Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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I'm looking for several ways to prove that $\int_{0}^{\infty }\sin(x)x^mdx=\cos(\frac{\pi m}{2})\Gamma (m+1)$ I'm looking for several ways to prove that $$\int_{0}^{\infty }\sin(x)x^mdx=\cos\left(\frac{\pi m}{2}\right)\Gamma (m+1)$$
for $-2< Re(m)< 0$
| Here is a fake proof: Let $x = \sqrt{t}$. Then
$$ \int_{0}^{\infty} x^{m} \sin x \, dx = \frac{1}{2} \int_{0}^{\infty} t^{m/2} \frac{\sin \sqrt{t}}{\sqrt{t}} \, dt. $$
Since
$$ \frac{\sin \sqrt{t}}{\sqrt{t}} = \sum_{n=0}^{\infty} \frac{\phi(n)}{n!} (-t)^{n} \quad \text{for} \quad \phi(n) = \frac{n!}{(2n+1)!} = \frac{\G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Frullani version of the classic $\log \left( 1 + 2\alpha \cos px + \alpha^2\right)$ integral. I read in a paper about Frullani integrals the following claim
$$
\begin{align*}
I & :=\int_0^\infty \frac{1}{x}\log\left(\frac{1 + 2\alpha \cos px + \alpha^2}{1 + 2\alpha \cos qx + \alpha^2}\right)\,\mathrm{d}x
=
\lef... | Untilize the Frullani-like integral
$\int_0^\infty \frac{\cos px -\cos qx}xdx
=\ln\frac qp$
to obtain
\begin{align}
I_{a^2<1}= &\int_{0}^{\infty} \frac{1}{x} \, \ln \frac{1+2a\cos px+a^{2}}{1+2 a\cos qx + a^{2}} \ dx\\
=&\> 2 \int_{0}^{\infty}
\sum_{k=1}^\infty \frac{(-a)^{k+1}}{k}\frac{\cos(k px)-\cos(k qx)}x \ dx\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Residue Computation How can we compute the residue of the function $\dfrac{(1+zx)^{k+1/3}}{z^2(z-a)} $ at $z=0$?? where $x$ is some fixed real number and $k$ is some fixed positive integer? I have tried computing laurent series of each but it gets too complicated to do.
| The residue of $\frac{f(z)}{z^2}$ at $z = 0$, where $f$ is holomorphic near $0$ is $f'(0)$, or, which sometimes is easier to find, the coefficient of $z$ in the Taylor expansion of $f$. Here we have
$$f(z) = \frac{(1+zx)^c}{(z-a)},$$
which is not totally trivial to differentiate (but not too hard either). We can develo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For what $x\in[0,2\pi]$ is $\sin x < \cos 2x$ What's the set of all solutions to the inequality $\sin x < \cos 2x$ for $x \in [0, 2\pi]$? I know the answer is $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi]$, but I'm not quite sure how to get there.
This is what I have so far: $\\
... | You reached the right inequality. A slightly different way of finishing goes as follows.
We are interested in where $(2\sin x-1)(\sin x+1)\lt 0$.
We find the places where $(2\sin x-1)(\sin x+1)=0$. This happens at $x=\frac{\pi}{6}$, $x=\frac{5\pi}{6}$, and $x=\frac{3\pi}{2}$.
These $3$ points divide the interval $[0,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.
Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer.
I've tried proving this by induction, but I've not succeed so far. What I did was:
Let'... | You want to try to write this expression in terms of the previous one. So, start by expanding out the terms:
$$
\begin{align*}
\frac{1}{5}(n+1)^5&=\frac{1}{5}\sum_{k=0}^{5}\binom{5}{k}n^k\\
&=\frac{1}{5}(n^5+5n^4+10n^3+10n^2+5n+1)\\
&=\frac{1}{5}n^5+n^4+2n^3+2n^2+n+\frac{1}{5}
\end{align*}
$$
Do the same thing for $(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Successive Differentiation and division I am working on successive differentiation. I have ran into some confusion and would like some help with the process of differentiating when dealing with division. Here is the problem that sparked my intentions to post here: $$y = \frac{x^2 + a}{x + a}$$ I am reading Calculus Mad... | $$ \dfrac{dy}{dx} = \frac{(x + a)(2x) - (x^2 + a)(1)}{(x + a)^2} =\dfrac{2x^2 + 2ax - x^2 - a}{(x+a)^2} = \frac{x^2 + 2xa - a}{(x + a)^2} $$
So far so good.
Now we need to differentiate again, using the quotient rule, as you did when finding $\dfrac{dy}{dx}$. It gets messy-looking, but in this case, simplifies relative... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem :
If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.
Plea... | HINT:
So, we have $$\frac p{2\sin\theta}=\frac{2\cos\theta}{\sin\theta+\cos\theta}$$
Apply Componendo and dividendo to get $$\frac{p+2\sin\theta}{p-2\sin\theta}=\frac{2\cos\theta+\sin\theta+\cos\theta}{2\cos\theta-(\sin\theta+\cos\theta)}=\frac{\sin\theta+3\cos\theta}{\cos\theta-\sin\theta}$$
Similarly, $$\frac p{2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| Put $y=\sqrt{x-1}$, getting $x=y^2+1$; thus the equation becomes
$$
\sqrt{y^2-4y+4}+\sqrt{y^2-6y+9}=1
$$
which should ring a bell.
It becomes $|y-2|+|y-3|=1$ that can be treated without resorting to squaring; divide it into cases:
If $y<2$, the equation becomes $2-y+3-y=1$, or $2y=4$, that means $y=2$, absurd.
If $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ al... | There is a new formula, that I've created. It can be written in a few different ways, below are two of them:
$\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\pi\int_{0}^{1} (1-u)\cot{\pi u}\left(1-\cos{2\pi n u}\right)\,du$
$\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\frac{\pi}{2}\int_{0}^{1} (1-u)\cot{\frac{\pi u}{2}}\left(1-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 5,
"answer_id": 4
} |
Order of $b$ when $ba=ab^3$ and $\operatorname{ord}(a)=4$.
Let $G$ be a non-abelian group and let $a,b \in G$ such that
*
*$\operatorname{ord}(a) = 4$, and
*$ba=ab^3$
Given that $\operatorname{ord}(b)$ is an odd prime number, compute its value.
So, $\operatorname{ord}(b)$ may be $3,5,7,11,13,...$ but I have no ide... | You can find out a multiple of the order of $b$ by shifting $a^4 = e$ past $b$, one $a$ after the other:
$$\begin{align}
b &= b\cdot e = b(a^4) = (ba)(a^3) = a(b^3a)a^2\\
&= a (b^2ab^3)a^2 = a(bab^6)a^2 = a^2 (b^9) a^2\\
&= a^3(b^{27}a) = a^4 b^{81}\\
&= b^{81}
\end{align}$$
So we have $b^{81} = b \iff b^{80} = e$, hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simplify a radical expression for a limit I'm supposed to find the slope of the tangent of the function $$D(p)=\frac{20}{\sqrt{p-1}}$$ at $p=5$.
