Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do you solve this trig/geometry question? In a quadrilateral $ABCD$, if
$\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right) + \sin\left(\frac{C+D}2\right)\cos\left(\frac{C-D}2\right) = 2$
then $\sin\left(\frac A 2\right) \sin\left(\frac B 2\right) \sin\left(\frac C 2\right) \sin\left(\frac D 2\right) = $
a) $... | For $w$, $x$, $y$, $z$ between $0$ and $\pi$, the relation
$$\sin w \;\cos x \;+\; \sin y \; \cos z \;=\; 2$$
holds if and only if each term ---and each factor of each term--- is itself $1$. Specifically,
$$w = y = \frac{\pi}{2} \quad\text{ and }\quad x = z = 0$$
Now, note that any two angles of a quadrilateral have a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the... | Multiplying
$$\frac{\sqrt{x^2+8}-3}{x+1}$$
by
$$\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\ \ (=1)$$
gives you
$$\begin{align}\lim_{x\to -1}\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}&=\lim_{x\to -1}\frac{\color{red}{(x+1)}(x-1)}{\color{red}{(x+1)}(\sqrt{x^2+8}+3)}\\&=\lim_{x\to -1}\frac{x-1}{\sqrt{x^2+8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Verify Demorgan's Law Algebraically If $\overline X \equiv \text { not }X$,
De Morgan's Laws are stated as:
*
*$ \overline{(A + B)}= \overline A\cdot \overline B$
*$ \overline{(A\cdot B)} = \overline A + \overline B$
Verify the above laws algebraically.
I can prove this using truth tables and logic gates but alge... | By Complementarity Law,
$$P + \overline P = 1 \space\text{ and }\space P \cdot \overline P = 0$$
(Note: I shall only be using $P + \overline P = 1$ as its dual is automatically true)
First Law::
DeMorgan's $1^{\text{st}}$ law states $\overline{X+Y} = \overline X \cdot \overline Y$
It is sufficient to prove that $(X ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/943164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What exactly happens in the algebraic steps here? $$ \frac{n(n+1)}{2} + (n+1) = (n+1)(\frac{n}{2} + 1) = \frac{(n+1)(n+2)}{2} $$
I don't understand what happens from the first to the second and from the second to the third one.
| For the first equality, $n+1$ is factored out. But it is easier if read from right to left:
$$
(n+1)\bigg(\frac{n}{2}+1\bigg) = (n+1) \cdot \frac{n}{2}+ (n+1) \cdot 1 = \frac{n(n+1)}{2}+(n+1)
$$
For the second equality, the sum $\frac{n}{2}+1$ is evaluated:
$$
(n+1)\bigg(\frac{n}{2}+1\bigg) = (n+1)\bigg(\frac{n}{2}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of a function containing inverse trigonometric functions Q. $$\int \sin\left\{2\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)\right\}dx$$
$\implies$ $$\int \sin\left\{\sin ^{-1}\left(\frac{2\left(\sqrt{\frac{3-x}{3+x}}\right)}{1+\left(\sqrt{\frac{3-x}{3+x}}\right)^2}\right)\right\}dx$$
$\implies$ $$\int \:\f... | Set $u=\arctan\sqrt{\dfrac{3-x}{3+x}}$. Then
$$
\sqrt{\dfrac{3-x}{3+x}}=\tan u
$$
and
$$
\dfrac{3-x}{3+x}=\tan^2u
$$
and, with straighforward computations,
$$
x=3\cos2u
$$
so
$$
dx=-6\sin2u\,du
$$
Thus your integral becomes
$$
\int-6\sin^{2}2u\,du
$$
that you should be able to carry over.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the least value of $n$ such that $n^{25}\equiv_{83}37$ Let $n\in\mathbb{N}$. Find the least value of $n$ such that $n^{25}\equiv_{83}37$.
I concluded that $0<n<83$. Then I wrote $n^{25}$ as $\left(n^5\right)^5$ and let $t=n^5$. Now, it is hard to solve $t^5\equiv_{83}37$. How to solve this?
| Working mod $83$, a prime number. Fermat's little theorem says:
$$a^{83} \equiv a$$ for any integer $a$ so
$$a^{m \cdot 82+1} \equiv a$$ for any $a$ and $m\ge 0$ integers
so in particular
$$37^{m \cdot 82+1} \equiv 37$$
Now try to find $m$ so that $m\cdot 82 + 1 = k \cdot 25$ thus getting $37^{m \cdot 82+1}= (37^k )^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integrating $\int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx$ We need to evaluate $\displaystyle \int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx$
and some solution to this starts as,
$\displaystyle\int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx =
\int_0^{\pi/2} {\{\sin(\pi/2 -x)\}^2 \over 1 + \sin (\pi/2 -x)\cos... | Using double-angle formula, we get
$$I=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2+\sin 2 x} d x$$
$$I\stackrel{x\mapsto\frac{\pi}{4}-x}{=} 2 \int_{0}^{\frac{\pi}{4}} \frac{1}{2+\cos 2 x} d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin 2 x}{2+\cos 2 x} d x$$
Since $\dfrac{1}{2+\cos 2 x}$ is even and $\dfrac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\cos(x+y)$ and $\sin(x+y)$ given that $\cos x + \cos y = a$ and $\sin x + \sin y = b$ If $\cos x + \cos y = a$ and $\sin x + \sin y = b$.
