Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Laurent series expansion with multiple taylor expansions Find the laurent series expansion for the following function: $$f(z)=\frac{e^z}{z^3-z^4}$$
What I've done is:
\begin{align}
f(z) & = \frac{e^z}{z^3-z^4} \\
& = \frac{e^z}{z^3(1-z)} \\
& = \frac{e^z}{z^3} .\frac{1}{1-z} \\
& = \frac{e^z}{z^3}\Big(1+z+z^2+\dot... | Here's how I would do it. First of all, I would compute the Taylor expansion of $\dfrac{e^z}{1-z}$:\begin{align}\frac{e^z}{1-z}&=\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)\left(1+z+z^2+z^3+\cdots\right)\\&=1+2z+\frac52z^2+\cdots+\left(\sum_{k=0}^n\frac1{k!}\right)z^n+\cdots\end{align}And then\begin{align}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use coordinate method to solve a pretty hard geometry problem Here is a hard geometry problem for my homework.
Let $D$ be a point inside $\Delta ABC$ such that $\angle BAD=\angle BCD$ and $\angle BDC=90^\circ$. If $AB=5,BC=6$ and $M$ is the midpoint of $AC$, find the length of $DM$.
(Taken from HK IMO Prelim 2012)
T... | After your calculations we have the system:
$$\begin{cases} b^2+c^2=36, \\ x^2+\left(b+y\right)^2=25, \\ \sqrt{11}x = 5y, \end{cases}\tag{1}$$
where $x>0,\;y>0$.
Substituting $x=\dfrac{5}{\sqrt{11}}y\;$ into $2$nd equation of $(1)$, we get:
$$
\dfrac{25}{11}y^2+(b+y)^2=25;
$$
$$
\dfrac{36}{11}y^2+2by+(b^2-25)=0;\tag{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many integral solutions does $2x + 3y + 5z = 900$ have when $ x, y, z \ge 0$? Solution: Let $2x + 3y = u.$ Then we must solve $\begin{align} u + 5z = 900 \tag 1 \\ 2x + 3y = u \tag 2 \end{align}$
For $(1),$ a particular solution is $(u_0, z_0) = (0, 180).$ Hence, all the integral solutions of $(1)$ are $\begin{case... | Since the generating function
$$ f(z)=\frac{1}{(1-z^2)(1-z^3)(1-z^5)} $$
has a triple pole at $z=1$ and simple poles at $9$ points of $S^1$, we may decompose it as
$$ f(z) = \frac{1}{30(1-z)^3}+\frac{7}{60(1-z)^2}+\sum_{k=1}^{8}\frac{R_k}{z-R_k} $$
and deduce that the coefficient of $z^n$ in $f(z)$, up to a bounded err... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How find the function $f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$ Find all function $f:R\to R$ and such
$$f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$$
Let $x=y=0$,then $$f(f(0))=f(0)$$
| Replace $x=0$ in the given identity, we get
$ f(f(0)) = f(0).$
If we let $f(0) = a$, then
$$f(a) = a\tag{1}.$$
Replace $y=0$ in the given identity, we get
$$f(f(x)) = f(x) + xf(0) = f(x) + ax.\tag{2}$$
Let $x=a$ in (2), we get
$$f(f(a)) = f(a) + a^2.\tag{3}$$
But from (2), $f(f(a)) = f(a) = a$. So (3) becomes
$$a = a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1}$ Given $x, y,z \ge 0$. Prove that
$$\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} $$
Attempt
Notice that $$(x^{3} + y^{3}+1) ... | It should be
$$ (ab + bc + ca + 2a\sqrt{bc} +2b\sqrt{ac} +2c\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2),$$
where $a\geq1$, $b\geq1$ and $c\geq1$, for which it's not so easy to find a proof.
By the way, the Karamata's inequality helps.
Indeed, let $f(x)=\sqrt{x+1},$ $x^3+y^3=c$, $x^3+z^3=b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $x^{2}\equiv-7\:\left(128\right)$? How to solve $x^{2}\equiv-7\:\left(128\right)$ ?
I know that $x^{2}\equiv a\:\left(2^{l}\right)$ is solvable iff $a\equiv1\:\left(8\right)$
which is the case here, but how can we find the solution?
| The general method would be to first solve mod 2 and discover that $x \equiv 1 \pmod{2}$. Then "lift" to mod $2^2$, which means that you observe that if $x \equiv 1 \pmod{2}$, then $x\equiv 1$ or $3 \pmod{4}$. Both of these are solutions. So lift them again to mod $2^3$ and get that $x \equiv 1, 3, 5,$ or $7 \pmod{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determinant of block matrix where all blocks are $n \times n$ I found the following formula on Wikipedia:
Suppose $A$, $B$, $C$, and $D$ are matrices of dimension $n \times n$, $n \times m$, $m \times n$, and $m \times m$, respectively.
$$\det \begin{pmatrix} A & B \\ C & D\end{pmatrix} = \det(D) \det \left(A - B D^{-... | Since all four blocks are square and the bottom ones do commute (the zero matrix commutes with everyone),
$$\det \left[ \begin {array}{ccc|ccc} 0&0&0&-4&0&0\\ 0&-1&0
&0&-5&0\\ 0&0&-2&0&0&-6\\ \hline 7&0&0&0&0
&0\\ 0&8&0&0&0&0\\ 0&0&9&0&0&0
\end {array} \right] = \det \left( \mathrm O_3 + \mbox{diag} (4 \cdot 7, 5 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top $ is not always true? If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$,
It is possible to prove that
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)... | Observe that
$A = \dfrac{1}{2}(A + A^T) \Longleftrightarrow A = A^T, \tag 1$
that is,
$A = \dfrac{1}{2}(A + A^T) \tag 2$
if and only if $A$ is a symmetric matrix. This is easily seen as follows: if
$A = A^T, \tag 3$
then
$A = \dfrac{1}{2}(2A) = \dfrac{1}{2}(A + A) = \dfrac{1}{2}(A + A^T); \tag 4$
going the other wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve the equations and find $x,y,z$ Given:
$$x^3+y^3+z^3=x+y+z$$
And:
$$x^2+y^2+z^2=xyz$$
Find all real and positive solutions to these equations, if any.
So most probably, we'll factorise in this way:
$$x^3+y^3+z^3-3xyz=x+y+z-3xyz
\rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)-3(x^2+y^2+z^2)$$
But I'm having trou... | Denote by $\sigma_i$ the elementary symmetric functions and by $p_r$ the power sums of the nonnegative variables $x$, $y$, $z$. We are told that
$$p_3=\sigma_1,\qquad p_2=\sigma_3\ .\tag{1}$$
Of course $x=y=z=0$ is a solution of $(1)$. We therefore may assume $\sigma_1>0$ in the sequel. Using the inequalities $$\left({... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Polynomial of 4th degree I would like to ask if someone could help me with the following equation.
\begin{equation}
x^4+ax^3+(a+b)x^2+2bx+b=0
\end{equation}
Could you first solve in general then $a=11$ and $b=28$.
I get it to this form but I stuck.
\begin{equation}
1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac... | You are almost there !
$$1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0$$
$$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^3}+\frac{x+1}{x^4}\right) = 0$$
$$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^2}\right)^2 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$ If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$
I know this is true for acute angle triangle.