Using the definition of slope of a tangent at a point, I get
$$\lim_{h \to 0}\frac{\frac{20}{\sqrt{(5+h)-1}}-10}{h}$$
Which, using "regular" (high school) algebra I simplified... | Multiplying by the conjugate, we obtain:
$$ \begin{align*}
\dfrac{10(2-\sqrt{h+4})}{h\sqrt{h+4}}
&=\dfrac{10(2-\sqrt{h+4})}{h\sqrt{h+4}} \cdot \dfrac{2+\sqrt{h+4}}{2+\sqrt{h+4}}\\
&=\dfrac{10(4-(h+4))}{h\sqrt{h+4}(2+\sqrt{h+4})}\\
&=\dfrac{-10h}{h\sqrt{h+4}(2+\sqrt{h+4})}\\
&=\dfrac{-10}{\sqrt{h+4}(2+\sqrt{h+4})}\\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $ I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!
Below is how I derive my approximation. I use mainly Stirling Approximat... | We will first approximate $$f(n,k)=\frac{n!}{(n-k)!}n^{-k} = \prod_{i=0}^{k-1} \left(1-\frac{i}{n}\right)$$
Actually, we will approximate the logarithm of this value.
$$\begin{align}\log f(n,k) &= \sum_{i=0}^{k-1} \log\left(1-\frac{i}{n}\right)\\&= -\sum_{i=0}^{k-1}\sum_{j=1}^\infty \frac{1}{j}\left(\frac{i}{n}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c$ are in Geometric Progression and$ a-b,c-a,b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to? If $a,b,c$ are in Geometric Progression and $a-b$,$c-a$,$b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to?
The statements I could get are
1)$b^2=ac$
2)$(c-a)^2=2(b-a)(b-... | The condition for a harmonic progression is that
$$ \frac{1}{a-b} + \frac{1}{b-c} = \frac{ 2}{ c-a}. $$
Clearing denominators, we get
$$ a^2 -2b^2 + c^2 + 2ab + 2 bc - 4ac = 0. $$
This gives us
$$(a+b+c)^2 = a^2 + b^2 +c^2 + 2ab + 2bc + 2ac = 0 + 3b^2 + 6ac =9b^2.$$
Hence, either $ a+b+c = 3b$ or $a+b+c = -3b$.
Now, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A natural number equation For what values of $n$ the equation $x^2 - (2n+1)xy + y^2 + x = 0$ has no solution in natural numbers ? (for $n=1$ it has a trivial solution).
| Letting $m=2n+1$ just for brevity, we see that:
$$(x-y)^2 = x((m-2)y-1)\\
(x+y)^2 = x((m+2)y-1)$$
Now, note that if $p|x$, then $p|x+y$ so $p|y$ and therefore $p\not\mid (m\pm2)y-1$.
Therefore, $x,(m-2)y-1$ are relatively prime, so must be perfect squares. Similarly, $(m+2)y-1$ must be a perfect square.
In the original... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to solve this : $ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $ How to solve this :
$$ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $$
We can write it as follows :
$$ \prod^{\infty}_{n=2}\frac{(n-1)(n+1)}{n^2} $$
How can we proceed further please guide.. thanks..
| HINT:
$$\text{If }T_r=\frac{(r-1)(r+1)}{r^2},$$
$$T_{r+1}=\frac{r(r+2)}{(r+1)^2}\text{ and } T_{r-1}=\frac{(r-2)(r)}{(r-1)^2}$$
Clearly, the denominator of any $n$th term will be cancelled by the numerators of the previous & the next term for $n\ge3$
Putting $n=2,3,\cdots,n-2,n-1,n$
$$\prod_{2\le r\le n}\frac{(1\cdot3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) ... | It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta $ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well ... | use the ptolemy inequality !
$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}(b+c)+c^2\cdot{b}(c+d)+d^2\cdot{c}(a+d)+a^2\cdot{d}(a+b)$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 2
} |
Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$ I found the following formula
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$
and it is cited that Euler proved the formula above , but how ?
Do there exist other proofs ?
Can we have ... | $$
\begin{align}
&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\
&=(k+1)\zeta(k+4)
+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}
\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\
&=(k+1)\zeta(k+4)
+\sum_{\substack{m,n=1\\m\ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "79",
"answer_count": 7,
"answer_id": 0
} |
How can I solve $\cos^2 x + \sin x +1 = 0$? The solution set of the equation $$\cos^2 x + \sin x +1 = 0$$ is? I haven't studied trigonometry, I'm kinda lost on this issue ...
| We can use the identity $\sin^2 x + \cos^2 x = 1$.
$$\begin{align} \cos^2 x + \sin x + 1 & = 0 \\ \\ (1 - \sin^2 x) + \sin x + 1 & = 0 \\ \\ \sin^2 x - \sin x - 2 & = 0\end{align}$$
Now, you have a quadratic in $\sin x$. For example, set $t = \sin x$, and you'll have a quadratic in $t$, find the roots, then find the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Prove that $|z| \le |z-1|+1$ and $|z-1| \le |z|+1$ Prove that for any $ z \in \mathbb{C}$:
*
*$|z| \le |z-1|+1$
*$|z-1| \le |z|+1$
I take any $z \in \mathbb{C}$ such that $z=a+bi$ for some $a,b \in \mathbb{R}$. I received:
*
*$\sqrt{a^2 +b^2} \le \sqrt{a^2+b^2 -2a+1}+1$
*$\sqrt{a^2+b^2 -2a+1}\le \sqrt{a^2+b^2... | We do it the ugly way, to show it can be done. But the right way is to use the Triangle Inequality. For example, we want to show that
$$\sqrt{a^2+b^2} \le \sqrt{(a-1)^2+b^2}+1.\tag{1}$$
(Note that we did not expand yet. Why bother, unless it is useful?)
Since both sides are non-negative, Inequality (1) is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
| Solution 1:
We usually do some tests by replacing x=0; x=1 and x=-1 in the equation (usual roots, you can take it as habit) here we can notice that -1 is a solution for the equation (root) so :
$$2x^2 - x - 3 =(x+1)\times 2 \times Q$$ and $$Q=ax+b;$$
Variables a and b can be extracted easily for this example (sometime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Solving for $\tan \theta$ given $\sin \theta/2$ QUESTION:
I'm having a hard time figuring this problem out. I've looked through my lectures and cannot find a problem that relates to this one. I do have my identities pulled up in front of me. I'm unsure where to start though. Can someone give me a kick in the right dire... | Since we're given $\sin\frac{\theta}{2}=\frac{3}{5}$, we can use the Half-Angle Formula for Sine to write $\tan\theta$ as a function of $\sin\frac{\theta}{2}$.
Now normally the Half-Angle Formula for Sine is $\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$, with the $\pm$ being dependent on which quadrant the an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Problem finding $\int\sqrt[3]{\frac{x+1}{x-1}}dx$ Given the integral $\int\sqrt[3]{\frac{x+1}{x-1}}dx$, I can't find its value. I can't one correct substitution. I tried $t=\sqrt[3]{x+1}$ but then $dt\cdot {3\sqrt[3]{x+1}^2}=dx$ but then I'm stuck with integrating $\frac{3t^3}{\sqrt[3]{t^3-2}}$.Almost the same when su... | Hint: Put $$\frac{x+1}{x-1}=u^3$$ Then, $$dx=\frac{3u^2(u^3-1)-3u^2(u^3+1)}{(u^3-1)^2}du=\frac{-6u^2}{(u^3-1)^2}du$$the integral becomes $$-\int \frac{6u^3}{(u^3-1)^2}du $$Then use partial fraction expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Congruence Equation $3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$ How to solve the following congruence equation?
$$3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$$
If $t_n$ be the $n$th triangular number, then
$$t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}$$
so if we solve this congruence equation we fin... | Work first modulo $2$. We can rewrite the congruence as $n^3+n\equiv 0\pmod{2}$. Note that this always holds.
