Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\lfloor a_n \rfloor$=? where $a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$ $\lfloor a_n \rfloor$=?, where $$a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$$
and the number of radicals is equal to n.
$\lfloor a_1]=\lfloor \sqrt{2005} \rfloor=44$, $\lfloor a_2 \rfloor=\lfloor \sqrt{2005+\sqrt{2005}}... | $\def\e{\varepsilon}$Here we bound the sequence by examining the sequence of successive differences.
Let $x=2005$.
We have the recursion
\begin{align*}
a_0 &= 0 \\
a_n &= \sqrt{x+a_{n-1}},\quad n\ge1.
\end{align*}
Note that $a_n\ge0$ for $n\ge 0$.
Let $d_n = a_n-a_{n-1}$ for $n\ge 1$.
One can verify that
\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition.
I have so far reduced it algebr... | Hint :
Observe that $y^2 \le x^2 +y^2$ so $y^4 \le (x^2 + y^2)^2$.
Also $3x^2 +y^2 \ge x^2 +y^2$ which implies that $\frac{1}{\sqrt{3x^2 + y^2}} \le \frac{1}{\sqrt{x^2+y^2}}$
Combining both we get $$\frac{y^4}{\sqrt{3x^2 + y^2}} \le \frac{(x^2+y^2)^2}{\sqrt{x^2+y^2}} = (x^2+y^2)\sqrt{(x^2+y^2)}$$
So for given $\epsilon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals.
Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$
The problem is from an inequality handout. Here is my attempt to solve the problem:
I first rew... | Another way.
By Holder $$\left(\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\right)^2\sum_{cyc}a^2(b+c)(b^3+c^3)\geq(a^2+b^2+c^2)^3$$ and it's enough to prove that:
$$4(a^2+b^2+c^2)^3\geq9\sum_{cyc}(a^2b+a^2c)(b^3+c^3)$$ or
$$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq0,$$ which is true by Schur, AM... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and... | From Cauchy-Schwartz inequality,
$a^2 + b^2 = (a^2 + b^2) ( \dfrac{1}{2} + \dfrac{1}{2} ) \ge (\dfrac{1}{\sqrt{2}} a + \dfrac{1}{\sqrt{2}} b )^2 = \dfrac{1}{2} (a + b)^2$
Hence $(a + b)^2 \le 2 (a^2 + b^2) = 8 $
Thus $|a + b| \le \sqrt{8} $
The last inequality is equivalent to $ -\sqrt{8} \le a + b \le \sqrt{8} \lt 4 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac... | As was mentioned in the comments, the integral $$\int_0^1\frac{dx}{1-x^2}$$
diverges, so you cannot split the integral $$\int_0^1\left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)dx$$
up into parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Solving the equation $2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$
Solve the equation $$2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$$
Here is my work:
$$2x^2-5x+4=(4x-7)\sqrt{x-1}$$
$$2(x-1)^2-(x-1)+1=(4(x-1)-3)\sqrt{x-1}$$
Using the substitution $\sqrt{x-1}=t$,
$$2t^4-t^2+1=t(4t^2-3)$$
$$2t^4-4t^3-t^2+3t+1=0$$
Here I plugge... | What you did is fine. Now, let\begin{align}p(t)&=2t^4-4t^3-t^2+3t+1\\&=2\left(t^4-2t^3-\frac12t^2+\frac32t+\frac12\right)\end{align}and let $r(t)$ be the depressed quartic of $t^4-2t^3-\frac12t^2+\frac32t+\frac12$, that is\begin{align}\left(t+\frac12\right)^4-2\left(t+\frac12\right)^3-\frac12\left(t+\frac12\right)^2+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \fr... | You can also do it using Pochhammer symbols
$$\frac{n(n+1)}{n(n+1)+2}=\frac{n(n+1)}{(n-a)(n-b)}$$
$$P_p=\prod_{n=1}^p\frac{n(n+1)}{(n-a)(n-b)}=\frac{\prod_{n=1}^p n(n+1) }{{\prod_{n=1}^p (n-a)(n-b)} }=\frac{\Gamma (p+1) \Gamma (p+2)}{(1-a)_p (1-b)_p}$$
Take logarithms and use Stirling approximation
$$\log(P_p)=((a+b) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r =... | Alternative proof:
The desired inequality is easily written as
$$\left(a + \frac{\sqrt2\, bc - 2 - \sqrt2}{2}\right)^2 + \frac{2 - c^2}{2}\left(b - \frac{2 + \sqrt2 - c - c\sqrt2}{2 - c^2}\right)^2 + \frac{c(c - 1)^2}{c + \sqrt2} \ge 0.$$
So, if $c < \sqrt2$, the desired inequality is true.
If $c \ge \sqrt2$, it suffic... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further ... | Yet another way.
Lemma 1:
$$ \int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n} \tag{1}$$
Proof of Lemma 1: by parity and De Moivre's formula
$$ \int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{2n}\,d\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What is the value of the $\measuredangle x$ in the figure below? For reference:
If ABCD is a square calculate "x" being "M" and "N" midpoints of CD and AD respectively.
My progress:
$\triangle ABN \cong \triangle BC (2:1- special~ right~ triangle) \implies \measuredangle ABN = \frac{53^o}{2}=\measuredangle MBC\\
\ther... |
WLOG, we assume $ABCD$ is a unit square. Then,
$BN = NP = \frac{\sqrt5}{2}$
Using power of point of $B$,
$BH \cdot BP = AB^2 \implies BH \cdot \sqrt5 = 1$
$BH = \frac{1}{\sqrt5}$
Now, $\frac{BQ}{BR} = \frac{HQ}{NR} = \frac{BH}{BN}$
$\frac{BQ}{3 / 2 \sqrt2} = \frac{HQ}{1/2\sqrt2} = \frac{1/\sqrt5}{\sqrt5/2}$
$BQ = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I thi... | You can use the conjugate quantity to get an alternative expression of the roots:
Suppose that $B > 0$. Then:
$\frac{-B + \sqrt{B^2-4AC}}{2A} = \frac{(-B + \sqrt{B^2-4AC})(-B - \sqrt{B^2-4AC})}{2A(-B - \sqrt{B^2-4AC})} = \frac{(-B)^2 - (\sqrt{B^2-4AC})^2}{2A(-B - \sqrt{B^2-4AC})} = \frac{2C}{-B - \sqrt{B^2-4AC}}$
If $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Frustrated because used standard R addition formula for mechanics problem, but the answer came out wrong. How to resolve this so doesn't happen again? In the middle of quite a lengthy mechanics problem, I got to:
$$\cos\alpha-\sin\alpha = 0.05\ $$
and I wanted to find the value of $\ \alpha,\ 0<\alpha<90^{\circ}\ $ tha... | \begin{align}
\cos\alpha - \sin\alpha &= \frac{1}{20}\\
\cos\alpha \cdot \frac{\sqrt 2}{2} -\sin\alpha\cdot \frac{\sqrt 2}{2}
&= \frac{\sqrt 2}{40}\\
\cos\alpha \cdot \cos\left(\frac{\pi}{4}\right)
-\sin\alpha \cdot \sin\left(\frac{\pi}{4}\right)
&= \frac{\sqrt 2}{40}\\
\cos\left(\alpha + \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving a curve has no curve point I am trying to show that the curve $x^2 + y^2 - 3 = 0$ has no rational point, where a rational point is defined as a solution $(x,y) \in \mathbb{Q}^2$ to this equation.
My attempt was to proceed by contradiction. Assume there is such a point $(x,y) \in \mathbb{Q}^2$ and let $x = \frac... | you want a solution to the equation:
$$x^2+y^2=3$$
you could try proving that if this is a point on a circle, it cannot have both coordinates be integer, so say:
$$x=\sqrt{3}\cos\theta,y=\sqrt{3}\sin\theta$$
for $x$ to even be rational $\cos\theta$ must be be a rational multiple of $\sqrt3$ and similar for $y$. Now sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}... | Consider a square of area $1$ and divide it into 3 congruent rectangles.
The area of each rectangle is $\frac{1}{3}$.
Select one of the three rectangles.