I want to know whether it is true for every real $A,B,C$ such that $A+B+C=\pi.$
| Yes, this inequality is true for any reals $A,$ $B$ and $C$ such that $A+B+C=\pi.$
Indeed, let $\cos(A-B)=x$ and $\cos(A+B)=y.$
Thus, we need to prove that:
$$x(\cos(A+B-2C)+\cos(A-B))\geq8(\cos(A+B)+\cos(A-B))(-\cos(A+B))$$ or
$$x(\cos3(A+B)+x)+8y(x+y)\geq0$$ or
$$x(4y^3-3y+x)+8y(x+y)\geq0$$ or
$$x^2+(4y^3+5y)x+8y^2\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me... | Let $X=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
$$
X=\begin{pmatrix} 2 & 4 \cr 1 & 2 \end{pmatrix}.
$$
For complex numbers there are se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cd... | The identity you used leads to a proof of the desired result by induction as follows. So suppose that $10^n$ divides $3^{10^n}-1$ for all $n=2,\ldots,N$. Applying the identity yields
$$
\frac{3^{10^{N+1}}-1}{3^{10^N}-1}=10 + \sum_{k=1}^9(3^{k\cdot 10^N}-1),
$$
and the right side is a multiple of $10$ since each term in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How can I find this limit applying squeeze theorem? How can I evaluate this limit applying squeeze theorem?
$$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}$$
I found this limit with standart way:
$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{x\to6}\frac{\left(1+\sqrt{ 3-\sqrt{x-2}} \right)\times \left( 1-\s... | First step let wlog $y=x-6\to 0^+$ thus
$$\lim_{x\to6^+}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{y\to0^+}\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y}$$
then by Bernoulli inequality
$$\sqrt{y+4} =2(1+y/4)^\frac12\le 2(1+y/8)=2+y/4$$
$$3-\sqrt{y+4}\ge 3-2-y/4=1-y/4$$
$$\sqrt{ 3-\sqrt{y+4}}\ge \sqrt{ 1-y/4}$$
$$1-\sqrt{ 3-\sqrt{y+4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proof regarding the minimum of a function I want to show that the function
$$f(x)= \frac{2x}{1+(\frac{1}{k-x})^a} + \frac{k-2x}{1+2\cdot(\frac{1}{k-x})^a+ (\frac{1}{k-2x})^a}$$
with
$$ x \in [0,\frac{k}{2}], a \in [0,\infty], k \in [0,1] $$
has the minimum
$$min(f(x)) = \begin{cases}
\frac{k}{1+3\cdot(\frac{1}{k})^... | Update
We can also prove the case when $a > \frac{11}{4}$.
We will use Fact 1 and 2 whose proof is simple and thus omitted.
Fact 1: If $a > \frac{11}{4}$, we have
$$\frac{2x}{1+(\frac{1}{k-x})^a} \ge \frac{k}{1 + (\frac{2}{k})^a}$$
for $\frac{k}{4} \le x \le \frac{k}{2}$.
Fact 2: If $a > \frac{11}{4}$, we have
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan ... | Your way is wrong since
$${\tan(\tan x) - \sin (\sin x)}\neq { \tan x - \sin x}$$
indeed this step
$$\frac{\tan x \tan (\tan x)}{\tan x}=\tan x\cdot \frac{ \tan (\tan x)}{\tan x}\color{red}{=\tan x \cdot 1}=\tan x$$
and the similar for $\sin x$ are not allowed (in general we can't evaluate limit only for a part or fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Variance of taking balls from a bag question A bag contains $2$ white balls and $2$ black balls.
Each instance a ball is taken from the bag, if it's white it's returned to the bag and if it's black it's replaced by a white ball. The game ends when no black balls remain in the bag. Let $Y$ be the number of instances of ... | Let $Y$ be the number of black balls taken. The probability distribution table:
$$\begin{array}{c|l}
Y&P(Y)\\
\hline
2&\frac1{2^3}\\
3&\frac1{2^4}+\frac3{2^5}\\
4&\frac1{2^5}+\frac3{2^6}+\frac{3^2}{2^7}\\
5&\frac1{2^6}+\frac3{2^7}+\frac{3^2}{2^8}+\frac{3^3}{2^9}\\
\vdots
\end{array}$$
Note:
$$P(Y=n)=\sum_{i=1}^{n-1} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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$\lim_{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}$. Did I choose my $\delta$ Properly? Preliminaries:
We know that, the limit as $ x \rightarrow a $ of a function $ f(x) $ is $ L$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0 $ such that $ 0 <|x-a|<\delta \implies |f(x) - L|<\epsilon$. Our function is $ f(x) ... |
Thanks to the property of triangle inequality, we have that:
$$
\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| \color{red}{\mathbf{-}} \left|\frac{1}{4}\right|
$$
This is not correct.
Instead, you could rewrite:
$$\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|=\frac{1}{4}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time... | Your error is going from
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8) (x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8) \\(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
to
$$x^3(x^1+x^2+x^3+x^4+x^5+x^6+x^7)(x^1+x^2+x^3+x^4+x^5+x^6+x^7) \\(x^1+x^2+x^3+x^4+x^5+x^6+x^7)$$
and you should have written an $x^0$ term as in
$$x^3(x^0+x^1+x^2+x^3+x^4+x^5+x^6+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
In multiplying 2 matrices, how do you know whether to operate on rows or columns? My understanding is that multiplying a matrix by a matrix on its left means operating on rows, and multiplying a matrix by a matrix on its right means operating on columns.
When there are 2 matrices next to each other to be multiplied, h... | I think this example will help, if we have any $3×3$ matrix $A$, and we multiply it by the permutation matrix $P_{2,3}$ from the right, here is what happens,
$$AP_{2,3}= \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \... | Prove that $$\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}<1$$ for any real $x$.
Now, $$\lim_{x\rightarrow-\infty}\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1,$$ which gives $$\sup\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1$$ and the answer $a\geq1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} +... | COMMENT.-These are the two real roots given by Wolfram. It is impossible that you can calculate them by simple means. The two non-real roots are equally complicated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$
Let $\sum_i^na_i=n$, $a_i>0$. Then prove that $$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$
I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. T... | You need to use another queue:
By Rearrangement, AM-GM and C-S we obtain: $$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$
I used the following.
$$\fr... | {
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ ... | let $t=x+3$ and work with the 'symmetric' form,
$$\sqrt[3]{t-1}+\sqrt[3]{t+1}=-\sqrt[3]{t}\tag{1}$$
Take the cubic power,
$$2t + 3(t^2-1)^{1/3}(\sqrt[3]{t-1}+\sqrt[3]{t+1})=-t$$
Simplify with (1),
$$(t^2-1)t=t^3$$
which leads to the only solution $t=0$, hence $x=-3$.
| {
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Pairs of perfect squares Two perfect squares are said to be friendly if one is obtained from the other by adding
the digit 1 on the left. For example, $1225 = 35 ^ 2$ and $225 = 15 ^ 2$ are friendly. Prove that there are
in finite pairs of friendly and odd perfect squares.
Maybe Any product of square-full numbers is sq... | Be $a, b$ belonging to integers, we want to show that for all $a, b$ belonging to integers, there is $ k $ belonging to the natural such that they are infinite.
$a ^ 2 - b ^ 2 = 10 ^ k$
$(a + b) (a-b) = 2^k. 5 ^ k $
$\begin{cases}
a + b = 2 ^ {k-1} \\
a - b = 2.5 ^ k
\end{cases}$
$2a = 2.5 ^ k + 2 ^ {k-1}$
$a = 5 ^ k... | {
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Solve for equations involving floor function
Solve for real $x$: $$\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = x - \lfloor x \rfloor + \frac{1}{3}$$
Hello! I hope everybody is doing well. I was not able to solve the above problem. And this problem becomes even more difficult with the fact that $x $ i... | I would approach the problem in this way.
Let's put
$$
{1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = x - \left\lfloor x \right\rfloor + {1 \over 3} = \left\{ x \right\} + {1 \over 3} = t
$$
Since the fractional part of $x$ has values in the range $[0, \, 1)$ we have the the... | {
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Show that $\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2 = \frac{2}{n+1}\binom{2n}{n}$
Show that $$\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \dfrac{2}{n+1}\dbinom{2n}{n}$$ where $n \in \mathbb{N}$
Consider $\dbinom{n}{r}=0 $ for $r<0 $ and $r>n$. Using $\displaystyle\su... | The given sum is equal to the constant term of $f(z)f(1/z)$, where $$f(z)=\sum_{k=0}^{n+1}\left[\binom{n}{k}-\binom{n}{k-1}\right]z^k=(1-z)(1+z)^n;$$ in other words, it is the coefficient of $z^{n+1}$ in $$z^{n+1}f(z)f(1/z)=-(1-z)^2(1+z)^{2n},$$ equal to $-\binom{2n}{n+1}+2\binom{2n}{n}-\binom{2n}{n-1}$ which reduces t... | {
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Find the distance travelled by an ant in the coordinate plane
An ant is standing on the origin. It begins by walking 1 unit in the positive $x$-direction and then turns $45$ degrees counterclockwise and walks $\dfrac{1}{2}$ units in that direction. The ant then turns another $45$ degrees and walks $\dfrac{1}{3}$ units... | The problem is amenable to some simple complex number analysis (you have to work in radian measure, note that $45^{\circ} = \frac{\pi}{4}$).
