{1+z^2}
&=\frac1{2+2w+w^2}\tag1\\[9pt]
&=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt]
&=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\
&=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by
$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left|
\begin{array}{c}
\frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\
\frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\fra... | By your definition, $\mathcal{C} = \{(x,y) \in \mathbb{R}^{2}: x^4 + y^4 = r^4\}$. Which can be parametrized as
\begin{align}
\mathcal{C} =
\begin{cases}
\left(+\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\
\left(+\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r\\
\left(-\sqrt{\cos (\theta )},+\sqrt{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Is there a special value for $\frac{\zeta'(2)}{\zeta(2)} $? The answer to an integral involved $\frac{\zeta'(2)}{\zeta(2)}$, but I am stuck trying to find this number - either to a couple decimal places or exact value.
In general the logarithmic deriative of the zeta function is the dirichlet series of the van Mangolt ... | By differentiating both sides of the functional equation$$ \zeta(s) = \frac{1}{\pi}(2 \pi)^{s} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s),$$ we can evaluate $\zeta'(2)$ in terms of $\zeta'(-1)$ and then use the fact that a common way to define the Glaisher-Kinkelin constant is $\log A = \frac{1}{12} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if ... | It's worth noting that you haven't actually avoided the quotient rule, at all. Rather, you've simply written out the quotient rule result in a different form. However, we can avoid the quotient rule as follows.
First, we clear the denominator to give us $$x^2(x-y)=x+y,$$ or equivalently, $$x^3-x^2y=x+y.$$ Gathering the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The value of double integral $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$? Given double integral is :
$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$$
My attempt :
We can't solve since variable $x$ can't remove by limits, but if we change order of integration, then
$$\int _0^1\int _0^{\frac{1}{x}... | $$\begin{align}\int_{0}^{1}\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\text{d}y&=
\int_{0}^{1}\left(\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{1}{1+y^2}\int_{0}^{\frac{1}{x}}x\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\left[x^2\right]_{0}^{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Division in finite fields Let's take $GF(2^3)$ as and the irreducible polynomial $p(x) = x^3+x+1$ as an example. This is the multiplication table of the finite field
I can easily do some multiplication such as $$(x^2+x)\cdot(x+1) = x^3 + x^2 + x +1 = x+1+x^2+x+2 = x^2$$
I am wondering how to divide some random fields s... | If we have a field $K = F[X]/p(X)$, then we can compute the inverse of $\overline{q(X)}$ in $K$ as follows.
Since $p$ is irreducible, either $q$ is zero in $K$ or $(p,q)=1$. By the Euclidean algorithm, we can find $a,b$ such that $a(X)p(X) + b(X)q(X) = 1$. Then $\overline{b(X)} \cdot \overline{q(X)} = \overline{1}$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds?
Three players each have a red card, blue card and green card. The players will play a game that consists of three rounds. In each of the three rounds each player randomly turns over one of his/her ... | The probability that the three players choose different colors on the first turn is $\frac{3}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$. Given that they do, consider the generating function for the colors played on the second turn: $(r+b)(b+g)(r+g)=\color{red}1b^2 2g+\color{red}1b^2 r+\color{red}1b g^2+\color{gre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Writing a summation as the ratio of polynomial with integer coefficients
Write the sum
$\sum _{ k=0 }^{ n }{ \frac { { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) }{ { k }^{ 3 }+9{ k }^{ 2 }+26k+24 } } $ in the form $\frac { p(n) }{ q(n) }$, where $p(n)$ and $q(n)$ are polynomials with integral c... | $\sum _{ k=0 }^{ n }{ \frac { { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) }{ { k }^{ 3 }+9{ k }^{ 2 }+26k+24 } } = \sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{2(k+2)} }-\sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{(k+3)} }+\sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{2(k+4)} }$
Now consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Behaviour a function when input is small If I have the function:
$P(N)=\frac{P_0N^2}{A^2+N^2}$, with $P_0, A$ positive constants
For small $N$, am I right in thinking that because $A$ dominates $N$ we have that
$P(N) \approx \frac{P_0N^2}{A^2}$
| More precisely, we can write for $A^2>N^2$
$$\begin{align}
P(N)&=P_0\left(\frac{N^2}{A^2+N^2}\right)\\\\
&=P_0\left(\frac{(N/A)^2}{1+(N/A)^2}\right)\\\\
&=P_0\frac{N^2}{A^2}\sum_{k=0}^\infty (-1)^k\left(\frac{N^2}{A^2}\right)^{k}\\\\
&=\frac{P_0N^2}{A^2}-P_0\frac{N^4}{A^4}+P_0\frac{N^6}{A^6}+O\left(\frac{N^8}{A^8}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0Let $I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0<x<\pi$ and $\tan I=\frac{2}{\sqrt3}$.Find $x$.