Now work modulo $3$. The congruence can be rewritten as $n+2\equiv 0\pmod{3}$. This holds precisely if $n\equiv 1\pmod{3}$.
Finally, work modulo $5$. The congruence can be rewritten as $3n^3-3n^2+3n-3\equiv 0\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Differentiating w. r. t. $x$
Differentiate $$ \text{arccot} \frac{1-x}{1+x} $$ with respect to $x$
After putting $x= \cos \theta$, I got $$\text{arccot} \left(\tan^2 \frac{\theta}{2}\right)$$
Then how do I reach the answer?
Thank you in advance
| Method $1:$
HINT:
Putting $x=\tan \theta \implies \theta=\arctan x$
$$\frac{1-x}{1+x}=\frac{1-\tan\theta}{1+\tan\theta}=\frac{\tan\frac\pi4-\tan\theta}{1+\tan\theta\tan \frac\pi4}\text{ as } \tan\frac\pi4=1$$
$$=\tan\left(\frac\pi4-\theta\right)=\cot\{\frac\pi2-\left(\frac\pi4-\theta\right)\}=\cot\left(\frac\pi4+\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Factoring $x^5 + x + 1$
Factor. $\displaystyle x^5 + x + 1$.
I'm trying my hand on these types of expressions. In the book, it mentions that for these types you should rewrite the $x$ term such as $2x-x$. But I don't know how to make any use of this technique.
What is this type of factoring called by the way?
| As Gerry suggests, we can add and subtract $\;+ x^2 - x^2 = 0 $ without changing the expression:
$$\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}$$
Now factor out the common factor...w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$ Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $$\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$$
I have got rid of $i$ in the denominator, leaving $$\frac{x^2-y^2-2ixy}{x^2+y^2}$$
but now... | It may be easier to note that our expression is equal to
$$\frac{1}{2}\left(\frac{1}{T}+T\right).$$
Substituting, we get
$$\frac{1}{2}\left(\frac{x+iy}{x-iy}+\frac{x-iy}{x+iy} \right).$$
Bring to the common denominator $x^2+y^2$. The top is then $(x+iy)^2+(x-iy)^2=2x^2-2y^2$, and we are finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial Fraction decomposition when denominator is in $x^2 + a$ form Why isn't the expansion of $$ \frac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} $$
in the form of
$$ \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+16} $$ since (as I recall) the denominator is of square power and you should decompose it (in the numerator) until t... | There is a fast way, which is seldom seen on a calculus textbook. Polynomials $s^2 + 1$ and $s^2 + 16$ are coprime and squarefree, so we can solve it in a manner similar to linear denominators.
The numerator on $s^2 + 1$ is the unique representative of
$$\frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16} + \left\langle s^2 + 1 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inverse laplace transform - with symbolic variables Transform: $$ F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)} $$
My steps:
$$ F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s+a)(s-a)(s-2b)} $$
$$ = \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b} + K $$
$$ K = 0$$
$$A = F(s) * (s+a) $$ at s = -a
$$ A = \fra... | The routine way is to use the way we get in partial fractions and you correctly noted that. If we set:
$$ \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)}=F= \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b}
$$ then by doing boring :-) handy calculations we can find $A,B$ and $C$. We have then:
$$\frac{-Asa+2Aab+As^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the simplest form How can we simplify the expression below:
For any real numbers $a$ and $b$,
$
\frac{\frac{\frac{a+b}{2}+ b}{3}+ b} {4}+ b \cdots $
This is same as asking what is $$\frac{a+ b\sum_{i=1}^ni!}{n!}$$ as $n$ approaches to $+ \infty$?
Thanks.
| Let's rewrite your sum like this:
$\dfrac{a+ b\sum_{i=1}^ni!}{n!}=\dfrac{a+ b \left( 1!+2!+ \cdots + \left( n-2\right)! + \left( n-1\right)! +n! \right)}{n!}$
$=\dfrac{a}{n!}+b \left( \dfrac{1!}{n!} +\dfrac{2!}{n!} + \cdots \dfrac{ \left( n-2\right)!}{n!} + \dfrac{\left( n-1\right)!}{n!} + \dfrac{n!}{n!} \right)$
$=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| I believe the general term after the first is
$$\frac{(2n+1)! \cdot 4 \cdot 2}{2^n \cdot n! \cdot 2^{n+2} \cdot (n+2)! } \ , $$
with $ \ n \ \ge \ 1 . $
The series then ought to be convergent. The series is clearly larger than
$$1 \ + \ \sum_{n=0}^{\infty} \frac{1}{2} \cdot \left( \ \frac{5}{8} \ \right)^n \ , $$
so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
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Help on finding the math behind the 68/95/99 rule for Normal distribution I also see on the section on the Central Limit Theorem that the normal distribution is used and gives the graph of the 68/95/99 rule. However, I haven't seen the actual math used to prove this or how it gets computed. Can anyone give me a link ... | If you don't want to just use numerical integration: for large $t$ the standard normal CDF has the asymptotic series (which can be established by integration by parts)
$$
\Phi(t) \approx 1 + \frac{\exp(-t^2/2)}{\sqrt{2\pi}} \left(- \frac{1}{t} + \frac{1}{t^3} - \frac{3}{t^5} + \frac{3 \cdot 5}{t^7} - \frac{3 \cdot 5 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove :$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$ For sequence $u_n$ satisfing : $$\begin{cases} u_1=\sqrt{2015}\\ u_{n+1}=u_n^2-2\end{cases}$$
How to prove : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$
| The recurrence relation may be expressed as
$$
{u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4}.
$$
Which allows telescopic cancellation,
$$
\begin{align*}
\left ({u_{1}}{u_{2}}\cdots{u_{n}}\right )^2 &=
\left (\frac{\color{#036}{{u_{2}}^{2} - 4}}{{u_{1}}^2 - 4} \right )
\left ( \frac{\color{#063}{{u_{3}}^{2} - 4}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
| To answer your question, "and then?", with details, this is a way the calculation you began could continue, and this is without calculus:
$z=(4\frac{y+3}{3})^2 + y^2 = 16\frac{(y+3)^2}{9} + \frac{9y^2}{9} = \frac{16(y^2+6y+9)+9y^2}{9} = \frac{25}{9}y^2 + \frac{32}{3}y + 16$.
Center of a parabola is at $y = \frac{-b}{2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
L... | Perhaps the easiest way to go about this involves noticing that $$\frac 14N^4 + \frac 12N^3 + \frac 14N^2=\frac {N^2(N+1)^2}4$$
If you add $(N+1)^3$ to each side of $P(N)$ you get$$1^3+2^3+\dots N^3+(N+1)^3=\frac {N^2(N+1)^2}4+(N+1)^3$$
The left-hand side is what you need for $P(N+1)$ and the task is to show that the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)} \right|$$ converge... | i think this is an easy answer
$$ \sum_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
$$ \frac{2n+3}{n(n+1)} = \frac{3(n+1) - n}{n(n+1)} = \frac{3}{n} - \frac{1}{n+1} $$
$$ \sum_{n=1}^{\infty} \left(\frac{3(-1)^n}{n} - \frac{(-1)^n}{n+1} \right) $$
$$ = \int_0^1 3\sum_{n=1}^{\infty} (-1)^n x^{n-1} - \sum_{n=1}^{\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.
| yes,use $Cuachy-Schwarz$ inequality,we have
$$\left[\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right][1+1]\ge\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2$$
use $Cauchy-Schwarz $ inequality,we have
$$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(a+b)\ge(1+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving that a sequence is between certain values at certain n I'm given that $a_1=1$, and for every $n \gt1, a_{n+1} = a_n + \frac{1}{a_{n}}$. I need to prove that $20 < a_{200} < 24$. I tried finding a limit at infinity setting both limits to $L$ ( for $a_n$ and $a_{n+1}$ but that gave me nothing useful. All I figure... | I've decided to submit my answer.