Each step can be characterised as a vector in the complex plane. The first step is $\displaystyle 1 = 1e^{i0}$, the second is $\displaystyle \frac 12e^{i\frac{\pi}{4}}$, and the $... | {
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Prove the given inequality if $a+b+c=1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such qu... | Also, by AM-GM
$$a^3+b^3+c^3+a^2b+a^2c\geq a^3+4\sqrt[4]{b^3\cdot c^3\cdot a^2b\cdot a^2c}=a^3+4abc.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{1}{a^2+4bc}\leq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\left(\frac{1}{a^2+4bc}-\frac{1}{4bc}\right)\leq\frac{a+b+c}{5abc}-\frac{a+b+c}{4abc}$$ or
$$\sum_{cyc}\frac{a^2}{b... | {
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Fourier series of a function in interval $[-\pi/2,\pi/2]$ I need to show that $$\frac{\cos x}{3}+\frac{\cos 3 x}{1 \cdot 3 \cdot 5}-\frac{\cos 5 x}{3 \cdot 5 \cdot 7}+\frac{\cos 7 x}{5 \cdot 7 \cdot 9}-\cdots=\frac{\pi}{8} \cos ^{2} x$$ on the interval $[-\pi/2,\pi/2]$. I have tried to find the Fourier coefficient $a_n... | The way to think about this problem is to notice that LHS has period $2\pi$ and satisfies $f(x+\pi)=-f(x)$, while RHS has period $\pi$ so it satisfies $g(x+\pi)=g(x)$. This means that you need to extend $\cos^2(x)$ to the full interval $[-\pi,\pi]$ in such a way that it satisfies the same property as LHS, so you consid... | {
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Find real parameter $a$, such that the solution of the linear system lies in the second quadrant
For which real parameter $a$ lies the solution of the system of equations
$$\begin{aligned} \frac{x}{a+1} + \frac{y}{a-1} &= \frac{1}{a-1}\\ \frac{x}{a+1} - \frac{y}{a-1} &= \frac{1}{a+1} \end{aligned}$$
in the second quad... | Adding and subtracting the equations we obtain
$$\frac{2x}{a+1} = \frac1{a-1}+\frac1{a+1}\implies x=\frac12 \frac{a+1}{a-1}+\frac12=\frac a{a-1}$$
$$\frac{2y}{a-1} = \frac1{a-1}-\frac1{a+1}\implies y = \frac12-\frac12 \frac{a-1}{a+1}=\frac 1{a+1}$$
then we need $y>0$ and $x<0$.
| {
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Fixed Point Theorems and Contraction Mappings I am trying to solve this following exercise.
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and
\begin{align*}
0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b).
\end{align*}
Use Theorem 1 ... | Pick $\lambda$ so that $0<\lambda F'<1$. E.g. take $\lambda=\frac1{2K_2}$. Then $0<\frac{K_1}{2K_2}\le\frac {F'}{2K_2}\le\frac{K_2}{2K_2}=\frac12<1$. Using that $f'=1-\lambda F'$ we obtain $0<1-\frac12=\frac12\le f'=1-\frac {F'}{2K_2}\le1-\frac{K_1}{2K_2}<1$. If $\mu=1-\frac{K_1}{2K_2}$ then $0<\frac12\le f'\le\mu<1$ (... | {
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$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{... | Clearly one of $a,b,c$ is $2^4$
We need to minimize $b+c$ with $bc=5^4$
Now $(b+c)^2-4bc=?\ge0$
So $(b+c)^2\ge4bc=?$ the equality occurs if $b=c$
Fortunately, that condition keeps $b,c$ natural
Else we need to resort to check for possible combinations
$$5^k,5^{4-k},0\le k\le4-k$$
| {
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Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$
, where $i> \dfrac{n}{2}$. How can I get the ge... | Assuming $n$ is even and $i<\frac{3}{2}n$,
$$\lfloor\frac{n^2+n-2i}{2n}\rfloor = \lfloor\frac{n^2-(n+2i)}{2n}\rfloor = \frac{n}{2}-\lfloor\frac{n+2i}{2n}\rfloor = \frac{n}{2}-1$$
So
$$S(n,i) = (\frac{n^2+n-2i}{n}-\frac{n}{2}+1)*(\frac{n}{2}) = (\frac{n}{2}-2\frac{i}{n}+2)*(\frac{n}{2})$$
So
$$S(n,i) = (\frac{n}{2})^2... | {
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} -... | Squaring both sides we get $$x+\sqrt{x}+x-\sqrt{x}-2\sqrt{x^2-x}=\frac{m^2x}{x+\sqrt{x}}$$
Simplifying we get
$$2x-\frac{m^2x}{x+\sqrt{x}}=2\sqrt{x^2-x}$$
simplifying we get
$$m^4x^2=-4x(x^2+x+2x\sqrt{x})$$ diviniding by $$x\neq 0$$ $$4x+4+m^4=-8\sqrt{x}$$ squaring again
$$16x^2-32x+8xm^4+8m^4+m^8+16=0$$
Can you solve ... | {
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a stat... | $$L=\lim_{n \rightarrow} \left(\frac{3n-2}{3n+1} \right)^{2n}= \lim_{n \rightarrow \infty} \frac{\left([1-2/(3n)]^{3n/2}\right)^{4/3}}{\left([1+1/(3n)]^{3n}\right)^{2/3}}=\frac{ e^{-4/3}}{e^{2/3}}=e^{-2}$$
| {
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Proving a much stronger version of AM-GM for three variables Here, @MichaelRozenberg stated the following inequality without proof:
Theorem. For all non-negative $a,b,c\in\mathbb R$,
$$a^6+b^6+c^6-3a^2b^2c^2\geq16(a-b)^2(a-c)^2(b-c)^2.$$
In my answer below I give a complete brute-force proof. However, more elegant ... | I meant the following reasoning.
Let $a=\min\{a,b,c\},$ $b=a+u,$ $c=a+v$ and $u^2+v^2=2kuv.$
Thus, by AM-GM $k\geq1$ and
$$\sum_{cyc}(a^6-a^2b^2c^2)=\frac{1}{2}(a^2+b^2+c^2)\sum_{cyc}(a-b)^2(a+b)^2\geq$$
$$\geq\frac{1}{2}(u^2+v^2)(u^4+v^4+(u^2-v^2)^2)=(u^2+v^2)(u^4-u^2v^2+v^4).$$
Id est, it's enough to prove that:
$$(u... | {
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Complex Numbers Identity Let $a,b$ be non-zero complex numbers with:
$$a2^{|a|}+b2^{|b|} = (a+b)2^{|a+b|}$$
Show that $a^6 = b^6$.
I gave it a go, but to no avail.
| First note that $a+b=0$ always gives a solution for which $a^6=b^6$.
If $a$ and $b$ are real, we can assume without loss of generality that $a=1$. We get $2+b2^{|b|}=(1+b)2^{1+|b|}$, so $2^{1-|b|}+b=2(1+b)$, so $2^{1-|b|}-b=2$. By piecewise analysis, we find $b=0$ or $b=-1$. But $a$ and $b$ are required to be non-zero,... | {
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If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2... |
Minimize $z=(x^2+y^2)^2$ s.t. $x^2+2xy-y^2=6$.
The contour curve is a circle with the center at the origin: $x^2+y^2=\sqrt z$.
To minimize $z$ implies to find the smallest radius of the circle. The smallest circle must touch the hyperbola. The slope of tangent to the hyperbola at $(x_0,y_0)$ is:
$$2x+2y+2xy'-2yy'=0 \... | {
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Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder,
Hey I would just like to know how to solve this as the questio... | Any textbook would tell you to do polynomial division. Personally, I think polynomial division is too opaque of a technique to use before you know how to solve a question like this without it. So here is an example using what basically amounts to long division, but much less mysterious.