$\int\frac{\sqrt{1+\cos t}}{17-8\cos t}dt=\int\frac{\sqrt2\cos\frac{t}{2}}{17-8\times\frac{1-\tan^2\frac{t}{2}}{1+\tan^2\frac{t}{2}}}dt=\... | HINT:
$$\frac{\sqrt{1+\cos t}}{17-8\cos t}=\dfrac{\sqrt2\left|\cos\dfrac t2\right|}{17-8\left(1-2\sin^2\dfrac t2\right)}$$
Set $\sin\dfrac t2=u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | It can be easily shown that the series on the left reduces to the sum of integers, multiplied by $-1$ if $k$ is even.
For even $k$:
$$\begin{align}
&1^2-2^2+3^3-4^2+\cdots+(k-1)^2-k^2\\
&=(1-2)(1+2)+(3-4)(3+4)+\cdots +(\overline{k-1}-k)(\overline{k-1}+k)\\
&=-(1+2+3+4+\cdots+\overline{k-1}+k)\\
&=-\frac {k(k+1)}2\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by
$$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?
| By Wilson's theorem, $12!\equiv -1\pmod{13}$.
By Wolstenholme's theorem, $H_{12}\equiv 0\pmod{13}$. Since:
$$ a = \sum_{k=1}^{12}\frac{23!}{k} + (12!)(14\cdot 15\cdot\ldots\cdot 23)+\sum_{k=14}^{23}\frac{23!}{k} $$
we have:
$$ a\equiv 0-10!+0 \equiv \frac{-12!}{11\cdot 12}\equiv \frac{1}{(-2)\cdot(-1)}\equiv\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:
Let $x$, $y$, $z$ be real
numbers. Prove that
$$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$
The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square ... | For real $x$,we have(or Use Cauchy-Schwarz inequality)
$$x^2+1\ge\dfrac{1}{2}(x+1)^2$$
the same we have
$$y^2+1\ge\dfrac{1}{2}(y+1)^2$$
$$z^2+1\ge\dfrac{1}{2}(z+1)^2$$
so
$$(x^2+1)(y^2+1)(z^2+1)\ge\dfrac{1}{8}[(x+)(y+1)(z+1)]^2$$
Use AM-GM inequality
$$\dfrac{1}{8}[(x+1)(y+1)(z+1)]^2+8\ge 2\sqrt{\dfrac{1}{8}[(x+)(y+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Geometric Inequality $(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)\leq (abc)^{2}$ everyone.
$a$,$b$,$c$ are three sides of a triangle.
Prove or disprove the following.
$(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)\leq 8(abc)^{2}$
I know two inequalities.
$8(s-a)(s-b)(s-c)\leq abc~$ , $~(a+b)(b+c)(c+a)\geq 8abc$
But for the above... | Use Heron's formula
$$(a+b-c)(b+c-a)(c+a-b)(a+b+c)=\dfrac{a^2b^2c^2}{R^2}$$
where $R$ be the center of the circumcircle of $\Delta ABC$.
your inequality can write as
$$8R^2\ge\dfrac{(a+b)(b+c)(a+c)}{a+b+c}$$
since
$$9R^2\ge a^2+b^2+c^2$$
it suffices to prove
$$8(a+b+c)(a^2+b^2+c^2)\ge 9(a+b)(b+c)(a+c)\tag{1}$$
use AM-G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\lim_{x\to+\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ Do you have any tips on how to solve the limit in the title? Whatever I think of doesn't lead to the solution. I tried using: $\sin{x}-\sin{y}=2\cos{\frac{x+y}{2}}\sin{\frac{x-y}{2}}$ and I got:
$$\lim_{x\to+\infty}\bigg(2\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}\sin{... | Notice, $$\lim_{x\to +\infty}\left(2\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{ \sqrt{x+1}-\sqrt x}{2}\right) \right)$$
$$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{1}{2(\sqrt{x+1}+\sqrt x)}\right)$$
$$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k ... | Notice
$$16k^4 +48k = 16k(k^3+3).$$
If $k$ is even, we are done. If $k$ is odd, then $k^3 +3$ is even, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$ Please help me to evaluate this integral:
$$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$
Using substitution $x=2\arctan t$ it can be transformed to:
$$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$
Then I t... | A Fourier analytic approach. If $x\in(0,\pi)$,
$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{2+\cos x}\right) &=& \text{Im}\log(2+e^{ix})\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,\sin(nx),\end{eqnarray*}$$
hence by Parseval's theorem:
$$ \int_{0}^{\pi}\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | It is actually very simple. Use nothing but AM-GM.