It's a little different from Oleg567's,
but the results are the same.
I first try to estimate the growth of $a_n$.
From the desired bounds,
it looks like it might be of order
$\sqrt{n}$.
Suppose
$a_n
=c\sqrt{n}
$.
Then
$c\sqrt{n}+1/(c\sqrt{n})
=c\sqrt{n+1}
=c\sqrt{n}\sqrt{1+1/n}
\appro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that if $a_{n+1} = a_n^2$, the last $n$ digits of $a_{n+1}$ are the same as the last $n$ digits of $a_n$. I have been working on this problem for a while. I know that I have to prove it using induction, but I'm unsure of the next step. The formula for the terms is: $a_{n+1} = 5^{2n}$ with $a_1 = 5$. The sequence ... | The correct formula for $a_{n+1}$ in terms of $a_1$ is
$$
a_{n+1}=a_1^{2^n}\tag{1}
$$
You have $a_1=5$, so what you want to show is that
$$
5^{2^n}\equiv 5^{2^{n-1}}\pmod{10^n}\tag{2}
$$
which is equivalent to
$$
10^n\mid(5^{2^{n-1}}-1)5^{2^{n-1}}\tag{3}
$$
Now,
$$
\begin{align}
5^{2^{n-1}}-1
&=(5^{2^{n-2}}+1)(5^{2^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Domain and Range of f(x) Find the domain and range of Function?
$$F(x) = \frac{1}{\sqrt{25-x^2}}$$
Domanin f(x) has real value , if
$$25-x^2\geqslant0$$
$$-x^2\geqslant-25$$
$$x^2\le25$$
$$-5\le x\le5$$
$$D_f= [-5,5]$$
let y= f(x)
$$y = \frac{1}{\sqrt{25-x^2}}$$
then what should i do to find range?
| As noted above, the domain is $(-5,5)$.
Another way to find the range is to find the values of $y$ for which the equation
$\displaystyle y=\frac{1}{\sqrt{25-x^2}}$ has a solution for x:
1) Since $\displaystyle\frac{1}{\sqrt{25-x^2}}>0,$ we have that $y>0$.
2) $\displaystyle y^2=\frac{1}{25-x^2} \implies 25-x^2=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}} \mathrm dx$ I am having real trouble evaluating this indefinite integral
$$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$
Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ ... | $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2$$
If we set $\int(\sin x+\cos x)dx=\sin x-\cos x,\sin2x=1-u^2$
$$\implies \int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx=2\int\frac{du}{2-(1-u^2)^2}=2\int\frac{du}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}$$
$$=\frac2{2\sqrt2}\int\frac{(\sqrt2+1-u^2)+(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding $a, b\in\mathbb{R}$ so that $\lim_{x\to \infty} \sqrt{ax^2 + bx} - \sqrt{x^2 + x + 1} = 1$ Find the values of real constants $a$ and $b$ such that
$$\lim_{x\to \infty} \sqrt{ax^2 + bx} - \sqrt{x^2 + x + 1} = 1$$
If I obtained $a = 1$ and $b=0$, is that the correct answer?
Or is $b$ actually some other value?... | Hint: $$\dfrac{ax^2 + bx - x^2 - x - 1}{\sqrt{ax^2 + bx} + \sqrt{x^2 + x + 1}} = \dfrac{(a-1)x + (b-1) - \frac{1}{x}}{\sqrt{a + \frac{b}{x}} + \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$ Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$
Prove that:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$
| $36 + 9y^2 = 4x^2$ Suggests we could use $y = 2 \tan A$ to get $x = \pm 3\sec A$. Given the objective, it is sufficient to consider $x = 3 \sec A, ~~ A \in [0, \frac{\pi}{2})$
Then $(2x - y - 4 )^2 = 4(3 \sec A - \tan A - 2)^2 $
It is easily checked that the minimum occurs when $3 \tan A = \sec A$ or when $\sin A = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding a set of representatives in a certain array made from a multiset Let $M$ be an $n \times n$ array whose entries are from the multiset consisting of a single one, three twos, five threes,..., and $2n-1$ $n$'s, with the additional condition that, for $1\leq k\leq n$, no row or column have more than $k$ $k$'s.
Is ... | It turns out that it is not the case that one can always find a set of distinct representatives for matrices as described in the question. Here is an example of a 9x9 matrix where no set can be chosen:
$$\begin{pmatrix}x & x & x & 6 & 5 & 5 & 5 & 5 & 5\\x & x & x & 5 & 6 & 6 & 6 & 6 & 5\\x & x & x & 6 & 6 & 6 & 6 & 6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| ( Motivation for this (late) answer is to demistify the derivation of the square factor, and show that it reduces to the routine factorization of an associated quadratic. )
Using that $\,x^7 + 1 = (x+1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)\,$:
$$
(x+1)^7 - x^7 - 1 = (x+1)\left((x+1)^6 - x^6 + x^5 - x^4 + x^3 - x^2 + x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
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Logic for decomposing a permutation into different products composed of transpositions I know that any permutation cycle can be decomposed into transpositions as follows:
$(a_1,a_2...,a_n) = (a_1,a_{n-1})...(a_1,a_2)$
But in my book there is an example of the following form
$(1,2,3,4,5) = (5,4)(5,2)(2,1)(2,5)(2,3)(1... | In his question and comments the OP keeps asking HOW did the author of his book come up with
$\tag a \sigma = (1,2,3,4,5) = (5,4)(5,2)(2,1)(2,5)(2,3)(1,3)$
But we have the transposition algebra rules:
\begin{align}(1) \quad (ab)(ab)&=1_{Id}\\
(2) \quad (ab)(cd)&=(cd)(ab)\\
(3) \quad (ab)(bc)&=(ca)(ab)\\
(4) \quad (ab)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
lim calculus problem with infinity $$\lim_{n→∞} \left( \left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n}\right)$$
I tried replacing n+5 with u and simplifying but the answer seems off. Help would be greatly appreciated.
| We need to do some rewriting:
$$\begin{align}
\left(\frac{n+3}{n+5}\right)^{n+4} &= \left(\frac{n+5-2}{n+5}\right)^{n+4} \\
&= \left(1-\frac{2}{n+5}\right)^{n+5-1} \\
&= \underbrace{ \left(1-\frac{2}{n+5}\right)^{n+5} }_{\to e^{-2}}\ \underbrace{ \left(1-\frac{2} {n+5}\right)^{-1} }_{\to 1}
\end{align}
$$
So the left t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A closed form for $\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx$ Let $\operatorname{erfi}(x)$ be the imaginary error function
$$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$
Consider the integral
$$I=\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sq... | My strategy here is to use Parseval's equality to express the integral in a simpler form. This requires a strategic splitting of the integrand into Fourier transforms.