Take $p(x) = x^3+2x^2-3x+5$. We ... | {
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$\sin 2x = 1/2, 2x = 150, 390$. But why not 30? $0<x<360$
So $\sin(2x) = \sin (30 + 360)$ or $\sin(2x) = \sin (180-30)$.
It derives from
\begin{align*}
\sin x - \cos x & =\frac{1}{\sqrt2}\\
\sin^2 x + \cos^2 x -2\sin x \cos x & = \frac{1}{2}\\
2\sin x \cos x & = \frac{1}{2}
\end{align*}
But why not $\sin (2x) = \sin (... | We have that
$$\sin 2x = \frac12=\sin 30° \iff 2x=30°+k360° \quad \lor \quad2x=180°-30°+k360°$$
therefore all the solutions are in the form
*
*$x=15°+k180°=15°,15°\pm 180°,15°\pm360°,\ldots$
*$x=75°+k180°=75°,75°\pm 180°,75°\pm360°,\ldots$
and for the range $0°\le x <360°$ the solutions are $15°$, $75°$, $195°$, $... | {
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Can you simplify an Euler formula expression when you are only interested in the real part? I was trying to find a way to simplify $cos(x) + cos(x+a)$ using Euler's formula:
f(x) = $ e^{ix} + e^{i(x+a)} $
= $cos(x) + cos(x+a) +i(sin(x) +sin(x+a))$
= $ 2 cos(x+ \frac{a}{2} )cos(\frac{a}{2}) + i(2sin(x+\frac{a}{2})cos(\... | For the step 2 : $f(x) = 2cos(\frac{a}{2})e^{ix}e^{i\frac{a}{2}}$ is correct. However, the real part of this is not $2cos(\frac{a}{2})cos(x)cos({\frac{a}{2}})$ (did you forget the $i^2$ term?).
Let's compute the expression $z := e^{ix}e^{i\frac{a}{2}}$,
we have
$$\begin{eqnarray}z & = & \Big(cos(x) + i\ sin(x)\Big) \c... | {
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Is this formula correct, series and Bernoulli numbers According to this page:
https://proofwiki.org/wiki/Definition:Bernoulli_Numbers
The series is :
$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}+\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$
where $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0, B_4=-\dfrac{1}{30}$
How ... | Note, the generating function for the Bernoulli numbers is defined as
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=0}^\infty\frac{B_nx^n}{n!}
\end{align*}
with $\frac{B_\color{blue}{n}}{n!}$ the coefficient of $x^\color{blue}{n}$.
Since we have $B_0=1, B_1=-\frac{1}{2}, B_2=\frac{1}{6}, B_3=0, B_4=-\frac{1}{30},\ldots$ we... | {
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Finding the complete Taylor expansion of $\frac{1}{1+z^2}$ around $z=0$. For an exercise, I need to find the complete Taylor expansion for $(1+z^2)^{-1}$ around $z=0$. I have tried decomposing first $(1+z^2)^{-1}$ into partial fractions. Since $1+z^2=0$ gives $z=\pm i$, the partial fractions are:
$$\frac{1}{1+z^2} = \f... | Since $\frac{1}{1+z^2} = \frac{1}{1-(-z^2)}$, the Taylor expansion around $z = 0$ (i.e., the Maclaurin series) would be the sum of an infinite Geometric series, i.e.,
$$\frac{1}{1+z^2} = \sum_{n=0}^{\infty}\left(-z^2\right)^n \tag{1}\label{eq1A}$$
with this being convergent for $|z^2| \lt 1$.
Regarding your work, since... | {
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Compute Taylor Series $\ln(x^2-x+1)$ Please share if you have a faster approach, but I’ll share what I have so far. I think it reduces to some combinatorial problem, I’d like to see a combinatorial argument if possible.
From computing for small $n$ in wolfram alpha, I think the answer is
\begin{align*}
\ln(x^2-x+1)&=\... | Tips:
$$\ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)$$
Answer:
$$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots \\ \ln(1+x^3)=x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3}-\dfrac{x^{12}}{4}+\cdots \\ \ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)=\sum_{k=1}^\infty a_kx^k \\ a_k=\begin{cases} \dfrac{2(-1)^{k+1}}{k} & \text{when }3|k \\ \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Checking a newtonian binomial expansion I am trying to check that the following expansion of $\sqrt{1-x}$ using Newton's Binomial Theorem "appears correct by squaring both sides"
$$(1-x)^{1/2}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5x^4}{128}+...$$
then to do the check, I got the following:
$$(\sqrt{1-x})^2=(... | Your last line is wrong. Expanding the first few terms on the right-hand side:
$$\left(1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\ldots\right)^2=1-2\cdot\frac{x}{2}+\frac{x^2}{4}-2\cdot\frac{x^2}{8}+\frac{x^4}{64}+2\cdot\frac{x^3}{16}-2\cdot\frac{x^3}{16}+\ldots$$
Everything cancels apart from $1-2\cdot\frac{x}{2}=1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can you go from the integral of x squared to the summation of x squared? To be specific, I was wondering if you could go from the indefinite integral of $x^2$, mainly
$\int$ $x^2$$dx = \frac{x^3}{3}$, to the summation of $x^2$, $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$?
| At first glance, $\frac{x^3}{3}$ and $\frac{x(x+1)(2x+1)}{6}$ seem like very different functions. Finding the sum of $n^2$ can be represented as a piecewise function: $f(x) = 0$ when $-1 ≤ x < 0$, $f(x) = 1$ when $0 ≤ x < 1$, $f(x) = 4$ when $1 ≤ x < 2$, $f(x) = 9$ when $2 ≤ x < 3$ and so on.
However, if we integrate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$ As the L.c.m contains $2^7$ as a factor and G.c.d ... | If $$\gcd(x,y)=16=2^4$$ and $$\operatorname{lcm}(x,y)=48000=2^7\cdot3\cdot5^3$$ then $$xy=2^{11}\cdot3\cdot5^3$$
So we can let
$$x = 2^\alpha\cdot3^\beta\cdot5^\gamma \qquad \text{and} \qquad
y = 2^{11-\alpha}\cdot3^{1-\beta}\cdot5^{3-\gamma}$$
We need
$$\begin{align}
\min(\alpha, 11-\alpha) &= 4 \\
\max(\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
r... | Use the componendo and dividendo rules, i.e.
$\frac ab = \frac cd \implies \frac{b-a}{b+a} = \frac{d-c}{d+c}$
$$r=\frac{1-x^2-y^2}{(1-x)^2+y^2}\implies
\frac{1-r}{1+r} =-x+\frac{y^2}{x-1}$$
Rearrange
$$y^2 = (x-1)\left(x+\frac{1-r}{1+r}\right)=
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{1}{1+r}\Big )^2
$$
where $u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would ca... | Shortest proof:
$$(5+i)^4(239-i)=114244+114244i.$$
Taking the arguments,
$$4\arctan \frac15-\arctan\frac1{239}=\frac\pi4.$$
Note that the computation avoids the fractions and immediately generalizes to other Machin-like formulas (https://en.wikipedia.org/wiki/Machin-like_formula#More_terms).
To perform the computatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^... | You have shown divisibility by 15.
To show divisibility by 30,
just note that the expression is even
(odd-odd+even).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Limit of $\frac{e^{-xy}}{1+x^2+y^2}$ in $x,y ≥ 0$ I am trying to figure out if $f(x,y)=\frac{e^{-xy}}{1+x^2+y^2}$ goes to zero, in the area x,y ≥ 0, when $x^2+y^2$ goes to infinity.
So I rewrote the function as $\frac{1}{e^{xy}}\frac{1}{1+x^2+y^2}$.