$$\frac{a^3}{b} + ab \geq 2a^2$$
$$\frac{b^3}{c} + bc \geq 2b^2$$
$$\frac{c^3}{a} + ac \geq 2c^2$$
$$LHS + (ab+bc+ac) \geq 2(a^2+b^2+c^2) \geq 2(ab + bc +ac)$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| Using long division,
$$ \frac{x^2+4}{x^2+6x+10} = 1 -6\frac{x+1}{x^2+6x+10}= 1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10}$$
Integrating $$1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10} $$ should be straight forward
Note that $x^2+6x+10= (x+3)^2 +1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$ The equation $x^2+bx+c=0$ has distinct roots .If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$
Let $\alpha$ and $\beta$ are the root... | So from equation we get quadratic which is $x^2-2x-1$ so roots if this are $1+\sqrt{2},1-\sqrt{2}$ so sum of roots is $-b/1=2,$ and product is $c/1=-1$ so $b^2+c^2+bc=4+1-2=3$ hope its clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem :
Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$
My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8... | Method $1$
$1.$ Use the identity
$$2\sin x \sum_{k=1}^{n}\cos(2k-1)x = \sin2nx$$
which can easily be verified by using
$$2\sin x \cos(2k-1)x = \sin 2k x - \sin 2(k-1) x$$
and the telescoping property of the sums.
$2.$ Use the above formula to get
$$\frac{\sin 2nx}{\sin x} = 2 \sum_{k=1}^{n}\cos(2k-1)x$$
$3.$ Integrate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
If $f(x)$ is a continous function such that $f(x) +f(\frac{1}{2}+x) =1; \forall x\in [0, \frac{1}{2}]$ then $4\int^1_0 f(x) dx =$ . Problem :
If $f(x)$ is a continous function such that $f(x) +f(\frac{1}{2}+x) =1; \forall x\in [0, \frac{1}{2}]$ then $4\int^1_0 f(x) dx = ?$
My approach :
$$f(x) +f(\frac{1}{2}+x) =1... | \begin{align*}
&4\int_0^1 f \\
=& 4\int_{0}^{0.5} f(t) dt + 4 \int_{0.5}^1 f(t) dt\\
=& 4\int_{0}^{0.5} f(t) dt + 4 \int_0^{0.5} f(x+\frac12) dx \quad\text{($x = t - \frac12$)}\\
=& 4\int_{0}^{0.5} (f(t) + f(t+\frac12)) dt\\
=& 4\int_{0}^{0.5} 1 dt\\
=& 4\cdot 0.5\\
=& 2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
| What we have is
$$\frac1{1^3} - \frac1{7^3} + \frac1{9^3} - \frac1{15^3} + \cdots $$
Imagine if we used negative summation indices (e.g., $n=0, -1, -2, \cdots$). Then we would have
$$\frac1{(-7)^3} - \frac1{(-1)^3} + \frac1{(-15)^3} - \frac1{(-9)^3} + \cdots$$
You should see then that the sum is
$$\frac12 \sum_{n=-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series represen... | Clearly the $r^{th}$ numerator is $1+2+3+...+r= \frac{r(r+1)}{2}$ .
And the $r^{th}$ denominator is $r!$.
Thus $$\displaystyle U_r=\frac{\frac{r(r+1)}{2}}{r!}=\frac{r(r+1)}{2r!}$$
Since the degree of the numerator is $2$ , use partial fractions to find $A,B,C$ such that (If you use partial fractions up to $(r-3)!$ , i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.