Begin by writing
$$\text{erfi}({\sqrt{x}})=\frac{2}{\sqrt{\pi}} \sqrt{x} \int_0^1 dt \, e^{x t^2}$$
and consider the following Fourier Transform:
$$\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
} |
How much caffeine is left in your system? The half-life of caffeine in your system is 6 hours. If you drink 100mg of caffeine every day, what percentage is left in your system after a week?
If you drink 100mg caffeine every day for a month (30d), how long does it take for the level of caffeine in your system to drop be... | The answer to both depends on when (and how, since one rarely drinks the coffee instantaneously) the coffee is consumed during the day.
For the purposes of this problem, however, I will assume that all the coffee is consumed at exactly $t_{\text{coffee}}$ every day and that we are looking for the quantity remaining at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$? Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?
It is said that this is derived from trigonometry, but I couldn't find why this is the case.
| Here's a proof that has essentially nothing to do with trigonometry:
Hold $y$ constant and differentiate the function $$f(x) = \arctan{\frac{x + y}{1 - xy}} - \arctan{x} - \arctan{y}$$ to find that
\begin{align}f'(x) &= \frac{1}{1 + \left(\frac{x + y}{1 - xy}\right)^2} \cdot \left(\frac{(1 - xy) - (x + y)(-y)}{(1 - xy)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Laplace transformation of $ \frac{\sin^2{t}}{t^2} $? I tried convolution theorem also tried this process..
$$Laplace(\frac{f(t)}{t}) = \int_s^\infty F(u) du $$
So,
$$Laplace(\frac{\sin{t}}{t} * \frac{\sin{t}}{t}) = \int_s^\infty F(u) du * \int_s^\infty F(u) du $$
where F(u) = Laplace(f(t))
I know this look's funny as ... | \begin{align*}
\mathcal L \left( \frac{\sin^2(x)}{x^2} \right ) &= \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x^2} dx \\
&= \left[e^{-sx} \sin^2(x) \int \frac 1{x^2}dx \right ]_0^\infty - \int_0^\infty \left( 2 e^{-s x} \sin (x) \cos (x)-s e^{-s x} \sin ^2(x) \right ) \frac{(-1)}{x} dx \\
&= \int_0^\infty e^{-sx} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find a closed form solution to a recurrence of the following form? I need to find the closed form solution to the following recurrence -:
$ T(n) = 8*T(n/2) + 0.25*n^2$ with $T(1) = 1$ and $n=2^j$ and this is what I have tried so far but just can't seem to get a pattern out for this recurrence. Letting 0.25 to be... | You may be interested to know that the recurrence
$$ T(n) = 8\;T(\lfloor n/2 \rfloor) + 1/4 \; n^2 $$
where $T(0) = 0$ and $T(1) = 1/4$ actually has a closed form solution for all $n$ and not just for $n$ a power of two. (We will show how to solve this with $T(1) = 1$ at the end of this post.)
This closed form is obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots ... | $n|\sum\limits_{k=1}^{n-1}k^2\iff n|\frac{n\cdot(n-1)\cdot(2n-1)}{6}$
$\iff 6|(n-1)\cdot(2n-1)$
$\impliedby n\equiv 1\pmod 6$
If $n\not\equiv 1 \pmod 6$, then since $(2n-1)$ is odd, $n\equiv 1\pmod 2\, \&\, 2n\equiv 1\pmod 3$
i.e., $n\equiv-1\pmod 2 \,\&\, n\equiv -1\pmod 3\implies n\equiv -1 \pmod 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Laurent series of a trigonometric function Find the Laurent Serie(and residue) aroud $z_0=0$ of the function
$f(z) = \frac{1}{1-\cos z}$.
Progress:
It looks very trivial but it seems to get complicated so I'll only try with 3-4 terms:
We can use that $\cos z = 1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0... | Factor out a $z^2/2!$ from the denominator to get
$$\begin{align}\frac{1}{1-\cos{z}} &= \frac{2}{z^2} \frac{1}{\displaystyle1-2 \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!}}\\ &= \frac{2}{z^2} + \frac{4}{z^2} \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!} + \frac{2}{z^2} \left ( \sum_{k=1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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second order ODE. I have this second order ODE:
$$-au''+bu'=x^2$$
where $u=u(x)$, $a=\text{constant}>0$, $b=\text{constant}\ge0 $
and $$u(0)=1 ,u(1)=0$$
I tried to solve it and I get:
$$u(x)=D+Ce^{bx/a}$$
where $D,C$ are constant.
Is my answer correct?
How I can find $D$ and $C$?
I realy need help because there are 3... | Hint
I'd guess particular solution as
\begin{align}
u_p' &= Ax^2+Bx+C \\
u_p'' &= 2Ax + B
\end{align}
Now, substitute it to the ODE given
$$
-2Aax - Ba + Abx^2 + Bbx + Cb = Abx^2 + (Bb - 2Aa)x + Cb - Ba = x^2
$$
from which you can find
\begin{align}
Ab &= 1 \\
Bb - 2Aa &= 0 \\
Cb - Ba &= 0
\end{align}
$$
A = \frac 1b; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Showing a recursive sequence is Cauchy let $a_1=1$, $a_2=2$, and define $a_n=\frac{1}{2}(a_{n-1}+a_{n-2})$. How can I show that this sequence is Cauchy?
I began with $|a_n-a_{n-1}|=\frac{1}{2}(a_n-a_{n-1})$
which goes to $\frac{1}{2}(a_{n-1}+a_{n-2})-\frac{1}{2}(a_{n-2}+a_{n-3})$
I am have been toying with this for qui... | You said yourself that
$|a_n-a_{n-1}| = \frac{1}{2}(a_{n-1}+a_{n-2}) - \frac{1}{2}(a_{n-2}+a_{n-3})$.
Cancelling the $a_{n-2}$ terms leaves $\frac{1}{2}(a_{n-1}-a_{n-3})$. Expanding $a_{n-1}$ gives
$\frac{1}{2}(a_{n-1}-a_{n-3}) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}+a_{n-3})-a_{n-3})\right) = \frac{1}{2}\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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How to evaluate the summation $S_b$ This question is from my notebook, not hw or else, only exercise to understand better. I tried by myself. However, since my trail are too trivial, I dont need to write here. i am confused a bit. I want to learn how to solve such question types. I guess I need to use residue. I am gra... | we'll use the integral $$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz $$
where $C_N $ is the square contour of side $ 2N + 1 $
By residues theory :
$$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz = 2\pi i\left( \sum_{-N}^N \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , n \right) + \mathrm{Res} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Since $\frac{1(1 +1)(2+1)}{6} = 1$, then base case holds. Suppose result holds for $n$. Then
$$\sum_{i=1}^{n+1} i^2 = \sum_{i=1}^{n}i^2 + (n+1)^2 = \frac{n(n+1)(2n + 1)}{6} + (n+1)^2 $$
$$ \frac{n(n+1)(2n + 1)}{6} + (n+1)^2 = \frac{(n+1)[n(2n+1) + 6(n+1)}{6} = \frac{(n+1)(2n^2 + 7n + 1)}{6} =_{*} \frac{(n+1)(n+2)(2n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x... | Hint: plot the graph of $f$, and notice that:
*
*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $-\infty$ to $0$
*$x=0$ is a vertical asymptote
*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $0$ to $+\infty$.