Divide one of the remaining two rectangles into three congruent rectangles, select only, as done before. Continue like this.
Irrespective of the number of iterations,... | {
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"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$ with Popoviciu's inequality. I want to prove the inequality $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$$, for positive reals. I have a hint that it is solved with Popoviciu'... | For positives $x$, $y$ and $z$ we need to prove that
$$\sum_{cyc}\frac{x+y+z}{x}-3\geq4\sum_{cyc}\frac{x+y+z}{x+y}-12$$ or
$$\sum_{cyc}\frac{1}{x}+\frac{9}{x+y+z}\geq4\sum_{cyc}\frac {1}{x+y}$$
or
$$\sum_{cyc}\frac{1}{x}+\frac{3}{\frac{x+y+z}{3}}\geq2\sum_{cyc}\frac {1}{\frac{x+y}{2}},$$which is Popovicui for the conve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Factoring cubic equation Trying to understand this passage where this:
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]=0$$
gets factored into this:
$$\left(\frac{1}{6}\beta-I\right) \left(\frac{11}{12}\beta-I\right) \l... | We have by expanding and simplifying
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]$$
$$=\left(\frac{11}{12}\beta-I\right)\left[\frac{4}{9}\beta^2-\frac{4}{3}\beta I+I^2-\frac{1}{8}\beta^2-\frac{1}{6}\beta^2+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there any other method to find the remainder when $x+x^3+x^5+\cdots+x^{2n-1}$ is divided by $x^3-x$? A method I got from somewhere.
\begin{align}
f(x)
&=x+x^3+x^5+\cdots+x^{2n-1}\\
&= x\Bigg[\Big(x^{2(n-1)}-1\Big)+\Big(x^{2(n-2)}-1\Big)+\cdots+\Big(x^{2\times 1}-1\Big)+\Big(x^{2\times 0}-1\Big)+n\Bigg]\\
&= x\Big(... | First, we can eliminate a factor of $x$ , and find the remainder of
$P(x) = 1+x^2+x^4+\cdots+x^{2n-2}$ divided by $x^2-1$
Write $P(x)$ as
$P(x) = (x^2-1)Q(x) + ax + b$
Now substitute the values $x=1$, $x=-1$ to obtain
$P(1) = a+b$
$P(-1) = b-a$
and note that $P(1)=P(-1) = n $, and solve the simultaneous equations...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why do we consider the lower bound in this Inequality? Given that $a,b,c,d,e$ are real numbers such that
$a+b+c+d+e=8$,
$a^2+b^2+c^2+d^2+e^2=16$.
Determine the maximum value of $e$.
Let $t$ be a real number. Now consider the expression $$ \sum_{cyc} (t-a)^2 = \sum_{cyc} a^2-2t \sum_{cyc} a
+ 5t^2$$
Now obviously $$(... | Because if $e>\min\limits_{t}f$ so for $t=\frac{6}{5}$ and $a=b=c=d=\frac{6}{5}$ we'll obtain a contradiction.
Another way.
By C-S
$$16=a^2+b^2+c^2+d^2+e^2\geq\frac{1}{4}(a+b+c+d)^2+e^2=\frac{1}{4}(8-e)^2+e^2,$$ which gives $0\leq e\leq\frac{16}{5}.$
The equality occurs for $a=b=c=d$, id est occurs, which says that $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $\varphi$ is an injective homomorphism and why it is surjective? Here is the question I want to solve:
Prove that the subgroup of $SL_2(\mathbb F_3)$ generated by $\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$ and $\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}$ is isomorphic to the quaternion group o... | You know that $A^2=B^2$, $A^4=\operatorname{Id}_2$, and that $AB=-BA$. So, since $\mathcal Q_8$ since the group generated by two generators $i$ and $j$ and by the relations $i^2=j^2$, $i^4=e$, and $ij=-ji$, there is a surjective group homomorphism $\varphi$ from $\mathcal Q_8$ onto the subgroup $G$ of $SL_2(\Bbb F_3)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | As noticed you went wrong multiplying for a quantity which can also be negative.
Using that for $A,B>0$ then
$$|A|<B \iff \left|\frac1{A}\right|>\frac1B $$
a very effective way to proceed in this case is the following
$$\left|\frac{7}{x+4}\right|<1 \iff \left|\frac{x+4}{7}\right|>1 \iff |x+4|>7$$
$$\iff x+4<-7 \quad \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
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How does $\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}}$ become $2 - \frac{1}{2^{k+1}}$? I'm studying induction here and throughout the proof we got from here:
$$ \begin{align}
&\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}} \tag1\\[0.75em]
=\; &2 - \frac{1}{2^{k+1}} ( 2-1) \tag2\\[0.75em]
=\; &2 - \fr... | They did this by adding up the two fractions:
$\frac{1}{2^k}$ - $\frac{1}{2^{k+1}}$
When dealing with the denominators, you can write $2^{k+1}$ as ${2^k}$$\cdot$${2^1}$
Then the $2^k$ cancels out with the first fraction's denominator and you are left with $2$$\cdot$$1$ which totals to 2. In the second fraction you are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simultaneous equations to solve unknown exponents Solved the following simultaneous equations for $a$ and $b$:
(1)
$$a + 2b = 2\\3a + b = 10$$
$$3a + 6b = 6\\3a + b = 10$$
$$5b = -4$$
$$b = \frac{-4}{5}\\a = \frac{18}{5}$$
(2)
How do we use the answers/information in (1) to solve the following equations for $x$ and $y$... | Note that $25^{x+50}=(5^2)^{x+50}=5^{2x+100}$ so the system can be written as
$$2^{x+2y} = 2^{20}\\5^{5x+y} = 5^{2x+100}$$
or
$$2^{x+2y} = 2^{20}\\5^{3x+y} = 5^{100}$$
and comparing exponents we obtain
$$x+2y=20$$
$$3x+y=100$$
This is very similar to the system
$$a + 2b = 2\\3a + b = 10$$
but the RHS is multiplied by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in th... | The generating function proof of Observation 1.
Write $$f(x,z)=\sum_{n=0}^{\infty}(1+x)^nz^n=\frac{1}{1-z(x+1)}$$ Then the coefficient of $x^a$ is $$h_a(z)=\sum_{k=a}^\infty \binom ka z^k$$
When $z=1/2,$ then $h_a(1/2)$ is your value.
But $$f(x,1/2)=\frac1{1/2-x/2}=\frac{2}{1-x}$$
So $h_a(1/2)=2,$ the result you want.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Here's another method.
Note that by re-arranging, we may write $a+b=c+d$ either as $a-c=d-b$ or as $b-c = d-a.$ Just keep that in mind as we manipulate the second equation:
$$\begin{align*}0 &= (c^2 + d^2) - (a^2 + b^2) \\ &= (d^2 - b^2) - (a^2 - c^2) \\ &= (d-b)(d+b) - (a-c)(a+c) \\ &= (a-c)(d+b) - (a-c)(a+c) \\ &= (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the correct general solution of $\sin x=-\frac{1}{2}$? My attempt:
$$\sin x=-\frac{1}{2}$$
$$x=n\pi+(-1)^{n}\sin^{-1}(-\frac{1}{2})$$
$$x=n\pi+(-1)^{n}(-\frac{\pi}{6})$$
$$x=n\pi-(-1)^{n}(\frac{\pi}{6})$$
My book's solution:
$$x=n\pi+(-1)^{n}(\frac{7\pi}{6})$$
Who is correct?
| We have that
$$\sin x=-\frac{1}{2} \iff x=-\frac \pi 6+2k\pi \quad \lor \quad \frac {7\pi} 6+2k\pi$$
which is equivalent to both solutions indeed
*
*for $n=2k$
$$x=2k\pi-(-1)^{2k}\left(\frac{\pi}{6}\right)=2k\pi-\frac{\pi}{6}$$
$$x=2k\pi+(-1)^{2k}\left(\frac{7\pi}{6}\right)=2k\pi+\frac{7\pi}{6}$$
*
*for $n=2k+1$
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the expected number of $\min\{i \mid a_i\geq a_{i+1}\}$? I am solving a computer programming problem.