If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
| $$((2k+1)(2l+1))^2-1=16k^2l^2+4k^2+16kl^2+16kl+4k+4l^2+16lk^2+4l.$$
Dropping all the terms with coefficient $16$,
$$4(k^2+k+l^2+l)=4(k(k+1)+l(l+1))$$ must be a multiple of $8$.
With a slightly simpler evaluation:
$$((2k+1)(2l+1))^2-1=(4kl+2k+2l)(4kl+2k+2l+2)=4(2kl+k+l)(2kl+k+l+1).$$
This is a multiple of $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 2
} |
Can we find $x_{1}, x_{2}, ..., x_{n}$? Consider this.
$$x_{1}+x_{2}+x_{3}+....+x_{n}=a_{1}$$
$$x_{1}^2+x_{2}^2+x_{3}^2+....+x_{n}^2=a_{2}$$
$$x_{1}^4+x_{2}^4+x_{3}^4+....+x_{n}^4=a_{3}$$
$$x_{1}^8+x_{2}^8+x_{3}^8+....+x_{n}^8=a_{4}$$
$$.............................$$
$$x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+.... | If you mean the diophantine-equation tag and the solutions are supposed to be integers, searching will be easy because high powers are spaced far apart. Because of the symmetry you can insist that $x_1 \ge x_2 \ge \dots \ge x_n$. You can focus just on the last equation $x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
\ge 1$.
My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of ... | By means of rearrangement inequality, it can be shown that, is $a_1\le a_2\le\cdots\le a_n$ and $b_1\le b_2\le\cdots\le b_n$, then $\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n}\ge \dfrac{a_1+a_2+\cdots+a_n}{n}\dfrac{b_1+b_2+\cdots+b_n}{n}$
Then, $\dfrac{a^2+3b^2+5c^2}{3}\ge\dfrac{a^2+b^2+c^2}{3}\dfrac{1+3+5}{3}$. Thus $a^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$
Do you know if there are other integer solutions to
$$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$
besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?
... | First, note that if $(a, b, c, d)$ is a solution, so are $(a, b, d, c)$, $(c, d, a, b)$ and the five other reorderings these permutations generate.
We can quickly dispense with the case that all of $a, b, c, d$ are positive using an argument of @dREaM: If none of the numbers is $1$, we have
$ab \geq a + b = cd \geq c +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+................ | Let us consider $$\Sigma=\frac{1}{6}+\frac{5}{6\times 12}+\frac{5\times8}{6\times12\times18}+\frac{5\times8\times11}{6\times12\times18\times24}+\cdots$$ and let us rewrite it as $$\Sigma=\frac{1}{6}+\frac 16\left(\frac{5}{ 12}+\frac{5\times8}{12\times18}+\frac{5\times8\times11}{12\times18\times24}+\cdots\right)=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| Hint At least for $(a, b) \neq (-1, 0)$, which is not realized by any value $t$, draw the line through $(-1, 0)$ and $(a, b)$ in the $xy$-plane. Writing the point $(a, b)$ of intersection of the line and the unit circle $x^2 + y^2 = 1$ in terms of the slope $t$ of the line gives exactly $$(a, b) = \left(\frac{1 - t^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Geometric progression in an inequality Problem: Show that if $a>0$ and $n>3$ is an integer then $$\frac{1+a+a^2 \cdots +a^n}{a^2+a^3+ \cdots a^{n-2}} \geq \frac{n+1}{n-3}$$
I am unable to prove the above the inequality.
I used the geometric progression summation formula to reduce it to proving $\frac{a^{n+1}-1}{a^2(... | The inequality holds trivially for $a = 1$, and it is invariant
under the substitution $a \to 1/a$. Therefore it suffices to
prove the inequality for $\mathbf{a > 1}$.
The hyperbolic sine is a convex function on $[0, \infty)$,
so for $a > 1$ the function
$$
f(x) = 2 \sinh \bigl(\log a \cdot \frac x2 \bigr)
= a^{x/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Indefinite integral $\int x\sqrt{1+x}\mathrm{d}x$ using integration by parts
$$\int x\sqrt{1+x}\mathrm{d}x$$
$v'=\sqrt{1+x}$
$v=\frac{2}{3}(1+x)^{\frac{3}{2}}$
$u=x$
$u'=1$
$$\frac{2x}{3}(1+x)^{\frac{3}{2}}-\int\frac{2}{3}(1+x)^{\frac{3}{2}}=\frac{2x}{3}(1+x)^{\frac{3}{2}}-\frac{2}{3}*\frac{2}{5}(1+x)^{\frac{5}{2}}+c... | You're correct.