Hence $x_1=f(x)$ can be considered as an independent variable on each... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to factor and find zeros of $2x^6+5x^4-x^2$ In the equation $y=2x^6+5x^4-x^2$, how can I factor it in order to find the zeros or x-intercepts? This is what I've gotten to, but I can't see what to do next: $x^2(x^2(2x^2+5)-1)$
| $$y=2x^6+5x^4-x^2 = x^2(2x^4 + 5x^2 - 1)$$
For the factor $\;2x^4 + 5x^2 -1,\;$ put $t = x^2$:
$$2x^4 + 5x^2 - 1 = 2t^2 + 5t - 1$$
Suggestion: Use the quadratic formula to find the roots (also zeros): $t_1, t_2$, then back substitute, knowing $t = x^2 \implies x = \pm \sqrt t$. So for each zero $t_i$, there are two co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/519143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x... | HINT: Notice that your congruences are equivalent to the following ones:
$$\left\{\begin{align*}
x\equiv-1\pmod6\\
x\equiv-1\pmod5\\
x\equiv-1\pmod4\\
x\equiv-1\pmod3
\end{align*}\right.$$
In other words, $x+1$ is divisible by $6,5,4$, and $3$. What’s the smallest positive integer with that property?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane.
Here is what I did.
Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\f... | You started out correctly, by plugging $z = 2$ into the equation and finding $\frac{x^2}{3} + y^2 + \frac{2}{5} = 1$, or
$$\frac{x^2}{3} + y^2 = \frac{3}{5}.$$
The next step is to put the equation in the standard form for an ellipse. (Recall that the standard form is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find the $(x^2_{1}+x_{1}x_{2}+x^2_{2})(x^2_{2}+x_{2}x_{3}+x^3_{3})(x^2_{3}+x_{3}x_{1}+x^2_{1})$ Let $x_1$, $x_2$, and $x_3$ be the roots of the equation
$$4x^3-6x^2+7x-8=0.$$
Find this value:
$$(x_1^2+x_1x_2+x_2^2)(x_2^2+x_2x_3+x_3^3)(x_3^2+x_3x_1+x_1^2)$$
My try:
$$x_1+x_2+x_3=\frac 3 2$$
$$x_1 x_2+x_2 x_3+x_1 x_... | Notice
$$
x_1^2 + x_1x_2 + x_2^2 = \frac{x_1^3 - x_2^3}{x_1 - x_2}
= \frac{(6x_1^2 -7x_1 + 8)-(6x_2^2 -7x_2 + 8)}{4(x_1-x_2)}\\
= \frac{6(x_1+x_2)-7}{4} = \frac{6(\frac{3}{2} - x_3) - 7}{4}
= \frac{1-3x_3}{2} = \frac32(\frac13 - x_3)
$$
and similar expression holds for $x_2^2 + x_2x_3 + x_3^2$ and $x_3^2 + x_3x_1 + x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequalities with absolute value question Solve: $ 5< \left\vert\dfrac{x+10}{x-10}\right\vert<6$
attempt at a solution:
Dividing into two:
$5<\left\vert \dfrac{x+10}{x-10}\right\vert $ And $\left\vert \dfrac{x+10}{x-10}\right\vert<6 $
For first we solve:
$ \dfrac{x+10}{x-10}<-5 $ or $ \dfrac{x+10}{x-10} >... | Given that $10$ isn't even in the domain of the given function, your solution to the first inequality cannot be correct. Without actually seeing the details, though, it's hard to say what went wrong.
As an alternative, note that $\frac{x+10}{x-10}>0$ for $x>10$ and for $x<-10,$ and $\frac{x+10}{x-10}<0$ for $-10<x<10$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of factors of a big number How to find the number of factors of $884466000$ without using a calculator?
| It's not too difficult to factor, since it's divisible by rather high powers of $2$ and $5$; right off the bat, we see that
$$884466000 = 1000 \cdot 884466 = 1000 \cdot 2 \cdot 442233$$
Now $442233$ has digit sum $4 + 4 + 2 + 2 + 3 + 3 = 18$, so it's divisible by $9$, giving
$$442233 = 49137 \cdot 9 = 16379 \cdot 27$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Con... | I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion,
$$
\frac{\log\left(1-x^4\right)}{1+x^2} = \displaystyle \sum_{k=1}^{\infty} x^{4k}\left(x^{2} -1\right)H_k,
$$
where $H_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 4
} |
If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightar... | $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii)$
$(i)-(ii)\Leftrightarrow (x-y).(x+y+1)=0$
$\Rightarrow x=y ; x+y=-1$
$+x=y$, now we solve $x,y$ then $xy=...$
$+x+y=-1 \Rightarrow xy=-2004$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Reflection across a line?
The linear transformation matrix for a reflection across the line $y = mx$ is:
$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$
My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how ... | You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be:
The line $y = mx$ is parametrised by $t \cdot \begin{pmatrix}1\\m\end{pmatrix}$. The line orthogonal to it is parametrised by $r \cdot \begin{pmatrix}-m\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 6,
"answer_id": 0
} |
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended u... | at lin3 col 4 is 0 not 1
det$\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}$
~det$\begin{bmatrix}
a+3 & a & 1 & 1 \\
0 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How can I prove by induction that $9^k - 5^k$ is divisible by 4? Recently had this on a discrete math test, which sadly I think I failed. But the question asked:
Prove that $9^k - 5^k$ is divisible by $4$.
Using the only approach I learned in the class, I substituted $n = k$, and tried to prove for $k+1$ like this:
... | $$\begin{align} 9\cdot 9^k - 5\cdot 5^k & = (4 + 5)\cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5 \cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5(9^k - 5^k)\\ \\ & \quad \text{ use inductive hypothesis}\quad\cdots\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Pascal's formula and Binomial Coefficients Prove Pascal's formula by substituting the values of the binomials coefficient as given in the equation
$$\binom nk= \frac{n!}{k!\,(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots1}.$$
I guess I just don't know what values I'm supposed to be substituting in.
| $$\binom nk=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}=\binom{n}{n-k}....[1]$$
$$\binom nk=\frac{n(n-1)!}{k(k-1)!(n-k)!}=\frac{ n}{k}\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}=\frac{n}{k}\binom {n-1}{k-1}...[2]$$
from [1] and [2] we have
$$\binom {n}{k}=\binom {n}{n-k}=\frac {n}{n-k}\binom {n-1}{n-k-1}=\frac{n-k+k}{n-k} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Property on glide reflections I need to prove that the conjugate of a glide reflection is a glide reflection.
What I have tried: Let $m: X= \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & -\cos \phi \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\begin{pmatrix... | Hint: The involved matrix, let's call her $M_{\psi}$, is invertible so simply solve $y=M_{\psi}x+b$ for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$? All is in the title:
If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.
| $\displaystyle (2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1=8b+1$ where $b$ is some integer
Now, $(2a+1)^4=(8b+1)^2=64b^2+16b+1\equiv1\pmod{16}$
This way we can show $(2a+1)^{2^n}\equiv1\pmod{2^{n+2}}$ for integer $n\ge1$
This is generalized as Carmichael function $$\lambda(2^{n+2})=2^n\text{ for }n\ge1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ $\displaystyle \lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ if $\lim_{n\rightarrow\infty}a_n=a$
Note that $\displaystyle\sum_{k=1}^n\frac{a_{n-k+1}}... | We will show that the limit is $a$.