The function $\frac{1}{1+x^2+y^2} $ definitley goes to zero. But what ... | If $x=0,$ then $\frac{1}{1+y^2} \rightarrow 0$ as $y\rightarrow \infty$. If $x$ is finite, then both $e^{-xy}$ and $\frac{1}{1+x^2+y^2}$ go to zero, as $y\rightarrow \infty$, so their product goes to zero. If both $x$ and $y$ approach $\infty$, first let $y$ go to $\infty$ then take the limit as $x \rightarrow \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that a function is negative over its domain I would like to demonstrate that the following function is negative
\begin{equation}
f(x)=-\frac{t}{4\sqrt{x}^3}\bigg[1-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg]+\bigg(\frac{1}{1-t}\bigg)\bigg(\frac{t}{2\sqrt{x}}\bigg)^2\bigg(1+\sqrt{x}\bigg)^{\frac{2-t}{t-1}}
\end{eq... | Here is an attempt to answer my question:
The function $f(x)$ has the same sign as the function $g(x)$, where
\begin{equation*}
g(x)=\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg[1+\frac{t}{1-t}\frac{\sqrt{x}}{(1+\sqrt{x})}\bigg]-1,
\end{equation*}
and so the problem is equivalent to showing that $g(x)<0$.
Note that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Riemann sum of $\int_1^2 {1\over x^2} dx$. I've spent quite a time solving the following problem:
Evaluate using Riemann's sum:
$$
I = \int_1^2{1\over x^2} dx
$$
I was first trying the following approach, which didn't work since the summation seems undoable to me:
$$
\Delta x = {1\over n}\\
I = \lim_{n\to\infty}\su... | There's no easy closed form for
$$\sum_{k = 1}^{n} \frac{1}{(n+k)^2}\,,$$
but since we're interested in a limit we can achieve our goal by approximating the terms of the sum in such a way that the approximation has an easy closed form. A very good approximation is obtained by
\begin{align}
\sum_{k = 1}^{n} \frac{1}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at $(a,b)$ Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at (a,b) when $a<b$
$a-b = ...$
Find gradient of the line, df(x) / dx
$\frac{(4x+3)3 - (3x-6)(4)} {(3x-6)^2}$
Which the same as $ -11/3$
$(12x+9 - 12x + 24 ) 3= -11(3x-6)^2$... | The equation $$-\frac{11}{3}x+16=\frac{4x+3}{3x-6}$$ must have only on e solution.
Compute the discriminant! It is $x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing ... | Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Clearly $x>0$
Let $\dfrac{3x}{\sqrt{x^2-9}}=y$ so that $x+y=\dfrac{35}4$
$$\implies9x^2=x^2y^2-9y^2\iff\dfrac19=\dfrac1{x^2}+\dfrac1{y^2}=\dfrac{(x+y)^2-2xy}{(xy)^2}$$
Use $x+y=\dfrac{35}4$ and $xy>0$ for find $xy=\dfrac{75}4$
So, $x,y$ are the roots of $$t^2-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Solve complex equation $\left(\frac{8}{z^3}\right) - i = 0$ from $$(a^3 + b^3) = (a+b)(a^2-ab+b^2)$$
I have $${(2/z)}^3 + i^3 =0$$
I have $$\left(\frac{2}{z} + i\right)\left(\left(\frac{2}{z}\right)^2-(2/z)(i)-1)\right) = 0$$
i.e.$ \left(\frac{2}{z}\right)+i = 0 $ or $ \left(\left(\frac{2}{z}\right)^2-(\frac{2}{z})(i)-... | $\dfrac8{z^3}=i\implies z^3=\dfrac8i=-8i$.
Since $2^3=8$ and $i^3=-i$, it is apparent that $z=2i, 2i\omega$, or $2i\omega^2$,
where $\omega$ is a primitive cube root of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x... | By standard limits we have
$$\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=\frac{1-\cos x +\cos x-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x^2}\frac{x }{\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}\frac{1+\sqrt{\cos{2x}}}{1+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $ Let $a,b,c$ be positive numbers. Prove that
$$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$
A... | A full expanding gives:
$$\sum_{cyc}(a^3b^3+2a^4bc-3a^3b^2c-3a^3c^2b+3a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+abc\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which true by Schur twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{r_1}{bc}+\frac{r_2}{ac}+\frac{r_3}{ab}=\frac1r-\frac1{2R}$ Prove $\dfrac{r_1}{bc}+\dfrac{r_2}{ac}+\dfrac{r_3}{ab}=\dfrac{1}{r}-\dfrac{1}{2R}$ where $a , b, c$ are sides of a triangle; $r $is the inradius; $R$ is the circumradius; the $r_i$ are the exradii; $s$ is semiperimeter; and $\triangle$ is area
Solv... | As $r_1=s\tan\dfrac A2$
$T(1,a)=\dfrac{r_1}{bc}=\dfrac{2Rs\sin A\tan\dfrac A2}{abc}=\dfrac{2Rs(1-\cos A)}{abc}=\dfrac{2R\triangle(1-\cos A)}{r\cdot4R\triangle}=\dfrac{1-\cos A}{2r}$
using $\triangle=rs=\dfrac{abc}{4R}$
$T(1,a)+T(2,b)+T(3,c) =\dfrac{3-(\cos A+\cos B+\cos C)}r$
Now use Prove trigonometry identity for $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$k \in \mathbb Z^{+}$. Prove that for all $n ≥ 2k^2$, $n! ≥ k^{n}$
Proposition
$k \in \mathbb Z^{+}$. Then for all $n ≥ 2k^2$, $n! ≥ k^{n}$
My attempt:
Lemma 1. $k \in \mathbb Z^{+}$. For all $n \in \mathbb N$, $(k^2 + n)! ≥ k^{2n}$
Proof: By induction.
Base case: $n = 0$.
$(k^2 + 0)! ≥ 1 = k^{2\cdot0} = k^{2n}$
I... | You have made a mistake at the end of the proof of the lemma
\begin{eqnarray*}
(k^2 + n + 1)! = (k^2 + n)! \cdot (k^2 + n + 1) & ≥ k^{2n} \cdot (k^2 + n + 1) \\
& ≥ k^{2n} \cdot \color{red}{k^2} \\
& = k^{2n+\color{red}{2}}.
\end{eqnarray*}
& in the base case of the proposition, you could be clearer by saying ... Usin... | {
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"source": "stackexchange",
"question_score": "1",
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Limit of power series, 2 variables In an assignment, I've run into the following problem.
$$\lim_{x\to\infty} \sum ^\infty_{n=3} \frac{n\cdot x^{n-2}\cdot(-1)^n\cdot(n-1)}{(2n)!}$$
I really hope someone can steer me the right way, thanks!
| We have that
$$\cos x=\sum ^\infty_{n=0} \frac{(-1)^n\cdot x^{2n}}{(2n)!}\implies \cos \sqrt x=\sum ^\infty_{n=0} \frac{(-1)^n\cdot x^{n}}{(2n)!}$$
$$-\frac{\sin \sqrt x}{2\sqrt x}=\sum ^\infty_{n=1} \frac{(-1)^n\cdot n x^{n-1}}{(2n)!}$$
$$\frac{\sin \sqrt x-\sqrt x \cos \sqrt x}{4\sqrt {x^3}}=\sum ^\infty_{n=2} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1.
2^5 ends with 2.
3^5 ends with 3.
And so on.
Do I simply add the ending digits to get my answer?
| Write the sum as
$$S=(1^5+99^5)+(2^5+98^5)+(3^5+97^5)+.....+ (49^5+51^5)+50^n$$ Then
$100=1+99,=2+98= 3+97,...$ is the common factor of all the baraketed terms. Because $(a^5+b^5)=(a+b)M(a,b)$. Next, $50^5$ is also divisible by 100. So the last two digits in $S$ are $00$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find derivative $\frac{dy}{dx}$, given $y(x)=\sin^{-1}\left(\frac{5\sin x+4\cos x}{\sqrt{41}}\right)$
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\dfrac{5\sin x+4\cos x}{\sqrt{41}}\bigg]$
My Attempt
Put $\cos\theta=5/\sqrt{41}\implies\sin\theta=4/\sqrt{41}$
$$
y=\sin^{-1}\big[\sin(x+\theta)\big]\implies\sin y=\sin(x+\... | Rewriting,
$$\sin y=\dfrac{5\sin x+4\cos x}{\sqrt{41}}=\sin(x+\theta)\tag 1$$
with $\theta=\tan^{-1}\frac4{5}$ and recognize $y\in[-\frac\pi2,\frac\pi2]$, hence $\cos y \ge 0$,
$$\cos y= \text{sgn}[\cos(x+\theta) ]\cos(x+\theta) $$
Take the derivative of (1),
$$\> y' = \dfrac{\cos(x+\theta)}{\cos y}
= \dfrac{\cos(x+\t... | {
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"url": "https://math.stackexchange.com/questions/3456149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to solve $2x^2-2\lfloor x \rfloor-1=0$ How do I solve $2x^2-2\lfloor x \rfloor-1=0$?