I wonder if an easy mathematical solution to the following problem exists or not.
Suppose we throw a fair die sufficiently number of times.
Let $a_i$ be the outcome of $i$-th roll.
What is the expected number ... | Let $X=\min\{i \mid a_i\geq a_{i+1}\}$
Then $P(X\ > i)=P(a_1 < a_2, a_2 < a_3 , ..., a_i < a_{i+1})$
Now, if we throw a $6-$faces die $k$ times, there are $\binom{6}{k}$ possible strictly increasing sequences.
Hence the above probability is
$$P(X\ > i) = \frac{\binom{6}{i+1}}{6^{i+1}}$$
for $i =1,2 \cdots 5$. And the e... | {
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Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ I have tried many ways but not success
$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$
or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1... | After squaring, you face the quartic equation
$$3 x^4+4 x^3-6 x^2-4 x-1=0$$
Since $\Delta=-94208$, there are two real roots. Now, forget the nasty formulae and use inspection or graphing; there are two roots "close" to $-2$ and $1.5$. So, use Newton method and obtain the following iterates
$$\left(
\begin{array}{cc}
n ... | {
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"source": "stackexchange",
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Spivak Proof: $\left| \frac{1}{x} \right| = \frac{1}{|x|}$ ..... if $x \neq 0$
Spivak's Proof:
$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$ so $\left| \frac{1}{x} \right| = \frac{1}{\left|x\right|} $
I am fine until the 'so'. Is the 'so'? $$ 1 \cdot \lef... | By this
$$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$$
we have shown that $\left| \frac{1}{x} \right|$ and $\left| x \right|$ are one the inverse of the other, that is $\left| \frac{1}{x} \right| = \frac{1}{\left|x\right|}$.
Edit
As an alternative
*
*for... | {
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Help with convergence sequence by cases I have to study the converge or diverge or this sequence:
\begin{eqnarray}
c_{n}=\left\{\left(\cos (\frac{a}{n})\right)^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}}\right\} \hspace{0.3cm} \text{for a,b > 0}
\end{eqnarray}
So I noticed that:
\begin{eqnarray}
(\cos (\frac{a}{n}))^{\frac{1... | HINT
Since $\cos \frac{b}{n}\to 1$ we can eliminate absolute value and we can use that
$$\frac{\ln\left(\cos \frac{a}{n}\right)}{\ln\left(\cos \frac{b}{n}\right)}=$$
$$= \frac{\ln(1+(\cos \frac{a}{n}-1))}{\cos \frac{a}{n}-1} \frac{\cos \frac{b}{n}-1}{\ln(1+(\cos \frac{b}{n}-1))}\frac{\cos \frac{a}{n}-1} {\left(\frac{a}... | {
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"source": "stackexchange",
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Issue solving an integral I'm having issues solving the following integral:
$$ \int{\frac{e^x}{e^{2x}-e^x-2}}dx $$
From what I can tell, I should substitute $u = e^x$ and $du = e^xdx$ and end up with
$$ \int{\frac{1}{u^2 - u - 2}}du $$
where I can then use partial fraction decomposition to simplify the integrand as fol... | Answer in the Question
Your answer
$$
\frac13\log\left(\frac{e^x-2}{e^x+1}\right)+C\tag1
$$
is partially correct; another possibility is
$$
\frac13\log\left(\frac{2-e^x}{1+e^x}\right)+C\tag2
$$
Which one is applicable depends on the initial value of $x$. If $x\gt\log(2)$, use $(1)$; if $x\lt\log(2)$, use $(2)$.
Wolfra... | {
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Finding all pairs of functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy $f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008}$ Find two functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy the condition:
$$
f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008}
$$
The first one I thought of was th... | Let find $y$ such that $x+y=xy\Rightarrow y=\frac{x}{x-1}$.
One can take $y=\frac{x}{x-1}$ for $x\neq 0$, $x\neq 1$, then
$$f\left(x+\frac{x}{x-1}\right)=g\left(\frac{1}{x}+\frac{x-1}{x}\right)\cdot\left(x\cdot \frac{x}{x-1}\right)^{2008}\Rightarrow$$
$$f\left(\frac{x^2}{x-1}\right)=g(1)\cdot \left(\frac{x^2}{x-1}\righ... | {
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"url": "https://math.stackexchange.com/questions/4274354",
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Solve the elliptic integral $\int \frac1{\sqrt{C-x^2+\frac{x^4}2}}$ as the solution to $\ddot x + x - x^3 = 0.$ It is said that the elliptic integral $\int \frac1{\sqrt{C-x^2+\frac{x^4}2}}$ (whose solution is in terms of cnoidal and snoidal functions) is the solution to $\ddot x + x - x^3 = 0.$ (Multiply the left side ... | First we have to express the radicand as a product of its roots:
The roots of the polynomial are
$$x_{1}= \pm \sqrt{1-\sqrt{2C+1}} \quad C>-\frac{1}{2}$$
$$x_{2}= \pm \sqrt{1+\sqrt{2C+1}} \quad C>-\frac{1}{2}$$
For simplicity denote $a=x_{1}^2, b=x_{2}^2$. Then
$$I= \int \frac1{\sqrt{C-x^2+\frac{x^4}2}}dx = \int \frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274467",
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$(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$ Problem: Let $a,b,c\ge0.$ Prove that: $$(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$$
This is strange form for me.
M... | We will prove the stronger inequality
$$
\sum (a^2 + b^2)\sqrt{a^2b^2 + c^2} \ge 2\sum(ab)^2 + |(a-b)(b-c)(c-a)|
$$
The inequality is equivalent to
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2} + 2(\sqrt{a^2b^2 + c^2} - ab)ab\right] \ge |(a-b)(b-c)(c-a)|
$$
However, this is true as
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2}\right... | {
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Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be writte... | A quick way of differentiation is to use logarithms on both sides of the equation:
$$
y = \left[ x + \sqrt{1 + x^2} \right]^{3 \over 2} \tag{1}
$$
We get
$$
\ln|y| = {3 \over 2} \ln\left| x + \sqrt{1 + x^2} \right| \tag{2}
$$
Differentiating both sides of (2) with respect to $x$, we get
$$
{1 \over y} \, {dy \over dx} ... | {
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Find closed formula for the exponential generating function of sequence counting permutations with 2 cycles. Find a closed formula for the exponential generating function of $\{c(n,2)\}_{n\geq2}$, that is the sequence counting permutations of length $n$ that consist of $2$ cycles.
My approach: The exponential generatin... | Your approach is correct, to finish notice that $\frac{1}{k(n-k)}=\frac{1}{n}\left (\frac{1}{k}+\frac{1}{n-k}\right )$ and adding over $k$ you get twice $H_{n-1}.$
For a combinatorial way, choose the elements that you want in the cycle containing one, say of size $k$ in $\binom{n-1}{k}$ ways. Notice that when you pick ... | {
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Write with one radical $\sqrt[4]{2^6}\cdot\sqrt{3^3}$
Write with one root (radical) $$\sqrt[4]{2^6}\cdot\sqrt{3^3}$$
In this lesson we have learnt that when the roots exist then $$\sqrt[n]{a}=\sqrt[nk]{a^k}$$
Using that here, we have $\gcd(4,2)=2,$ so
$$\sqrt[4]{2^6}\cdot\sqrt[2\cdot2]{\left(3^{3}\right)^2}=\sqrt[4]{... | The author of the solution notices that $6=2\cdot 3$, so $$\sqrt[4]{2^6}=\sqrt[4]{2^{2\cdot 3}}=\sqrt[4]{\left(2^3\right)^2}=\sqrt[4]{8^2}$$
By your provided equality, if we let $a=8, n=k=2$, then we have
$$\sqrt{8}=\sqrt[4]{8^2}$$
Now, $8=2^3$, so we can conclude
$$\sqrt[4]{2^6}\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}$$
Also... | {
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Series $\sum_{n=1}^\infty \frac{\cos n}{n^2}$ This is the series I want to find:
$$\sum_{n=1}^\infty \frac{\cos n}{n^2}$$
WolframAlpha gives the answer of $\frac14 + \frac16 (\pi - 3)\pi$.