$u = x$, $dv=\sqrt{x+1}dx$ $\Rightarrow$ $v=\frac{2}{3}(x+1)^{3/2}$, $du=dx$.
&
$$\int udv=uv-\int vdu$$
$$\Downarrow$$
SOLUTION:
$$\int x \sqrt{x+1}dx=x\frac{2}{3}(x+1)^{3/2}-\int\frac{2}{3}(x+1)^{3/2}dx=\bbox[5px,border:2px solid #F0A]{x\frac{2}{3}(x+1)^{3/2} -\frac{4}{15}(x+1)^{5/2}+C}$$
VERIFIC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$.Find $a,b,c$.
The... | Use polar coordinate.
Let $(x_1,y_1)=(3+6\cos\theta_1,4+6\sin\theta_1), (x_2,y_2)=(3+6\cos\theta_2, 4+6\cos \theta_2)$.
Then by your equation, we have
$$(2+6\cos \theta_1)(2+6\cos \theta_2)+(2+6\sin\theta_1)(2+6\sin\theta_2)=0$$
Using sum and difference formula, this gives us
$$18\cos(\theta_2-\theta_1)=-4-3(\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove a function is uniformly continuous
Prove the function $f(x)=\sqrt{x^2+1}$ $ (x\in\mathbb{R})$ is uniformly continuous.
Now I understand the definition, I am just struggling on what to assign $x$ and $x_0$
Let $\epsilon>0$ we want $|x-x_0|<\delta$ so that $|f(x)-f(x_0)|<\epsilon$
Could anyone help fill in th... | You have $$\begin{aligned}\vert f(x)-f(y) \vert &= \left\vert \sqrt{x^2+1}-\sqrt{y^2+1} \right\vert \\
&= \left\vert (\sqrt{x^2+1}-\sqrt{y^2+1}) \frac{\sqrt{x^2+1}+\sqrt{y^2+1}}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&= \left\vert \frac{x^2-y^2}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&\le \frac{\vert x-y \vert (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
integrate $\int \cos^{4}x\sin^{4}xdx$
$$\int \cos^4x\sin^4xdx$$
How should I approach this?
I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$
| Continuing from where you left off:
$$\sin^2x\cos^2x=\left({1-\cos2x\over 2}\right)\left({1+\cos2x\over 2}\right)$$
$$\sin^2x\cos^2x=\frac{1}{4}(1 - \cos^2(2x))$$
$$\sin^2x\cos^2x=\frac{1}{4}(\sin^2(2x))$$
Squaring both the sides:
$$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$
This can be integrated in two ways:
Method 1:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim_{n\to \infty} \frac{\frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$ Calculate
$$\lim_{n\to \infty} \frac{\displaystyle \frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$$
I have messed around with this task for quite a while now, but I haven't succ... | Combining $$\frac{1}{1-1/n}>1+\frac1n \ \ \ n>1 $$ with the fact that for all positive $n$ we have $$ \left(1+\frac1n\right)^n<e<\left(1+\frac{1}{n}\right)^{n+1}, $$ and after rewriting your sequence $a_n$ as $\frac{1}{n^2}\sum_{k=1}^n k(1+1/k)^k$, we find the general term $s_k$ of the sum satisfies $$k\left(e-\frac ek... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Formulae for Catalan's constant. Some years ago, someone had shown me the formula (1).
I have searched for its origin and for a proof.
I wasn't able to get true origin of this formula but I was able to find out an elementary proof for it.
Since then, I'm interested in different approaches to find more formulae as (1).
... | A more natural proof.