Let $\varepsilon>0$. First, $(a_n)$ is bounded, so $|a_n|<M$ for some $M>0$, for all $n\in\mathbb N$. There exists $n_0\in\mathbb N$ such that, for all $n\geq n_0$, $|a_n-a|<\frac{\varepsilon}{6}$. Also, there exists $n_1\in\mathbb N$ with $n_1>n_0$ such that, if $n\geq n_1$, then $\... | {
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"source": "stackexchange",
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Express $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've been trying with no success expressing this functions.
a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\s... | I mean just use the trig identities: $$\cos \frac{ PI }{4} = 1 - 2 \sin^2 \frac{ PI}{8} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given n = 4k. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$.
Given $n = 4k$, where $k$ is a non-negative integer. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$.
I have no idea how to prove this equation.
| Let $i=\sqrt{-1}$, and note that $(1+i)^4=(1-i)^4=-4$. Also, for $j \ge 0$, we have
$$\frac{1+(-1)^j}{2} = \begin{cases}1 &\text{if $2|j$}\\0 &\text{otherwise}\end{cases}.$$
Then
\begin{align}
\sum_{j\ge 0}\binom{4k}{2j}(-1)^j
&= \sum_{j\ge 0}\binom{4k}{2j}i^{2j} \\
&= \sum_{j\ge 0}\frac{1+(-1)^j}{2}\binom{4k}{j}i^{j}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^... | Multiplying through out the given Equation with 2
$$ \frac{2 Cos(x)Cos(\frac{y}{2})}{Cos(x-\frac{y}{2})} + \frac{2 Cos(y)Cos(\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $\implies$
$$\frac{Cos(x+\frac{y}{2})+Cos(x-\frac{y}{2})}{Cos(x-\frac{y}{2})}+\frac{Cos(y+\frac{x}{2})+Cos(y-\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving inequality with non-negative reals Let $a_1 , a_2 , ... , a_n$ be $n$ non-negative real numbers all less than $1$ and satisfying
let $$a=\sqrt {\frac1{n}\sum_{i=1}^na^2_i}≥\dfrac{\sqrt{3}}{3}$$ then how do we prove that $$\frac{a_1}{1-a^2_1} + \frac{a_2}{1-a_2^2}+....+\frac{a_n}{1-a_n^2}≥\frac{na}{1-a^2}$$
| let $a^2_{i}=x_{i},1\le i\le n$,then
if $x_{1},x_{2},\cdots,x_{n}<1$, are non-negative real numbers,such
$$\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}=r\ge\dfrac{1}{3}$$
then we have
$$\dfrac{\sqrt{x_{1}}}{1-x_{1}}+\dfrac{\sqrt{x_{2}}}{1-x_{2}}+\cdots+\dfrac{\sqrt{x_{n}}}{1-x_{n}}\ge\dfrac{n\sqrt{r}}{1-r}$$
Proof
Appl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$? Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$?
I think if $y=x^5$, then $P(y)=y^{4} +y^{3}+ y^{2}+y+1 =\displaystyle \frac{y^5-1}{y-1}$, $5$ is prime, then $P(y)$ is irreducible in $\mathbb{Q}[y]$ or $\mathbb{Q}[x^5]$ , how do I show ... | Let $f(x) = x^{20}+x^{15}+x^{10}+x^5+1 \in \mathbb{Z}[x]$, we have
$$\begin{array}{rrcl}
& (x^5-1)f(x) &=& x^{25} - 1\\
\implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1\\
\implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1 \pmod 5\\
\iff & ((x^5+1)-1)f(x+1) &=& (x^5+1)^5 - 1 \pmod 5\\
\iff & x^5 f(x+1) &=& (x^{25}+1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| This comes from my answer here. Let
$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$
Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value. We evaluate this integral by once again appealing to the residue theorem, but this time, we consider
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| Express all of the terms in terms of the symmetric polynomials $s_1=x+y+z, s_2=xy+yz+zx$ and $s_3=xyz$.
You get $s_1^3-3s_1s_2+3s_3=x^3+y^3+z^3=x+y+z=s_1$ and $s_1^2-2s_2=x^2+y^2+z^2=xyz=s_3$.
Putting the second equation into the first gives $s_1^3-3s_1s_2+3s_1^2-6s_2=s_1$, which simplifies to $s_2=(s_1^3+3s_1^2-s1)/(3... | {
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"source": "stackexchange",
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How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$
My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63... | For sure, L'Hopital's rule would be useful. But you could also use each Taylor expansion of $\sin(x)$ and $\tan(x)$ to expand $\sin (\tan (x))$ and $\tan (\sin (x))$. This would probably be tedious but it is doable (I made it).
Using what you wrote, you should arrive to $$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Pre Calculus Synthetic Division two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
| Hint:
Let $f(x) = 2x^{3} - 3x^{2} + px + q$.
Since there are $3$ roots to a cubic equation, and $3$ and $-2$ are roots of $f(x)$, we can factor $f(x)$ as
$f(x) = (2x - k)(x-3)(x+2)$
where $k/2$ is the third root of $f(x)$.
Since $f(x) = 2x^{3} - 3x^{2} + px + q$, we can expand the previous factorization of $f(x)$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating a determinant How do I calculate the determinant of the following matrix:
$$\begin{matrix}
-1 & 1 & 1 &\cdots & 1 \\
1 & -1 &1 &\cdots &1 \\
1 & 1 & -1 &\cdots &1\\
\vdots & \vdots & \vdots& \ddots &\vdots \\
1&1&1&\cdots&-1\\ \end{matrix}$$
Thanks in advance!
| Recall the formula for the determinant of a matrix $A$
$$\det(A) = \sum_{\pi\in S_n} \sigma(\pi) \prod_{i=1}^n A_{i,\pi(i)}.$$
Now the sign $\sigma(\pi)$ of a permutation $\pi$ is given by
$$\sigma(\pi) = \prod_{c\in\pi} (-1)^{|c|-1}$$
where the product is over the cycles of the disjoint cycle composition of $n.$
We wi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this equation to prove that $aRb \iff a^2 - b^2 = 1$ is antisymmetric correct?
Over $\mathbb{R}$, $aRb \iff a^2 - b^2 = 1$.
I tried determining if it was antisymmetric. I seem to have done it, but while doing the equation, I stumbled upon a scenario that always made me doubt my decisions:
Have
$$aRb \land bRa$$
$$... | The deduction you have does not work, as $a^2 = b^2$ does not imply $a = b$. It is actually much simpler. Assuming $aRb$ and $bRa$, you get (third line of equations) $-a^2 + b^2 = -1$ and $b^2 - a^2 = 1$. So $b^2 - a^2$ must be equal to $1$ and $-1$ at the same time, which is not possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim\limits_{n\to \infty}\left\{(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})\right\}^{\frac{1}{n}}$ Question is to evaluate :
$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$
I tried to do something like this but it... | Consider the log of $[(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}$:
$$\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right).$$
The limit of this sum is equal to $\int_0^1\ln(1+x)dx$, so
$$
\begin{aligned}
\lim_{n\to\infty} [(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}
&= \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$
$H_n$ denotes the harmonic numbers.