I do not know anything about this problem other than to prove that it is convergent.
Thanks.
| Here's a solution using Fourier series.
For $a\in\mathbb C-\mathbb Z$, consider the $2\pi$-periodic function $f$ defined as $f_a(x)=\cos (ax)$ for $x\in [-\pi,\pi)$.
Let's expand it as a Fourier series (since it's periodic). The Fourier coefficients are
$$a_n = \int_{-\pi}^{\pi} \cos(ax)\cos(nx)dx = \frac{(-1)^{n+1}}{\... | {
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | $4x^2-2xy-4x+3y-3 = 4x^2+(-2y-4)x+3y-3=4x^2+[-2(y-1)-2(3)]x+3(y-1)=(2x-y+1)(2x-3).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6... | In general
$$ \int_{0}^{\infty} \frac{1}{1+x^s} dx = \frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) \quad \Re(s)>1$$
$$\int_{0}^{\infty} \frac{1}{1+x^s}dx = \frac{2}{s}\int_{0}^{\infty} \frac{w^{\frac{2}{s}-1}}{1+w^2}dw \quad (w^2 \mapsto x^s)$$
Recall the integral representation of $\displaystyle \sec z$:
$$ \sec z = \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^2}}$, starting from the integral representation of $\arctan$ Starting from the integral representation of the arctan function, $$\arctan x = \int_0^x \frac{dt}{1+t^2},$$how does one prove the (half-angle) identity? $$\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^... | If you differentiate$$2\arctan\left(\frac t{1+\sqrt{1+t^2}}\right),$$then you get $\frac1{1+t^2}$. So\begin{align}2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)&=2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)-2\arctan\left(\frac0{1+\sqrt{1+0^2}}\right)\\&=\int_0^x\frac{\mathrm dt}{1+t^2}\\&=\arctan(x).\end{align}
| {
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"source": "stackexchange",
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Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \u... | For example, if you are interested in solutions on intervals $I\subseteq (0, \infty)$ write
$$
\frac{d}{dx}(xy(x)) =x \sqrt{y(x)} = \sqrt{x}\sqrt{xy(x)}.
$$
Then substitute $z(x):=xy(x)$ and solve $z'(x)=\sqrt{x}\sqrt{z(x)}$ by separation of variables (initial values $>0$ are adequate here). Then $y(x)=z(x)/x$ is a sol... | {
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"question_score": "1",
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Show that $|f(x)| \leq 2 \pi^2/9$ The problem is stated as:
Show that $|f(x)| \leq 2 \pi^2/9$ for $ |x| \leq \pi/3$ where $f(x) = \int_{0}^{x}\frac{t}{cos(t)}dt$
Solution:
So, first of all, I try to form a Maclaurin expression around $x = 0$. Without digging into the calculations too deep, I get the following:
$f(0) ... | Here is a simpler way:
For $|x| \leq \frac{\pi}{3}, \cos x \geq \frac{1}{2}$ or $\frac{1}{\cos x} \leq 2$
So, $ \displaystyle |f(x)| = \left|\int_{0}^{x}\frac{t}{cos(t)} ~ dt \right| \leq \int_{0}^{|x|} 2t ~ dt = x^2$.
As $ \displaystyle |x| \leq \frac{\pi}{3}, |f(x)| \leq \frac{\pi^2}{9} \leq \frac{2\pi^2}{9}$
| {
"language": "en",
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and ... | Well I have another solution which is very helpful for exams based on mcq pattern, like the one we give it in india.
I have learnt and derived a property in conic sections which says that in a ellipse, the product of perpendiculars from both foci of ellipse to tangent is equal to b$^2$ ,b is semi minor axis.
equation o... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solve $|2x−3| < |x+5|$ I've tried solving $$|2x-3|=|x+5|$$ but the same method does not look applicable here.
$$2x-3 = x+5, \text{if}\; x>3/2$$
$$2x-3= -(x+5), \text{if} \; x<3/2$$
This method was not working here.
| There are two nodes $x=3/2$ and $x=-5$, so we have to consider the in-equation $|2x-3|< |x+5|$ in three regions three regions
I: $x\le -5 \implies 3-2x<-(x+5) \implies x>8$ (A contradiction).
II:$-5 <x\le 3/2 \implies 3-2x<x+5 \implies x>-2/3 \implies -2/3<x\le 3/2$
III: $x>3/2 \implies 2x-3 <x+5 \implies x<8 \implies ... | {
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Alternative proof of an intereting identity of Catalan's Numbers and central binomial coefficients Some time ago i got from Polya's Urn Scheme that for the n-th Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$ and the central binomial coefficient takes place the identity
$$\sum_{n = 0}^\infty\frac{C_{n+k}}{4^n} = 2\bi... | The Catalan numbers generating function
power series multiplied by $x$ is
$$ \frac{1-\sqrt{1-4x}}2 = \sum_{n=0}^\infty C_n x^{n+1}. \tag{1} $$
Substitute $\,y\,$ for $\,x\,$ and subtract both equations to get
$$ \frac{1-\sqrt{1-4y}}2- \frac{1-\sqrt{1-4x}}2 =
\sum_{n=0}^\infty C_n (y^{n+1}-x^{n+1}). \tag{2} $$
Divide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sign of $P(Δ)$ with $P(x)=ax²+bx+c$ and $Δ=b²-4ac$ Let $P(x)=ax^2+bx+c$ be a polynomial with $a≥0$ and $b≥\frac{1}{8a}$, and $Δ=b^2-4ac$. Show that $P(Δ)≥0$.
My answer: We have $Δ=b^2-4ac$. If$Δ≤0$, then $P(Δ)≥0$. If $Δ>0$, there are two cases:
If $c≥0$, $P(Δ)=aΔ^2+bΔ+c≥0$. If $c<0,Δ=b^2-4ac$ then $Δ>b^2$ and $Δ>-4ac$,... | Completing the square differently, this is non-negative when $\,a \gt 0\,$, $\,\Delta \ge 0\,$, $\,2b - \dfrac{1}{4a} \ge 0\,$:
$$
\begin{align}
\frac{1}{a} P(\Delta) &= \Delta^2+\frac{b}{a}\Delta \color{red}{-2 \frac{b}{a}\Delta + \frac{b^2}{4a^2} + 2 \frac{b}{a}\Delta - \frac{b^2}{4a^2}}+\frac{c}{a}
\\ &= \Delta^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)... | We can write given Polynomial in Quadratic equation in (y-1) as-
f(x,y)=$x^2−2xy+6y^2−12x+2y+41$
=$x^2−2xy+y^2+36−12x+12y+5y^2-10y+5$
=${(x-y)}^2$ + ${6}^2$ -2.6.(x-y) + 5${(y-1)}^2$
=${(x-y-6)}^2$ + 5${(y-1)}^2$
=${[(x-7)-(y-1)]}^2$ + 5${(y-1)}^2$
=6${(y-1)}^2$ - 2(x-7)(y-1) + ${(x-7)}^2$
which is Quadratic equation i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Right... | HINT.-Equalizing to $t$ each of the expressions in first line we have
$$3(x^2+1)=tx\\4(y^2+1)=ty\\5(z^2+1)=tz$$ we deduce
$$x=\frac{t\pm\sqrt{t^2-36}}{6}\\y=\frac{t\pm\sqrt{t^2-64}}{8}\\z=\frac{t\pm\sqrt{t^2-100}}{10}$$Now taking into account the second line of the proposed system we have the solution given by Wolfram:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to determine the positive and negative nature of $a\sec\theta-b\tan\theta$?
Given $a,b\in\mathbb R^+$ and $a\gt b$, find the minimum value of $a\sec\theta-b\tan\theta$.