\begin{align}
\beta&=\sqrt{3}-1\\
J&=\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}dx\\
&\overset{\text{IBP}}=\left[\arctan\left(\frac{x(1-x)}{2-x}\right)\ln x\right]_0^1-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\
&=-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have... | If $y=\cos A\cos B\cos C,$
$2y=\cos C[2\cos A\cos B]=\cos C\{\cos(A-B)+\cos(A+B)\}$
As $A+B=\pi-C,\cos(A+B)=-\cos C$
On rearrangement we have $$\cos^2C-\cos C\cos(A-B)+2y=0$$
As $C$ is real, so will be $\cos C$
$\implies$ the discriminant $$\cos^2(A-B)-8y\ge0\iff y\le\dfrac{\cos^2(A-B)}8\le\dfrac18$$
The equality occu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 0
} |
Evaluation of Real-Valued Integrals (Complex Analysis) How to get calculate the integration of follwing:
$$\int_{0}^{2\pi} \frac{dt}{a + cos t} (a>1)$$
My attempt:
let, $z=e^{it}$
$\implies dt = \frac{dz}{it}$
and
$$\cos t = \frac{z + \frac{1}{z}}{2}$$
On substituting everything in the integral I got:
$$\frac{2}{i}\in... | HINT: The denominator: $z^2+2az+1$ can be easily factorize using quadratic formula as follows $$z^2+2az+1=(z+a-\sqrt{a^2-1})(z+a+\sqrt{a^2-1})$$
hence, $$\frac{1}{z^2+2az+1}=\frac{A}{z+a-\sqrt{a^2-1}}+\frac{B}{z+a+\sqrt{a^2-1}}$$
$$=\frac{1}{2\sqrt{a^2-1}}\left(\frac{1}{z+a-\sqrt{a^2-1}}-\frac{1}{z+a+\sqrt{a^2-1}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability that $2^a+3^b+5^c$ is divisible by 4
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.
For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=... | Observe that
$$2^a+3^b+5^c \equiv 2^a+(-1)^b+1 \pmod{4}$$
So for this to be $0 \pmod 4$, we have the following scenarios
*
*$a \geq 2$, $b$ is odd and $c$ is any number.
*$a=1$, $b$ is even and $c$ is any number.
The number of three tuples $(a,b,c)$ that satisfy the first case =$(4)(3)(5)=60$ and the number of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Find prob. that only select red balls from $n$ (red+blue) balls There are 4 blue balls and 6 red balls(total 10 balls). $X$ is a random variable of the number of selected balls(without replacement), in which
$$P(X=1)=0.1$$
$$P(X=2)=0.5$$
$$P(X=3)=0.2$$
$$P(X=4)=0.1$$
$$P(X=10)=0.1$$
Then, what is probability of only se... | Since there is $6$ red balls, if $10$ balls are selected, probability of selecting only red balls is $0$ and we only have to consider selecting $1,2,3,4$ balls.
Let $R$ be number of red balls selected.
$$\begin{align}P(R=i|X=i)&=\frac{_6P_i}{_{10}P_i}\\
P(R=X)&=\sum\limits_{i=1}^4P(X=i)\cdot P(R=i|X=i)\\
&=0.1\cdot \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
General solution for the series $a_n = \sqrt{(a_{n-1} \cdot a_{n-2})}$ Hey I'm searching a general solution for this recursive series:
$a_n = \sqrt{(a_{n-1}\cdot a_{n-2})}$
$\forall n \geq 2$
$a_0 = 1$,
$a_1 = 2$
| Elaborating on what Wojowu has mentioned,
$$a_n^2=a_{n-1}\cdot a_{n-2}$$
$$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}$$
That is, $a_n^2\cdot a_{n-1}=$ constant is invariant.
Hence $$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}= \ldots = a_1^2a_0 = 4$$
or, $$a_n^2=\frac{4}{a_{n-1}}=\frac{4}{\frac{2}{\sqrt{a_{n-2}}}}=2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfra... | Without Loss of Generality, Let us assume that $c$ is the maximum of $a,b,c$
Notice that $$\sum _{ cyc }^{ }{ \frac { a }{ b } } -3=\frac { a }{ b } +\frac { b }{ a } -2+\frac { b }{ c } +\frac { c }{ a } +\frac { b }{ a } -1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}$$
However, since $(a-b)^2,(c-a)(c-b)\ge 0$, $$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to find Laurent expansion I have been presented with the function $g(z) = \frac{2z}{z^2 + z^3}$ and asked to find the Laurent expansion around the point $z=0$.
I split the function into partial fractions to obtain $g(z) = \frac{2}{z} - \frac{2}{1+z}$, but do not know where to go from here.
|
The function $$g(z)=\frac{2}{z}-\frac{2}{1+z}$$ has two simple poles at $0$ and $-1$.
We observe the fraction $\frac{2}{z}$ is already the principal part of the Laurent expansion at $z=0$. We can keep the focus on the other fraction.