I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfl... | $$2x^2-1=2[x]~~~~(1).$$ $[x]$ is GIF/ Floor function
We have $2[x]+1=2x^2 \ge 0 \implies [x] \ge 0 ~~~(2)$
Let $x=n+q$, $n$-is integer $n\ge 0$ and $0 \le q <1$, then we get
$$2n^2+2q^2+4nq-1=2n~~~~(3)$$
$$\implies 2n^2-2n-1=-2q^2-4nq \le 0 \implies 2n^2-2n-1 \le 0 \implies \frac{1-\sqrt{3}}{2} \le n \le \frac{1+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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sum of series $\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$ The difference of
$$\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$$
$1). $less than Zero
$2). $Greater than $1$
$3). $less than $\frac{1}{2^{100}101}$
$4). $greater than $\frac{1}{2^{100}101}$
Solution I tried: we know that $$\log2 = 1-\displaystyle \frac{1}{2}+\frac{1}{3... | $$\log2-\sum^{100}_{n=1}\frac{1}{2^nn} = \sum^{\infty}_{n=101}\frac{1}{2^nn}$$
You can put bounds on this:
$$\frac{1}{2^{101} 101} \lt \sum^{\infty}_{n=101}\frac{1}{2^nn} \lt \sum^{\infty}_{n=101}\frac{1}{2^n 101}= \frac{1}{2^{100} 101}$$
so between about $3.9 \times 10^{-33}$ and about $7.9 \times 10^{-33}$; in fact ... | {
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"url": "https://math.stackexchange.com/questions/3459869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $f(x)=x^2+a\cdot x + b$ prove $b\leq -\frac{1}{4}$ Equation $f(f(x))=0$, for $f(x)=x^2+a\cdot x + b$, have four real solutions and sum of two of them is $-1$. Prove that $b\leq-\frac{1}{4}$.
My progress so far:
Let $x_{1}$ and $x_{2}$ be real solutions for $f(x)=0$. We know that discriminant of this equation is pos... | We can bound the sum of roots and improve the bound to $b\le\frac{-2\phi^{2}-2\phi-1}{4\phi\left(\phi+2\right)}=−0.405$, without needing Vieta's formula.
By the quadratic formula, the roots of $f\circ f(x)=f^{2}+af+b$ are $\underbrace{f(x)}_{x^{2}+ax+b}=\frac{-a\pm\sqrt{a^2-4b}}{2}$. Then, by applying the formula again... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Volume of ellipsoid in Cartesian co ordinates ( w/o changing to spherical or cylindrical systems) I tried triple integrating over the $$\int_{-a}^{a} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} {d}z {d}y {d}x... | Let $V$ denote the volume. You can simplify things from the start by taking advantage of symmetry:
$$\begin{align*}
V&=\int_{-a}^a\int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}\,\mathrm dz\,\mathrm dy\,\mathrm dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Difference of two sums of three squares I have proved that every integer is the difference of two sums of three squares, i.e.,
$n = (a^2 + b^2 + c^2) - (d^2 + e^2 + f^2)$
Is this result publishable?
| Any integer not congruent to $2$ modulo $4$ can be written as the difference of two integer squares. Hence the sum of two differences of two integer squares can always be made to represent any number. For example, let $n \ge 1$ be the number, and let $a \ge 1$ and $b \ge 1$ be any two numbers such that $a,b \not\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the infinite sum of the series $\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}$ WolframAlpha outputs the answer as
$$-\frac14 \pi\left(-\coth(\pi) + \pi \operatorname{csch}^2(\pi)\right)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Base... | First, we note that $$\frac{n^2}{(1+n^2)^2}=\color{blue}{\frac1{1+n^2}}-\color{orange}
{\frac{1}{(1+n^2)^2}}.$$
So $$\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}-\color{orange}{\sum_{n=1}^\infty
\frac{1}{(1+n^2)^2}}.$$
Using the identities $$\color{blue}{\sum_{n=1}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebysh... | Below I'll show how to find
(1) the minimal polynomial for $\omega+\omega^{-1}$,
(2) the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$, and
(3) the minimal polynomial for $\sin\left(\dfrac{2\pi}{5}\right)$.
(1) First note that $\omega$ is a root of $x^5-1$, but not of $x-1$. Hence $\omega$ is a root of $\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y... | $$\sin y=s,\cos y=c$$
$$(c+s+1)^2=c^2+s^2+1+2c+2s+2cs=2\underbrace{(1+c)}+2s\underbrace{(1+c)}=?$$
Another way
As $s^2+c^2=1$
$$(s+(c+1))^2=s^2+(c+1)^2+2s(c+1)=1-c^2+(c+1)^2+2s(c+1)=(c+1)[1-c+c+1+2s]=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How many positive integers have less than $90000$ have the sum of their digits equal to $17$? How many positive integers have less than $90000$ have the sum of their digits equal to $17$?
I tried to write the number as $ABCDE$ and use some math with that (so we need $A + B + C + D + E = 17$), and I tried to use stars a... | You need to find the number of solutions of $$A+B+C+D+E=17$$
where $0\leq A\leq 8$ and all other variables lie between $0$ and $9$. Check that it is coefficient of $x^{17}$ in the following expression:
$$(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
$$\times(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)^4$$
Now, apply GP formula and simp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$\int \frac{\tan x}{\cos^2x}dx$ $\int \frac{\tan x}{\cos^2x}dx=\int\frac{\frac{\sin x}{\cos x}}{\cos^2 x}dx=\int \frac{\sin x}{\cos^3x}dx$
$\cos x=t, -\sin xdx=dt \Rightarrow \sin xdx=-dt$
$\int \frac{-dt}{t^3}=-\int t^{-3}dt=\frac{1}{2}t^{-2}dt=\frac{1}{2\cos^2x}+C $
$\int \frac{\tan x}{\cos^2x}dx\\\tan x=t,\frac{dx}{... | Both are correct, recall $\tan^{2}(x)+1=\sec^{2}(x)$, so that your two answers differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\fr... | You may just apply $t=\tan x$, $dx = \frac{dt}{1+t^2}$ and the resulting expression is quite manageable,,
$$I=\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{1-\frac12\sin^2(2x)}
= \int \frac {\frac{1-t^2}{1+t^2}\cdot\frac{dt}{1+t^2}}{1-\frac12\left(\frac{2t}{1+t^2}\right)^2}=\int \frac{1-t^2}{1+t^4}dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving a first order ordinary differential equation that is not linear and not separable Problem:
Solve the following differential equations.
$$ ( 3x - y + 1 ) dx - ( 3x - y) dy = 0 $$
Answer:
I am going to use the substitution $z = 3x - y + 1$.
\begin{align*}
\frac{dz}{dx} &=3 - \frac{dy}{dx} \\
dz &=3 \, dx - dy \\
... | Making $u = 3x-y$ we have $\frac{du}{dx}=3-\frac{dy}{dx}$ and also
$$
u+1-u\frac{dy}{dx} = u+1-u\left(3-\frac{du}{dx}\right)=0
$$
and this DE now is separable
$$
\frac{u du}{2u-1} = dx
$$
or
$$
\frac 14\left(\ln(2u-1)+2u-1\right)=x + C_0
$$
or
$$
(2u-1)e^{2u-1}=C_1 e^{4x}
$$
and now using the Lambert function
$$
2u-1=W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Hard inequality for positive numbers The problem is to prove that for $a,b,c>0$ we have
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+... | The cyclically symmetric inequality is equivalent to:
$$\color{red}{\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)} + \color{blue}{\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1 \right)}+ \color{green}{\left(\frac{9abc}{4(a^3+b^3+c^3)}-\frac34\right)} \geqslant 0$$
$$\iff \color{red}{\frac{(a^2-b^2)^2}{a^2b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Factoring $x^4-2x^3+2x^2+x+4$
I need to show that the polynomial is not irreducible and I am trying to factor the polynomial
$$x^4-2x^3+2x^2+x+4$$
I checked from a calculator that it has a factor but how do I get it by myself?