My approach:
Let $$x=a\sec\theta-b\tan\theta\\\implies x+b\tan\theta=a\sec\theta\\\implies x^2+b^2\tan^2\theta+2bx\tan\theta=a^2(1+\tan^2\theta)\... | Your working shows you that the range of the function is $\mathbb{R}$ excluding the set $\{x:-\sqrt{a^2-b^2}<x<\sqrt{a^2-b^2}\}$
The local minimum is $\sqrt{a^2-b^2}$ and the local maximum is $-\sqrt{a^2-b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can i evaluate this integral with or without using CAS $\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$ $$\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$$
Firstly i used substitution $\pi x=t; \mathrm dx=\frac{\mathrm dt}{\pi}$
$$\pi \int_0^{\infty}\frac{\mathrm dt}{(t^2+\pi^2)\cosh t}$$
writing $\cos... | Recall the following two results:
$$\int_0^\infty e^{-bt}\cos\left(xt \right)\,dt=\frac{b}{b^2+x^2}\tag{1}$$
$$\int_0^\infty \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx=\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \tag{2}$$
$$
\begin{aligned}
\int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4321734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing there exists a unique $\theta$ for 2D rotation matrix I'm trying to solve some exercises in the book I'm reading. Take a look at this question:
Suppose A is a $2\times2$ rotation matrix. In other words $A^T A = I$
and $\det A = 1$. Show that there exists a unique $\theta$ such that
$A$ is of the form $$ A = \b... | The question says, or rather, should say:
Suppose we are given a matrix $A$ with the properties: $A^T A = I$ and $\det A = 1$.
Show that there exists a unique $\theta\in [0,2\pi)$ such that $A$ is
of the form $$ A = \begin{bmatrix} \cos\theta & -\sin\theta \\
\sin\theta &\cos\theta\end{bmatrix}. $$
$$$$
Let $$A = \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating the volume of the solid defined by $x^2+y^2+z^2 \leq 2a^2$ and $z \leq \frac{x^2+y^2}{a}$, with $a>0$
I have to calculate the volume of solid geometry for $a>0$
$$T=\{(x,y,z)\in \mathbb{R}; x^2+y^2+z^2 \leq 2a^2; z \leq \frac{x^2+y^2}{a}\}$$
I know that first formula is inside of sphere with radium $\sqrt... | Using cylindrical coordiantes, you have two surfaces:
Surface 1: $r^2 + z^2 = 2 a^2$
Surface 2: $ z = r^2 / a $
We want to find where these two surfaces intersect. Plugging the second equation into the first,
$ r^2 + r^4 / a^2 = 2 a^2 $
whose solution is by inspection, $r = a$
So, now if $ r \le a $ then $ r^2 / a \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left... | It is because when you expand $\cos(x)$ and $\ln(\cos(x))$, you need to consider more fourth-order term. Specifically,
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+o(x^5).$$
Then
$$\ln\left(\cos(x)\right)=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{\left(-\frac{x^2}{2}+\frac{x^4}{24}\right)^2}{2}+o(x^5)=-\frac{x^2}{2}-\frac{x^4}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Minimizing $(x+y)(z+t)(\frac{a}{x} + \frac{c}{z})(\frac{b}{y} + \frac{d}{t}).$ Let $a,b,c,d$ be positive constants and $x,y,z,t$ be positive variables. Find the minimum value of
\begin{equation}
(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right).
\end{equation}
I came up with this ni... | Using algebra
$$F=(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right)$$ Compute the partial derivatives
$$\frac{\partial F}{\partial x}=\frac{(t+z) (b t+d y) }{t x^2 y z}\color{red}{\left(c x^2-a y z\right)}=0$$
$$\frac{\partial F}{\partial y}=\frac{(t+z) (a z+c x)}{t x y^2 z} \color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving continuity of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ Using only the epsilon delta criterion, show that the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ is continuous over its entire domain of definition.
Definition of the epsi... | We have
\begin{align}
|x^3(1+y^2)- y^3(1+x^2)| & = |(x-y)(x^2+x y + y^2 + x^2 y^2)|\\
&\le |x-y|(x^2 + \frac{1}{2}(x^2+y^2) + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(x^2 + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(1 + x^2)(1+y^2)\\
\end{align}
Hence
\begin{equation}
\frac{|x^3(1+y^2)- y^3(1+x^2)|}{(1 + x^2)(1+y^2)}
\le \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When is the projection of an ellipsoid a circle? Consider an ellipsoid in the three dimensional Euclidean space, say
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
where $a$, $b$, $c$ are positive reals.
I'm counting the number of planes through the origin so that the image is a perfect circle. There may be ... | For convenience, we use $$ax^2+by^2+cz^2=1.$$
Consider the parametric lines of constant direction $(u,v,w)$ through some point $(x,y,z)$ on the ellipsoid.
$$a(x+tu)^2+b(y+tv)^2+c(z+tw)^2=1$$
or, after simplification,
$$(au^2+bv^2+cw^2)t^2+2(aux+bvy+cwz)t=0.$$
We will obtain the silhouette line on the ellipsoid by expre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
| Simpler:
$(y+1)dx + (x+1)dy = 0$
$ydx + xdy = -(dx+dy)$
$d(xy) = d(-(x+y))$
$xy = -(x+y) + c$
$xy + x + y = c$, which is equivalent to your given answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes). Source : Problem $4.2.$ , $n° 4 $
http://www.gycham.vd.ch/favre/2M/asymptotes/Asymptotes.pdf
The question I'm trying to solve is " determine, in case it exists, the equation of the hor... |
what is supposed to be the power of the denominator, in case the denominator is a polynomial placed under a radical sign?
Informally, we can think of this by reasoning "if $x$ is very large, then $9$ is insignificant compared to $x^2$, so $x^2+9$ acts like $x^2$ and $\sqrt{x^2+9}$ acts like $\sqrt{x^2}=|x|$. This doe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{... |
We can write the series as
\begin{align*}
\color{blue}{\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)}
&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{2n}{n}\left(\frac{1}{4}\right)^n\tag{1}\\
&=\frac{1}{2}\sum_{n=1}^{\infty}C_n\left(\frac{1}{4}\right)^n\tag{2}\\
&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator? The expression:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
Way 1:
If I punch the above expression in my calculator, I get $\frac{24}{25}$.
Way 2:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\si... | Notice that your formula to simplify the arctan gives a positive answer on the left hand side
$$2\arctan \frac{4}{3} > 0$$
but a negative on the right
$$\arctan \frac{\frac{8}{3}}{1-\left(\frac{4}{3}\right)^2} < 0$$
The formula itself is the mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the point in a tetrahedron that minimizes the sum of distances between vertices of a tetrahedron
Tetrahedron $ABCD$ has $AD=BC, AC=BD,$ and $AB=CD$. Find the point $X$ that minimizes the sum of the distances to the four vertices (i.e. it minimizes $f(X) = AX+BX+CX+DX$).
I found the following proof online, but I ... | Hint: the desired point $X$ is required to have the following property - from the point $X$ each of the faces $ABC, ABD, ACD, BCD$ of the tetrahedron should be viewed by the same solid angle (which is equal to $\pi$). This is equivalent to the following property of the required $X$: if one draws unit vectors from $X$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second functi... | Note that
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1+a}\right)=\frac{a\ln\left(\frac{a}{1+a}\right)+a\ln(1+x)}{1-ax},$$
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1-a}\right)=\frac{a\ln\left(\frac{a}{a-1}\right)+a\ln(1-x)}{1-ax}.$$
On subtracting, we have
$$\frac{d}{dx}\left(\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Solve the equation over the integers $2^a b + 2^b a = a^2+b^2$
Clearly all possible solutions are
$(0,0), (0,1), (1,0)$
What is a proper method of solving it?
Thank you.
EDIT:
Following Dietrich Burde advice, I will make an attempt:
For every $a, b \geq 0, 2^a>a$
Without loss of generality and since the equation is tot... | Following an observation by Dietrich Burde in the comments, we will prove the following:
Lemma: For all integers $a, b \geq 2,$ we have
$$2^{a} \cdot b + 2^{b} \cdot a > a^{2} + b^{2}.$$
Proof: We say a pair of integers $(a, b)$ is good if $a, b \geq 2$ and
$$2^{a} \cdot b + 2^{b} \cdot a > a^{2} + b^{2}.$$
Claim: if $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus. My thinking:
By Arithmetic and Geometric mean inequality (AGM):
$ab\le \left(\frac{a+b}{2}\right)^2$
We know $ab=2$
$\rightarrow$ $2\le \left(\frac{a+b}{2}\right)^2$
$\righ... | Substitute the given restriction into the proposed expression.