Since we want to find a Laurent expansion with center $0$, we look at the other pol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix}
1 & 2 & 1 \\
1 & 3 & 1 \\
1 & 4 & 1
\end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ .
I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix... |
Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
1) $A_1 + A_2 + A_3 = 0$
2) $2A_1 + 3A_2 +4 A_3 = 0$
Close! Instead of rows they should be columns because matrix multiplication would take the dot products of the row vectors of the first matrix with the column vectors of the second.
By elementa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is a proof of this limit of this nested radical? It seems as if $$\lim_{x\to 0^+} \sqrt{x+\sqrt[3]{x+\sqrt[4]{\cdots}}}=1$$
I really am at a loss at a proof here. This doesn't come from anywhere, but just out of curiosity. Graphing proves this result fairly well.
| For any $2 \le n \le m$, let $\phi_{n,m}(x) = \sqrt[n]{x + \sqrt[n+1]{x + \sqrt[n+2]{x + \cdots \sqrt[m]{x}}}}$. I will interpret the expression we have as following limit.
$$\sqrt{x + \sqrt[3]{x + \sqrt[4]{x + \cdots }}}\;
= \phi_{2,\infty}(x) \stackrel{def}{=}\;\lim_{m\to\infty} \phi_{2,m}(x)$$
For any $x \in (0,1)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 ... | HINT... consider the series for $\ln(1+x)$ and $\ln(1-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$
We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me
| If this problem can be solved without computation it is reducible; we assume this.
$f(x)=512x^9+576x^+216x^3-512x+219=0$ has two change-sign and $f(-x)$ has three ones so $f(x)$ has at least $9-5=4$ non real roots. We try to find a quadratic factor using the fact that $219=3\cdot73$ and $512=2^9$; this factor could co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
The number of ordered pairs $(x,y)$ satisfying the equation The number of ordered pairs $(x,y)$ satisfying the equation $\lfloor\frac{x}{2}\rfloor+\lfloor\frac{2x}{3}\rfloor+\lfloor\frac{y}{4}\rfloor+\lfloor\frac{4y}{5}\rfloor=\frac{7x}{6}+\frac{21y}{20}$,where $0<x,y<30$
It appears that $\frac{x}{2}+\frac{2x}{3}=\fra... | Note that
$$\Bigl\lfloor\frac x2\Bigr\rfloor\le\frac x2\ ,$$
and likewise for the other terms. So we always have $LHS\le RHS$, and the only way they can be equal is if
$$\Bigl\lfloor\frac x2\Bigr\rfloor=\frac x2\ ,$$
and likewise for the other terms. So
$$\frac x2\ ,\quad \frac{2x}3\ ,\quad \frac y4\ ,\quad\frac{4y}5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$
If n is a positive integer I need to show that
$2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$
My guess: Somehow I need two equivalent binomial expression whose coefficients I need to compare.But which two binomial expressions? I know not!
P.S... | Here is a generating function approach
$$
\begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{2}\\
&=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=k}^\infty\binom{n-k}{k}(2x)^n\tag{3}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find general solution to the equation $xy^{'}=2x^2\sqrt{y}+4y$ This is Bernoulli's differential equation:
$$xy^{'}-4y=2x^2y^{\frac{1}{2}}$$
Substitution $y=z^{\frac{1}{1-\alpha}},\alpha=\frac{1}{2},y^{'}=z^{'2}$ gives $$xz^{'2}-4z^2=2x^2z$$
Is this correct? What is the method for solving this equation?
| $$xy'(x)=2x^2\sqrt{y(x)}+4y(x)\Longleftrightarrow$$
$$xy'(x)-4y(x)=2x^2\sqrt{y(x)}-4y(x)\Longleftrightarrow$$
$$\frac{y'(x)}{2\sqrt{y(x)}}-\frac{2\sqrt{y(x)}}{x}=x\Longleftrightarrow$$
Let $v(x)=\sqrt{y(x)}$; which gives $v'(x)=\frac{y'(x)}{2\sqrt{y(x)}}$:
$$v'(x)-\frac{2v(x)}{x}=x\Longleftrightarrow$$
Let $\mu(x)=e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| $2^2=4\equiv1\pmod 3$, so $4^k\equiv1\pmod3$ for all integers $k$. And so for any odd number $2k+1$, we get $2^{2k+1}+1 = 4^k\cdot 2+1\equiv 2+1\equiv0\pmod3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.