| Since $$\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n=\frac12\ln^2(1-x)$$
divide both sides by $x$ then $\int_0^y$
$$\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}y^n=\frac12\int_0^y\frac{\ln^2(1-x)}{x}dx$$
Next multiply both sides by $-\frac{\ln(1-y)}{y}$ then $\int_0^1$ and apply that $-\int_0^1 y^{n-1}\ln(1-y)dy=\frac{H_n}{n}$
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
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Solve: $\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$ Solve: $$\dfrac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
My try:
Conditions identify: $\left\{ \begin{array}{l} x-1>0\\1-\sqrt[3]{2-x}>0\end{array} \right.\Leftrighta... | Ok, let me give you a hint.
$$\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
so
$$\log_{\frac{1}{2}}\left(x-1\right)>2\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
$$\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}[\left(1-\sqrt[3]{2-x}\right)]^2$$
$$\left(x-1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\d... | Using power means we have
$x+y+z\le 3\cdot {(\dfrac {x^n+y^n+z^n} {3})}^{\frac 1 n}=3$
Let $x\le y\le z$ then we have $z\ge 1$ so
$\dfrac x {x+yz}+\dfrac y {y+zx} \ge \dfrac 2 {z+1} \Leftrightarrow \\$
$\dfrac {(z-1)z(x-y)^2} {(x+yz)(y+zx)(z+1)} \ge 0$ is true
Hence it is sufficies to prove
$\dfrac 2 {z+1} +\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556519",
"timestamp": "2023-03-29T00:00:00",
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Probability Problem on Divisibility of Sum by 3 From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem. I would welcome a good solution.
| Using the Polya Enumeration Theorem and supposing we are choosing the three-element subset from $\{1,n\}$ we see that we have the following combinatorial class (not labelled):
$$\mathcal{Q} = \mathfrak{P}_3(\mathcal{Z}+\mathcal{Z}^2+\cdots+\mathcal{Z}^n).$$
Now the cycle index for the set construction is given by
$$Z(A... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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$x^{311} \equiv 662 \mod{713}$ We have to find $x$ such that $x^{311} \equiv 662 \mod{713}$
The example given in my notes has a few typos and the professor is unavailable.
| $$713=23 \cdot 31 $$
By Chinese Remainder Theorem, you need to solve
$$x^{311} \equiv 662 \pmod{23} \\
x^{311} \equiv 662 \pmod{31}$$
or equivalently
$$x^{311} \equiv 18 \pmod{23} \\
x^{311} \equiv 11 \pmod{31}$$
It follows that $x$ is invertible in both arithmetics, and then by Fermat Little Theorem we have
$$x^3\equi... | {
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Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$ I have to prove that
$$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$
The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I did the following steps:
$$\begin{align}
\frac{a+b}{2} &\ge ... | Actually, you don't need to square both sides, since $\frac{a+b}{2} \ge \sqrt{ab} \Leftrightarrow \frac{a+b-2\sqrt{ab}}{2} \ge 0
\Leftrightarrow \frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$. And for any real number x, $x^2 \ge 0$ is always true, so $\frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$ is true.
As for your last equation, $... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of a fraction So far I have,
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$
I have no idea how to keep going with this, every way I try I get stuck and can't do anything with it.
| More generally,
consider
$$\lim_{x\to a} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{a^2+1}}}{x-a}.$$
I will use just algebra.
If $x \ne a$,
$\begin{align}
\frac{\frac{x}{\sqrt{x^2+1}} - \frac{a}{\sqrt{a^2+1}}}{x-a}
&=\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\\
&=\left(\frac{x\sqrt{a^2+1}-a\sq... | {
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} |
A limit problem related to $\log \sec x$
If $$f(x) = \dfrac{{\displaystyle 3\int_{0}^{x}(1 + \sec t)\log\sec t\,dt}}{(\log\sec x)\{x + \log(\sec x + \tan x)\}}$$ then prove that $$\lim_{x \to {\pi/2}^{-}}f(x) = \frac{3}{2}$$ and $$\lim_{x \to 0}\frac{f(x) - 1}{x^{4}} = \frac{1}{420}$$
Looking at the integral sign in ... | Here is my work for the first limit: Let $f(x) = \frac{num(x)}{den(x)}$. We get $\infty/\infty$ so we can use L'Hopital:
\begin{align*}
\frac{num'(x)}{den'(x)}&=\frac{3(1+\sec(x))\log \sec x}{(\tan(x))(x + \log(\sec x + \tan x)) + (\log \sec (x)) (1 + \sec(x)) }\\
&= \frac{3(1+\sec(x))\log \sec x}{(\tan(x))[x + (\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \... | OK, here's another way to do it:
Let $a\circ b=\dfrac{a+b}{1-ab}$.
Prove that this operation is associative.
Show that $a\circ b\circ c = \dfrac{a+b+c-abc}{1-ab-ac-bc}$.
That implies that if $a+b+c=abc$, then $a\circ b\circ c=0$.
It's easy to show that $0\circ 0=0$, so we have
\begin{align}
0 & = 0\circ 0 = (a\circ b\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
| Since
\begin{align*}
p \cos(x+y+z) &= \cos x + \cos y + \cos z \\
p \sin(x+y+z) &= \sin x + \sin y +\sin z
\end{align*}
\begin{align*}
p e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz}
\end{align*}
Multiplying throughout by $e^{-i(x+y+z)}$, we get
\begin{align*}
p = e^{-i(y+z)} + e^{-i(z+x)} + e^{-i(x+y)}
\end{align*}
Since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displays... | Easiest and fastest solution:
Basically, try small values of $n$ and you'll see a pattern:
$\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+\frac{4}{4^4+4^2+1}\dots$
$\implies \frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\frac{4}{273}$
$\implies \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)+\frac{1}{4}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$ How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?
What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$
for values $x>1$. $C = 3/2, k = 1$
Is this right?
| Hint: The degree of the numerator is $4$, while the degree of the denominator is $3$. The degree of $O(x)$ is $4-3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Interval of convergence homework $$ \sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n $$
First, I start with the ratio test:
$$\lim_{n\to \infty} \left| \left(\frac{-1}{4}\right)^{n+1} \frac{(5(n+1))^{n+1}}{(n+1)!} (x-1)^{n+1} \div \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n \right| =$... | You made a small algebra error in solving the inequality at the end, interchanging the $4$ and the $5$: you should have
$$1-\frac4{5e}<x<1+\frac4{5e}\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of a function I have to solve a probability problem, and at a final step I arrive at the following integral:
$$\int_z^\infty xe^{-x^2/a^2}\displaystyle\frac{1}{\sqrt{x^2-z^2}}dx$$
I tried to integrate by parts, taking $f(x)=\displaystyle\frac{1}{\sqrt{x^2-z^2}}$ and
$g'(x)= xe^{-x^2/a^2}$, but this choice le... | $\int_z^\infty xe^{-\frac{x^2}{a^2}}\dfrac{1}{\sqrt{x^2-z^2}}dx$
$=\int_z^\infty\dfrac{e^{-\frac{x^2}{a^2}}}{2\sqrt{x^2-z^2}}d(x^2)$
$=\int_{z^2}^\infty\dfrac{e^{-\frac{x}{a^2}}}{2\sqrt{x-z^2}}dx$
$=e^{-\frac{z^2}{a^2}}\int_{z^2}^\infty e^{-\frac{x-z^2}{a^2}}~d(\sqrt{x-z^2})$
$=e^{-\frac{z^2}{a^2}}\int_0^\infty e^{-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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.