Then your problem reduces to study a single-variable real-valued function:
\begin{align*}
\frac{3}{2(a + b)^{2}} & = \frac{3}{2\left(a + \dfrac{2}{a}\right)^{2}}\\\\
& = \frac{3a^{2}}{2(a^{2} + 2)^{2}}\\\\
& = \frac{3}{2}\left(\frac{a}{a^{2} + 2}\right)^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Consider the non-homogeneous linear recurrence relations $a_n = 2a_{n−1} + 2^n$ find all solutions. I can show that $a_n^{(h)}$characteristic equation $r-2=0 \to a_n^{(h)}=\alpha2^n$
But I'm stuck on $a_n^{(p)}$ characteristic equation $A2^n = 2A2^{n-1} + 2^n$
Simplifies to $-A = 2$
$A = 2A + 2$
| Let $A(z)=\sum_{n=0}^\infty a_n z^n$ be the ordinary generating function. Then
\begin{align}
A(z)
&= a_0 + \sum_{n=1}^\infty (a_{n-1} + 2^n)z^n \\
&= a_0 + z \sum_{n=1}^\infty a_{n-1} z^{n-1} + \sum_{n=1}^\infty (2z)^n \\
&= a_0 + z A(z) + \frac{2z}{1-2z},
\end{align}
so
\begin{align}
A(z)
&= \frac{a_0}{1-z} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does the theorem 2 of algebra of limits contradict in this Q? Q: $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
Solution in my textbook using theorem 2 of algebra of limits I.e $\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}$ .
$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} &=\li... | Let,
$$f(x) = \frac {x^{15}-1}{x^{10}-1}$$
Well, I agree with you! that the value of function $f(x = 1)$ is $ \frac {1^{15}-1}{1^{10}-1}$
But,
The value of limit of function $f(x)$ as $x \to 1$ is $\frac 32$
*
*Limit of function: This means you take the value of the function as $x\to a$ doesn't mean $x = a $ it means... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$ I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$.
My work so far
By partial fraction decomposition we can get that:
$$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(z - 4)}{(z + 1)^2} + \frac{\frac 1 9 (z - 4)}{z - 2}$$
Now deriving L... | Hint:
*
*Compute the Laurent expansion of $\frac1{1+z}$
*Notice that $\frac{d}{dz}\frac{1}{1+z}=-\frac{1}{(1+z)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trouble integrating square of a sine I have the following expression which I am trying to evaluate:
$$ \frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx$$ where $k = \frac{\pi}{L}$.
According to my calculator, the answer should be $\frac{1}{6}$ but I can't seem to get this result by manual integration.
Here's what... | There is a trick. Integrate by parts
$$
\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx)~\mathrm{d}x = -\frac{1}{k}\cos(kx)\sin(kx) \bigg \vert^{\frac{L}{3}}_{\frac{L}{6}} + \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x = \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x
$$
So:
$$
\left(\frac{L}{3}-\frac{L}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Singapore Math 6th grade probability problem Karen has 12 cards, each written with a letter from the word PENNSYLVANIA. She picks a card at random. Without replacing she picks another card at random. Find the probability she picks two vowels.
I was sure the answer is $\frac{5}{12}\times\frac{4}{11}$ :-)
| With Y as vowel:
There are 5 vowels, picking a vowel first try gives you $\frac{5}{12}$ and then we have 11 letters and 4 vowels left, giving you $\frac{4}{11}$. Therefore, the answer would be $\frac{5}{12}\times\frac{4}{11}$, or $\frac{5}{33}$.
Without Y as vowel:
There are 4 vowels, picking a vowel first try giv... | {
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"source": "stackexchange",
"question_score": "1",
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Finding $\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{\pi^2+\ln^2\left(\frac{x-1}{2}\right)}\text{d}x$ Prove the integral
$$\int_{1}^{\infty} \frac{1}{1+x^2}
\frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{
\pi^2+\ln^2\left ( \frac{x-1}{2} \right ) }\text{d}x
=... | $$I=2\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x$$
Consider
$$f\left( z \right) =\frac{\mathrm{Li}_2\left( z \right)}{1+\left( 2z-1 \right) ^2}\frac{1}{\ln z}$$
Use a key-shaped contour with $0$ and $1$ as the keyholes:
contour
When $x<0$, we have
... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
| HINT
Start with rearranging the proposed expression as follows:
\begin{align*}
\sqrt{(n + a)(n + b)} - n & = \frac{(n + a)(n + b) - n^{2}}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b)n + ab}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b) + ab/n}{\sqrt{(1 + a/n)(1 + b/n)} + 1}
\end{align*}
Now it remains to take t... | {
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"question_score": "1",
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How many four-digit natural numbers not exceeding the number $4321$ can be formed using the digits $1, 2, 3, 4$ if repetition is allowed?
How many four-digit natural numbers not exceeding the number $4321$ can be formed using the digits $1, 2, 3, 4$ if repetition is allowed?
This is the question and I am solving it l... | There are actually $27$ numbers larger than $4321$. They are:
Numbers of the form $44\square\square$: There are $1 \cdot 1 \cdot 4 \cdot 4 = 16$ such numbers.
Numbers of the form $433\square$ or $434\square$: There are $1 \cdot 1 \cdot 2 \cdot 4 = 8$ such numbers.
Numbers of the form $432\square$: There are $3$ suc... | {
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and... | Without using the formula $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),$ I thought of the Gamma Function.
$$ \begin{aligned}
\text { Let } t &=\frac{1}{x^{n}+1}, \text { then } x=\left(\frac{1}{t}-1\right)^{\frac{1}{n}} \\
d x &=\frac{1}{n}\left(\frac{1}{t}-1\right)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Basic rotations in $4-D$ Basic rotations in $3-D$ are represented by the matrices:
$$
R^3_{xy}(\theta)=
\begin{bmatrix}
\cos \theta &-\sin \theta & 0\\
\sin \theta &\cos \theta & 0\\
0& 0 & 1\\
\end{bmatrix}\quad \text{rotation in the plane}\; (xy)
$$
$$
R^3_{xz}(\theta)=
\begin{bmatrix}
\cos \theta&0 &\sin \theta\\
0&... | In $4D$ there are ${4\choose 2}=6$ basic rotation matrices, each one characterized by the two out of four axes that they keep fixed while rotating vectors in the plane perpendicular to those axes. Alternatively, we can characterize them by the two out of four basis vectors they rotate. We know from $3D$ that the right-... | {
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"question_score": "2",
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Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positiv... | There are various variants of such equations. For example like this.
https://artofproblemsolving.com/community/c3046h2465588_a_class_of_diophantine_equations_in_three_variables
When solving such equations.
$$\frac{ 1 }{ a^2 } -\frac{ 1 }{ b^2 } = \frac{ 1 }{ c^2 } - \frac{ 1 }{ d^2 } $$
Pythagorean triples can help us.... | {
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"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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Find $T^*\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ in an inner product space $V=\mathbb{R^3}$
$f$ is an inner product ,$f\Bigg (\begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} ,\begin{pmatrix}
y_{1} \\
y_{2} \\
y_{3}
\end{pmatrix... | If it is the word "orthonormal" that is bugging you then you can do it by taking a basis, say $v_{1},v_{2},v_{3}$. Where you can pick them to be any basis you like. Say $(0,1,1),(1,0,1),(1,1,0)$. Or heck, you might even take the standard basis for $\mathbb{R}^{3}$ , $e_{1},e_{2},e_{3}$. They are not orthonormal wrt the... | {
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"source": "stackexchange",
"question_score": "2",
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Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$? Given $ \varphi$ is golden ratio, how do I prove this:
$ \displaystyle \tag*{}\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$
... | The point is the functional equation $$Li_2(1-z)+Li_2(1-1/z)=-\frac12 \log^2 z$$
of the dilogarithm $$Li_2(z)=\sum_{k\ge 1} \frac{z^k}{k^2}$$ which follows from $z \, Li_2'(z)=-\log(1-z)$,
and that $$1-z= (1-\varphi)e^{3i\pi/5} \implies z=e^{i\pi/5}$$
So $$\sum_{k\ge 1} \frac{(1-\varphi)^k}{k^2}\cos{\frac{3k\pi}{5}}=\f... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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For a natural number $b$, let $N(b)$ denote the number of natural numbers $a$ for which the equation $x^2+ax+b=0$ has integer roots.