I tried grouping but It didnt work I got
$x^2(x^2-2x+2)+x+4$ And I dont know how should... | Actually, if you just try to solve this like a general quartic equation you get a break: let $x=t+1/2$. Then
$$\begin{align}x^4-2x^4+2x^2+x+4&=\left(t+\frac12\right)^4-2\left(t+\frac12\right)^3+2\left(t+\frac12\right)^2+\left(t+\frac12\right)+4\\
&=t^4+\frac12t^2+2t+\frac{77}{16}\\
&=\left(t^2+at+b\right)\left(t^2-at+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Choosing group of size 3 in a roll of 5 unique dice Suppose you have $5$ fair dice (with sides $1-6$) colored red, green, yellow, orange and blue.
In how many ways can you roll them such that the resulting set of numbers rolled is of size $3$? For instance, the case where we roll: $1,2,3,1,2$ is accepted but $1,2,3,4,1... | You can either have three of one number and two singletons or two of two numbers and one singleton. For the first, you have $6$ ways to choose the number there will be three of, $5 \choose 3$ ways to choose the dice with that number, $5$ ways to choose number on the first of the other dice and $4$ ways to choose numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron who... | A tetrahedron is never a parallepiped. It is a pyramid with a triangular base.
All six faces of a parallelopiped are parallelograms but a tetrahedron has only four faces and all are triangles.
In short, what you are measuring is very unlike what you were asked to measure.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find $p$ and $q$ such that $x^2+px+q
Find $p$ and $q$ such that $$x^2+px+q<x$$ iff $$x \in (1,5)$$
I tried the following:
$$x^2+px+q = (x+\frac{p}{2})^2+q-\frac{p^2}{4}$$
where the global minimum is $$q-\frac{p^2}{4}$$ if $$x = \frac{-p}{2}$$
However this doesn't seem to help with the problem at hand...
| So $1$ and $5$ are the roots of $x^2+px + q = x$:
$$\begin{align*}
(x-1)(x-5) &= 0\\
x^2 - 6x + 5 &= 0\\
x^2 - 5x + 5 &= x\\
p &= -5\\
q&= 5
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\foral... | Use Banach fixed-point theorem, for $a_{n+1}=f(a_n)$ where $f(x)=\frac{x+2}{3x+2}$ and $f'(x)= \frac{-4}{(3 x + 2)^2}$. In fact, for positive $x$
$$1>\frac{x+2}{3x+2} > \frac{1}{3}$$
and, starting with $n\geq 1$, $a_n\in\left(\frac{1}{3},1\right)$. As a result, using MVT
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
minimum value of $\sum (x+1/x)^{10}$ expression
If $x,y,z>0$ . Then prove that $\displaystyle \bigg(x+\frac{1}{x}\bigg)^{10}+\bigg(y+\frac{1}{y}\bigg)^{10}+\bigg(z+\frac{1}{z}\bigg)^{10}\geq \frac{10^{10}}{3^{9}}.$
What i try
Let $\displaystyle f(x)=\bigg(x+\frac{1}{x}\bigg)^{10}.$ Then $\displaystyle f'(x)=10\bigg(x... | If it means that $x+y+z=1$ so by Jensen for $f(x)=x^{10}$ and by C-S we obtain:
$$\sum_{cyc}\left(x+\frac{1}{x}\right)^{10}\geq\frac{1}{3^9}\left(\sum_{cyc}\left(x+\frac{1}{x}\right)\right)^{10}=\frac{1}{3^9}\left(1+(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)^{10}\geq$$
$$\geq\frac{1}{3^9}\left(1+9\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Is $x^2 \geq \alpha(\alpha-1)$?
If $\alpha$ is a nonnegative real and $x$ is a real satisfying $(x+1)^2\geq \alpha(\alpha+1),$ is $x^2 \geq \alpha(\alpha-1)$?
The answer is yes. Consider two cases: $1) \, x < -1$ and $2)\, x\geq -1.$
In case $1,$ taking the square root of both sides of the inequality gives $-(x+1) \... | Here's another method.
Suppose $\alpha(\alpha+1) = (x+1)^2$ and $x\geq 0$. Then since $\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right)=(x+1)^2-\frac{1}{4},$ we have that $\alpha > x+\frac{1}{2}.$ Hence $\alpha(\alpha-1)=\alpha(\alpha+1)-2\alpha <(x+1)^2-2x-1=x^2.$
If $\alpha(\alpha+1) < (x+1)^2$ and $x\geq 0,$ the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the integral by first principle In my math course homework, I encountered this problem:
Find $$\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}$$ by first principle
$$\begin{split}\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_0\frac{dt}{1+e^{t^2}}-\int^{x^2}_0\frac{dt}{1+e^{t... | Compute $$\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}
$$
without using the fundamental theorem of calculus or the mean value theorem.
If $x^2 < t < (x+h)^2$, then
$$
x^4 < t^2 < (x+h)^4
\\
1 + \exp(x^4) < 1+\exp(t^2) < 1+\exp((x+h)^4)
\\
\frac{1}{1 + \exp(x^4)} > \frac{1}{1+\exp(t^2)} > \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Multinomial theorem member that does not contain an irrational number Hey I am supposed to solve the following problem:
Specify a development member that does not contain an irrational number:
$$\left (\sqrt{5} - \sqrt[3]{2} +2 \right )^{6}.$$
So I used multinomial theorem:
$$\sum_{i=0}^{6}\binom{6}{n_{1},n_{2},n_{... | No, your answer is not correct. Since $n_1\in\{0,2,4,6\}$ and $n_2\in\{0,3,6\}$ with $n_1+n_2\leq 6$, we find that the required number is
$$N:=\sum_{(n_1,n_2)\in S}\binom{6}{n_{1},n_{2},6-n_1-n_2}\left ( \sqrt{5} \right )^{n_{1}}\left ( - \sqrt[3]{2} \right )^{n_{2}}\left ( 2 \right )^{6-n_1-n_2}$$
where $S:=\{(0,0), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving Cauchy Schwarz inequality Taken from Titu Andreescu and Bogdan Enescu's Mathematical Olympiad Treasures on page 9 Problem 1.19, to prove,
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geqslant \frac{3}{2}$.
It is easy to see why $LHS$ may yield
$\geqslant\frac{(a+b+c)^2}{2(ab+bc+ca)}$ from Cauchy.