For a natural number $b$, let $N(b)$ denote the number of natural numbers $a$ for which the equation $x^2+ax+b=0$ has integer roots. What is the smallest value of $b$ for which $N(b)=20$... | First, observe that, for all $a, b \in \mathbb{Z}$, the polynomial $X^{2} +a X +b \in \mathbb{Z}[X]$ has integer roots if and only its discriminant $\Delta = a^{2} -4 b$ is the square of an integer:
*
*if there exists $\delta \in \mathbb{Z}$ such that $\Delta = \delta^{2}$, then $\delta$ and $a$ have the same parity ... | {
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"source": "stackexchange",
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Convert $f(x) = (\cos x+\sin x)(1-\sqrt{3}\tan x)$ into the form of $k(1+\tan x)\cos(x+a).$ I need your help to convert the following into the form of $k(1+\tan(x))\cos(x+a).$
$$f(x) = (\cos(x)+\sin(x))(1-\sqrt{3}\tan x)$$
Here is what I've done so far.
\begin{aligned}
f(x) &= (\cos(x)+\sin(x))(1-\sqrt{3}\tan x)\\
f(x)... | We have to find $k$ and $a$ such that
$k(1+\tan x)\cos(x+a)=(\cos x+\sin x)(1-\sqrt3\tan x)$
or
$k(\cos x+\sin x)\dfrac{\cos(x+a)}{\cos x}=(\cos x+\sin x)(1-\sqrt3\tan x)$
or
$k(\cos a-\sin a\tan x)=(1-\sqrt3\tan x).$
Now the similarity is obvious. Can you take it from here?
| {
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"answer_id": 0
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Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 ... | Let $\,\omega\,$ be a primitive $\,(2n+1)^{th}\,$ root of unity, so $\,\omega^{2n+1}=1\,$ and $\,\omega^{2n+1-k} = \dfrac{1}{\omega^k}=\overline{\omega^k}\,$. Then:
$$
P(x)=\sum_{k=0}^{2n} x^k = \prod_{k=1}^{2n}(x-\omega^k) = \prod_{k=1}^n\big((x-\omega^k)(x-\overline{\omega^k})\big) = \underbrace{\prod_{k=1}^n(x-\omeg... | {
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"source": "stackexchange",
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Finding $\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx$ Let $f$ be a analytic function in the closed unit circle with its center
at the point $\alpha\in\mathbb{R}$, then:
\begin{equation*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+... | Integrate as follows
\begin{align}
&\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos x+p^2} dx\\
=&\int_0^\pi\frac{2\Re f\left(\alpha+e^{ix}\right)}{1+2p\cos x+p^2} dx
= 2\int_0^\pi\frac{\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} \cos(kx)}{1+2p\cos x+p^2} dx\\
=& \> 2\sum_{k=0}^\infty \fr... | {
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"answer_count": 2,
"answer_id": 0
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How to prove $2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$
Prove $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
I got this question when I was going through some basic trigonometric identities as follows
$2(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}) =1\tag1$
very much straightforward to recognise
$2(\cos\... | Let $z=e^{\frac{\pi i}{2 n+1}}$ , then $$
\begin{aligned}
\sum_{k=1}^{n}(-1)^{k-1} z^{k} &=\frac{z\left[1-(-z)^{n}\right]}{1+z} \\
&=\frac{e^{\frac{\pi i}{2 n+1}}\left(1+(-1)^{n+1} e^{\frac{n \pi i}{2 n+1}}\right)}{1+e^{\frac{\pi i}{2 n+1}}}
\end{aligned}
$$
Multiplying both the numerator and denominator by $e^{-\frac{... | {
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"source": "stackexchange",
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Probability of winning a game by rolling the die first Two persons are playing a game where they take turns rolling a die (so A rolls first, then B, then A again and so on). The first person to roll a $6$ wins the game. What is the probability that the person who started the game (rolled the die first) wins?
This was t... | The probability of winning can be seen as a sum $P = P_1+P_2+P_3+\dots$. Here $P_1$ is the probability of winning at the first turn. Since you are the first to throw the dice, we simply have
$$P_1=\frac{1}{6}\,.$$
Then $P_2$ is the probability of winning at the second term. For this you need to have rolled something wh... | {
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How do you prove: $\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})} = 8\sin(\frac{\pi}{18})$ I have to prove:
$\displaystyle \tag*{} \alpha={\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})}}= 8\sin(\frac{\pi}{18})$
I tried to make a common denominator of $\alpha$ and us... | Good question. For me, I am more customized to write in degrees. Therefore, we can have
$${\frac{1}{\sin ^2 (40^\circ)} - \frac{1}{\sin ^2(80^\circ)}}= 8\sin(10^\circ)$$
Or we can write it in a different way
$${\frac{\sin ^2(80^\circ)}{\sin ^2 (40^\circ)}} - 1= 8\sin(10^\circ)\sin ^2(80^\circ)$$
Notice that $$\sin 20^\... | {
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"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{n\to\infty} \sqrt{n}\int_{0}^{\pi/2} \sin^{n} x dx$ Question:
Compute $I_n=\lim\limits_{n\to\infty} \sqrt{n}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx$.
Attempt:
We know the famous result that $\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=0$,
and by dividing it into two parts $\int_{0}^{\f... | Proceeding along your approach, without using the gamma function
Let
\begin{align*}
J_n &:= \sqrt{n}\int_{\pi/2 - n^{-1/3}}^{\pi/2} \sin^n x \,\mathrm{d} x, \\
K_n &:= \sqrt{n}\int_0^{\pi/2 - n^{-1/3}} \sin^n x \,\mathrm{d} x.
\end{align*}
We have
$$J_n
= \sqrt{n} \int_0^{n^{-1/3}} \cos^n y \,\mathrm{d} y
= \int_0^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392229",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Why can't I use trig substitution for this integral? $
\int \frac x {\sqrt {1-x^2}}dx
$
I was attempting to solve this integral, and it would appear the solution to it is $-\sqrt{1 -x^2}+C$. When I attempted to solve it, however, I attempted to let $x = \sin\theta$, making $dx=\cos{\theta}d{\theta}$
$
\int \frac {\sin\... | Method 1
Let $x=\sin \theta$, where $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}, $ then
\begin{aligned}
I &=\int \frac{\cos \theta d \theta}{\sqrt{1-\sin ^{2} \theta}} \\
&=\int \frac{\sin \theta \cos \theta d \theta}{\sqrt{\cos ^{2} \theta}}
\end{aligned}
$\text {Since } \cos \theta \geqslant 0 \text { f... | {
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Does there exists another approach to solve for the product of such expression? Problem: if the real roots of $x^3-3x+1$ are $\alpha , \beta $ and $\gamma,$ then what is the value of cyclic $(\alpha^2-\gamma)\;?$
Here is my approach, using trigonometry. Is my work correct?
| The problem can be brute-forced algebraically, but the calculations involve polynomial resultants and are laborious to do by hand, though easily computed using a CAS.
Let $\,x_1=\alpha, x_2=\beta, x_3=\gamma\,$, then the polynomial in $\,y\,$ with roots $\,x_i^2\,$ is:
$$\text{res}(x^3-3x+1, y-x^2, x)=y^3 - 6 y^2 + 9 y... | {
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$f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$.
Let $f(x)$ and $g(x)$ be two monic cubic polynomials, and let $r$ be a real number. Two of the roots of $f(x)$ are $r+1$ and $r+7$. Two of the roots of $g(x)$ are $r + ... | Your approach is fine. Note that\begin{multline}f(x)-g(x)=\\=(4-p+q)x^2+(2 r p+8p-12 q-2 q r-4 r-20)x-p r^2+q r^2-8 p r+12 q r-7 p+27 q.\end{multline}So, $4-p+q=0$; in other words, $p=q+4$. Replacing $p$ with $q+4$ in the coefficient of $x$ in $f(x)-g(x)$, we get that $4(3-q+r)=0$; in other words, $q=r+3$. And if repla... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$ Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$,
$$
I=\int\frac{1}{1+\sin ^{6} x} d x.
$$
As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get
$$
\begin{aligned}
I... | Answering by mistake the linked post instead of your, my shortest result for $n=4$ is
$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan
^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$
For $n=6$, doing the same
$$\frac {1}{1+\sin^6(x)}=\frac{2}{(a+1) (b+1... | {
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"source": "stackexchange",
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Factorial Number System with Repetitions Good Day
I know about how Factorial Number System is useful to find permutations and also find the the index of a permutation.
For example, permutation $[1, 3, 4, 0, 2]$ has index ($0$ - based) $1 \cdot 4! + 2 \cdot 3! + 2 \cdot 2! + 0 \cdot 1! + 0 \cdot 1! = 40$ where the const... | The index of a binary string can be found using the combinatorial number system.
For example, $N=100=1100100_2$ can be considered as an element of the k-combination set $\binom{7}{3}$.
As such it can be written $\{6,5,2\}$, where each element is a set bit in $N$. We then have $100=2^6+2^5+2^2$.
Its (zero-based) index i... | {
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Finding the maximum of $ \sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}$ for $0\le x \le 13$ using the Cauchy-Schwarz inequality gives two different answers I am trying to find the maximum of the expression $$\sqrt{x+27} + \sqrt{13-x} + \sqrt{x} \qquad \text{for } 0 \le x \le 13.$$
Clearly using Cauchy is the way to go here,... | Your approach is fine. You start by writing $$(\sqrt{x+27}+\sqrt{13-x}+\sqrt{x})^2\le[(x+27)+a(13-x)+bx]\left(1+\frac1a+\frac1b\right)$$
The factors $a$ and $b$ are to be determined. Since you want the right hand side to be an absolute maximum, that does not depend on $x$, you have
$$1-a+b=0$$
You also know that the Ca... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does $\frac{k^2(k+1)^2}{4} + (k+1)^3$ become $\frac{(k+1)^2(k+2)^2}{4}$? As part of an induction proof, the authors of a beginner combinatorics text reduce/factor a polynomial as follows, but do not show the minutiae of their algebraic steps:
$$\frac{k^2(k+1)^2}{4} + (k+1)^3 \tag1$$
manipulated to:
$$(k+1)^2\left(\... | For (1), we can separate the $\frac{k^2}{4}$ from $(k+1)^2$ and move out a factor of $k+1$ from $(k+1)^3$, giving us
$$\frac{k^2}{4}(k+1)^2+(k+1)(k+1)^2$$
We can then collect like terms (specifically, the $(k+1)^2$ terms), giving us
$$(k+1)^2\left(\frac{k^2}{4}+k+1\right)$$
Which is (2).
From there, we can divide and m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m... | Let $E$ middle of $BC$. Then triangles DEC, AFC and BAC are similar. Then $$\frac{CD}{EC}=\frac{AC}{FC}=\frac{BC}{AC}=k$$
$$EC=\frac{CD}{k}=\frac{1}{k}, AC=k FC=k, BC=k AC=k^2$$
$$BC=2 EC \Rightarrow k^2 = \frac{2}{k} \Rightarrow k=\sqrt[3]{2}$$
$$AC=k=\sqrt[3]{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding second derivative of $x^{x^x}$ If $f(x) = x^{x^x} $, then find $f''(1)$.
My attempt:Now,
$\begin{align}y =x^{x^x} \\&\implies ln(y) = x^x \cdot ln(x)
\\&\implies ln(ln(y))= x\cdot ln(x)+ln(ln(x))\\&\implies\frac{y'}{y\cdot ln(y)}
=ln(x) +1+\frac{1}{x\cdot ln(x)}\\&\implies y'=(x^{x^x})\cdot (x^xln(x))\cdot [ln... | Proceeding along your approach
Let $y = x^{x^x}$. We have $y(1) = 1$.
We have
$$y' = y x^x (\ln^2 x + \ln x + x^{-1}).$$
We have $y'(1) = 1$.
We have
\begin{align*}
y'' &= y' x^x (\ln^2 x + \ln x + x^{-1}) + y \cdot x^x(\ln x + 1) \cdot (\ln^2 x + \ln x + x^{-1})\\
&\qquad + y x^x (2x^{-1}\ln x + x^{-1} - x^{-2} ).
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Proving that $-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$ Show $-\ln(x-\sqrt{x^2-1})=ln(x+\sqrt{x^2-1})$
Can somebody verify that my solution is correct (or incorrect!) and also show me alternative methods to do this? Thanks!
Let $r=-\ln(x-\sqrt{x^2-1})$ and $s=ln(x+\sqrt{x^2-1})$
$\rightarrow e^r=(x-\sqrt{x^2-1})^{-1}$ ... | Your proof looks fine to me. You can also do this: $\ln(x-\sqrt{x^2-1})+\ln(x+\sqrt{x^2-1})=\ln\left((x-\sqrt{x^2-1})(x+\sqrt{x^2-1})\right)=\ln 1 =0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
... | It may qualify as trivial, but note that there are infinitely many such series.
Consider a divergent series with positive decreasing terms $\sum_{n=1}^\infty u_n$, such that $\sum_{n=1}^\infty u_n^2$ is convergent to $L$, then it's a known theorem that we can change the signs of $u_n$ so as to get a series convergent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. A... | $$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\Re \,2^n\int_0^\frac{\pi}{4}\frac{e^{i(n-2)x}}{(e^{ix}+e^{-ix})^n}dx=\Re \,2^{n-1}\int_0^\frac{\pi}{2}\frac{e^{-it}}{(1+e^{-it})^n}dt$$
$$=\Re \,2^{n-1}(-i)\int_{-\frac{\pi}{2}}^0\frac{e^{it}}{(1+e^{it})^n}\,i\,dt$$
Consider the contour in the comples plane: from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\g... | Hint: The induction step is just $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})\leq\frac{2}{(k+1)^2}$, which is equivalent to
$$
\frac{1}{k^2}\frac{k}{k+1}\leq\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\iff\frac{k+1}{k}=1+\frac{1}{k}\leq 1+\frac{1}{\sqrt{k+1}}
$$
which boils down to $k\geq\sqrt{k+1}$, for integers $k\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Sum of series $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$ I first derived the Fourier series for $f(x) = abs(x)$ to be:
$\frac{\pi}{2} - \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{cos((2k-1)x)}{(2k-1)^2}$
Now, I want to use this at x=0 to find $\sum_{k=1}^{\infty} \frac{1}{k^2}$. This is my work:
$abs(x) = 0 = \fr... | Let $A=\sum_{k=1}^{\infty}{\frac{1}{k^2}}$. Then we can see that $$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots=\sum_{k=1}^{\infty}{\frac{1}{(2k)^2}}=\frac{1}{4}\sum_{k=1}^{\infty}{\frac{1}{k^2}}=\frac{A}{4}.$$ It then follows that $$\frac{\pi^2}{8}=\sum_{k=1}^{\infty}{\frac{1}{(2k-1)^2}}=\sum_{k=1}^{\infty}{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ Find $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ where $C$ is the positively oriented circle $|z| =4$
How should I go about finding this? We know that we have simple poles at the seventh roots of unity, i.e. $z = e^{\frac{n \pi i}{7}}$ and we also note that these poles are interi... | It is easier to compute this integral using one resudue at infinity, due all other singularities are in interior of $C$.
$\int\limits_{|z|=4} \frac{8z^{11}-3z^6+1}{z^7-1} dz = -2 \pi i \mathop{\mathrm{Res}}_{z=\infty} \frac{8z^{11}-3z^6+1}{z^7-1} = (1)$
Using $\mathop{\mathrm{Res}}_{z=\infty} f(z) = -\mathop{\mathrm{Re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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