Yet how do... | Because $$\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}$$ it's
$$(a+b+c)^2\geq3(ab+ac+bc)$$ or
$$a^2+b^2+c^2+2ab+2ac+2bc\geq3(ab+ac+bc)$$ or
$$a^2+b^2+c^2\geq ab+ac+bc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Estimation of sum of series with cosine Prove the following:
$$
\sum_{{\large j = 1} \atop {\large j \neq k}}^{n}
{1 \over \left\vert\cos\left(k\pi/ n\right) -
\cos\left(j\pi/n\right)\right\vert} \leq cn^{2}
\qquad\mbox{where}\quad 1 \leq k \leq n\quad\mbox{is fixed.}
$$
I was able to get the upper bound to be $cn^{3}$... | By sum-to-product formulae, the sum $S(n)$ equals
$$S(n)=\frac12\sum^n_{j=1}\left|\csc\frac{(j+k)\pi}{2n}\csc\frac{(j-k)\pi}{2n}\right|$$
By Cauchy-Schwarz inequality,
$$S(n)\le\frac12\sqrt{
\underbrace{\sum^n_{j=1}\csc^2\frac{(j+k)\pi}{2n}}_{S_1}
\cdot\underbrace{\sum^n_{j=1}\csc^2\frac{(j-k)\pi}{2n}}_{S_2}}$$
Regardi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove a trigonometric inequality Let $a,b,c$ reals with sum 0. Prove that
$\left|\sin{a}+\sin{b}+\sin{c}\right| \leq \frac{3\sqrt{3}}{2}.$
My idea was to set $z_1=\cos{a}+i\sin{a}$ and also $z_2,z_3$. Then I tried replacing the sines in the inequality with these complex numbers. However, I didn’t get to anything intere... | By the triangle inequality, C-S and AM-GM we obtain:
$$|\sin{a}+\sin{b}+\sin{c}|=|\sin{a}+\sin{b}-\sin{a}\cos{b}-\cos{a}\sin{b}|=$$
$$=|(1-\cos{b})\sin{a}-\sin{b}\cos{a}+\sin{b}|\leq|(1-\cos{b})\sin{a}-\sin{b}\cos{a}|+|\sin{b}|\leq$$
$$\leq\sqrt{((1-\cos{b})^2+(-\sin b)^2)(\sin^2a+\cos^2a)}+|\sin{b}|=$$
$$=\sqrt{2-2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Question on Integration solution I am looking through the solution of integrating: $$\int_{}^{} \frac{x^4+1}{x^3+1} dx$$
I understand it but for the following part:
My problem is limited to the first and penultimate column. I would very much appreciate it if someone could explain to me these, especially the penultimat... | Solution of integral will be by partial fraction method
$$I=\int (x+\frac{1-x}{x^3+1})dx$$
$$I = \frac{x^2}{2}+\int (\frac{1-x}{(x+1)(x^2-x+1)})dx$$
Where you can solve for
$$I_1=\int (\frac{1-x}{(x+1)(x^2-x+1)})dx$$
Using partial fraction like done here
https://www.mathsdiscussion.com/discussion-forum/topic/indefinit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating an integral using Leibniz parametric integration formula Let there be $a, b > 0$, find:
$\int_0^{\pi/2}$$\frac{dx}{(a\cos^2(x)+b\sin^2(x))^2}$
Hints given:
1. Remember: $\cos^2(x) + \sin^2(x) = 1$, use it to make the integral a sum of two parts.
2. Define $f(x,y)=\frac{-1}{(y\cos^2(x)+b\sin^2(x))^2}$, and c... | $$f(a,b)=\int_0^{\pi/2}\frac{dx}{a\cos^2(x)+b\sin^2(x)}=\int_0^{\pi/2}\frac{sec^2(x)}{a+b \tan^2(x)}dx=\frac{1}{\sqrt{ab}}arctan\left(\sqrt{\frac{b}{a}}tan(x)\right)_0^{\pi/2}$$
$$\color{red}{f(a,b)=\int_0^{\pi/2}\frac{dx}{a\cos^2(x)+b\sin^2(x)}=\frac{\pi}{2\sqrt{ab}}}$$
$$\frac{\partial f(a,b)}{\partial a}+\frac{\part... | {
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"url": "https://math.stackexchange.com/questions/3507989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2... | Hint : Using homogeneity, WLOG we may set $a+b+c=3$, then note $f(x) = \dfrac{x}{(3-x)^2}$ is convex and use Jensen’s inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
... | As @Andrei already answered, you face a geometric sum, that is to say that
$$\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=\sum_{k = 1} ^ n \left(e ^{\frac{1}{n^2}}\right)^k=\frac{e^{\frac{1}{n^2}} \left(e^{\frac{1}{n}}-1\right)}{e^{\frac{1}{n^2}}-1}$$ Now, expanding the exponentials as Taylor series, multiplying for the numerat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence... | You could have done everything a bit faster. Starting from what you wrote
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$ let $x=\frac 1 n$, use the binomial expansion or Taylor series, replace $x$ by $\frac 1n$ to get
$$a_n=(2-a) n^2-\frac{1}{4}-\frac{5}{64 n^2}+O\left(\frac{1}{n^4}\r... | {
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"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $a$, $b$ for $z = \frac{1}{(a + ib)^3}$ to be on negative real axis
For what values of $a$ and $b$ does
\begin{equation}
z = \frac{1}{(a + ib)^3}
\end{equation}
lie on the negative real axis. Hence, or otherwise, find an expression in terms of $a$ only for $\lvert z \rvert$ when $\operatorname{arg}(z) = - \pi... | For $z$ to lie on the negative real axis, we have
$$z = \frac{1}{(a + ib)^3}=-r=re^{i(1+2n)\pi}$$
where $r>0$. Rearrange to get
$$a+bi= \frac1{\sqrt[3]r} e^{-i\frac{(1+2n)\pi}3}
=\frac1{\sqrt[3]r} \left( \cos\frac{1+2n}3\pi - i\sin\frac{1+2n}3\pi \right)
$$
with $n=0,1,2$. Then,
$$a= \frac1{\sqrt[3]r} \cos\frac{1+2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
... | My short solution:
$$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left[\left(\frac{1}{2}\right)^n-\frac{1}{4^n}\right]=1-\frac13=\frac 23$$
PS: I have applied a criteria of a series geometric when $|x|<1$. See the link http://mathworld.wolfram.com/GeometricSeries.html formula number (9).
| {
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"url": "https://math.stackexchange.com/questions/3512982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finite Derivative of $\frac{x^3-1}{x+1}$ I have the following qeustion:
Evaluating the front, back and central derivative I just need to plug it into the corresponding:
$$D_{+}(x)=\frac{f(x+h)-f(x)}{h},D_{-}(x)=\frac{f(x)-f(x-h)}{h},D_{c}(x)=\frac{f(x+h)-f(x-h)}{2h}?$$
Without choosing some point or value for $h$?
Fo... | You do the calculations indicated, as tedious as they might be!
Since $f(x)= \frac{x^3- 1}{x+ 1}$, $f(x+ h)= \frac{(x+ h)^3- 1}{x+ h+ 1}= \frac{x^3+ 3x^2h+ 3xh^2+ h^3- 1}{x+ h+ 1}$.
So $f(x+ h)- f(x)= \frac{(x+ h)^3- 1}{x+ h+ 1}- \frac{x^3- 1}{x+ 1}= \frac{x^3+ 3x^2h+ 3xh^2+ h^3- 1}{x+ h+ 1}- \frac{x^3- 1}{x+ 1}$.
To d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find sides of triangle, and radius of inscribed circle
$AD = x, \angle A = 60^{\circ}$
I was asked to find $BC$ when $BD=4, CF=2$
I tried using law of cosines,
$(4+x)^2 + (2+x)^2 - 2(4+x)(2+x) \cos 60^{\circ} = BC^2$
I get $x^2+6x+12=BC^2$
But the answer is $BC= 6, x^2+6x-24=0$
Also i was asked to find $\frac{\tria... | In the standard notation: $$BD=\frac{a+c-b}{2}$$ and $$CF=\frac{a+b-c}{2}.$$
Thus,
$$BC=BD+CF=6.$$
Also, $$\frac{S_{\Delta ADF}}{AG\cdot AE}=\frac{\frac{1}{2}AD\cdot AF\cdot\sin60^{\circ}}{AF^2}=\frac{\sqrt3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equations reducible to homogeneous form Solve the following differential equation:
$$(2x+y-3)dy=(x+2y-3)dx$$
I have tried this & I found:
$$(y-x)^3=c(y+x+2)$$
But in my book the answer is:
$$(y-x)^3=c(y+x-2)$$
Please tell which one is correct..
| $$ (x+2y-3)dx=(2x+y-3)dy$$
The two constants of $-3$ are keeping the equation from being homogeneous. We can eliminate the constants by an appropriately chosen substitution:
$$X=x+a,\quad Y=y+b$$and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 3)dX = (2X-2a + Y - b - 3)dY $$
It's ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| $$I=\int \frac{x}{x^4+4}dx= \frac{1}{2} \int \frac{dt}{4+t^2}=\frac{1}{2} \tan^{-1} t/2+C=\frac{1}{4} \tan^{-1} x^2/2+C$$ Here, we took $x^2=t \implies 2xdx=dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.