Find $\cos(x+y)$ and $\sin(x+y)$.
I only need some hints to start as I am not able to get any way to go forward to.
| HINT:
Let $u = \cos x + i \sin x$, $v = \cos y + i \sin y$. Then
\begin{eqnarray}
u + v = a+ i b \\
\frac{1}{u} + \frac{1}{v} = a -i b
\end{eqnarray}
so
$$u v = (a+ib)/(a-ib)$$
But $uv = \cos(x+y) + i \sin(x+y)$. Now take real and imaginary parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proof irrationality $n\sqrt{11}$
Prove that $\sqrt{11}$ is irrational, subsequently prove that $n\sqrt{11}$ is also irrational for every $n \in \mathbb{N}$. You are allowed to use that if $p$ is prime, and $p | a^2$, then $p|a$.
Can't you also prove that $n\sqrt{11}$ is irrational simply by saying:
Assume $n\sqrt{1... | I guess that your post also asked for a proof that $\sqrt{11}$ is irrational? Here it is.
Assume, for contradiction, that $\sqrt{11}$ is rational. Let $a$ and $b$ be relatively prime, positive integers. Then
\begin{align*}
\sqrt{11} &= \frac{a}{b} \\
11 &= \frac{a^2}{b^2}\\
11 b^2 &= a^2 .
\end{align*}
This means that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebra problem solve for a,b,c and d? Can anyone find the values of these integers: a,b,c and d?
$$1+\sqrt{2}+\sqrt{3}+\sqrt{6} = \sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
a+b+c+d = ?
Thank you.
| $$F=1+\sqrt2+\sqrt3+\sqrt6=(1+\sqrt2)(1+\sqrt3)$$
$$F^2=a+\sqrt{b+\sqrt{c+\sqrt d}}$$
Squaring we get $$F^2=(1+\sqrt2)^2(1+\sqrt3)^2=(3+2\sqrt2)(4+2\sqrt3)=12+8\sqrt2+6\sqrt3+4\sqrt6\implies a=12$$
$$F^4=a^2+b+\sqrt{c+\sqrt d}+2a\sqrt{b+\sqrt{c+\sqrt d}}$$
$$F^4=(3+2\sqrt2)^2(4+2\sqrt3)^2=(17+12\sqrt2)(28+16\sqrt3)$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral ${\large\int}_0^1\left(-\frac{\operatorname{li} x}x\right)^adx$ Let $\operatorname{li} x$ denote the logarithmic integral
$$\operatorname{li} x=\int_0^x\frac{dt}{\ln t}.$$
Consider the following parameterized integral:
$$I(a)=\int_0^1\left(-\frac{\operatorname{li} x}x\right)^adx.$$
Can we find a closed form f... | In reference to Kirill's answer, I will show that indeed $$ I(3) = \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{xyz} \frac{dx \, dy \, dz}{x+y+z-2} = \frac{7}{2}\zeta(3).$$
I will make the same change of variables I made in my answer to your other more recent question.
$$ \begin{align} I(3) &= \int_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small... | We need to prove that
$$\prod_{cyc}\frac{a^2+b^2-c^2}{2ab}\leq\frac{1}{8}$$ or
$$a^2b^2c^2\geq\prod_{cyc}(a^2+b^2-c^2)$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)\geq0,$$
which is Schur.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How find the matrix $X$ such $e^{X}$ is give it
Question:
let matrix $$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix} ,e^{X}=\begin{bmatrix}
-1&2\\
0&-1
\end{bmatrix}$$
and such $a+d=0$,
Find the matrix $X$
my idea
$$e^{Tr{(X)}}=det{(e^{X})}$$
so
$$1=1$$ is true
I think we can find the matrix function: $e^X$,but... | Here's a sketch. Since $d = -a$, you can write $X = \left(\begin{array}{rr} a & b \\ c & -a\end{array}\right)$, which gives
$$X^2 = \left(\begin{array}{cc} a^2 + bc & 0 \\ 0 & a^2 + bc\end{array}\right) = (a^2+bc)I.$$
This allows you to write a general expression for $X^n$: $X^{2n} = (a^2+bc)^n I$ and likewise $X^{2n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to evaluate the limit $\lim_{x\to 0} ((1+2x)^{1/3 }-1)/ x$ without using the l'Hospital's rule? $$\lim_{x\to 0} \frac{(1+2x)^{\frac{1}{3}}-1}{x}.$$
Please do not use the l'hospital's rule as I am trying to solve this limit without using that rule... to no avail...
| Just in the same spirit as Paul, remember that, for small values of $y$, $$(1+y)^n=1+n y+\frac{1}{2} n(n-1) y^2+\frac{1}{6}n (n-1) (n-2) y^3+O\left(y^4\right)$$ So, replace $n$ by $\frac{1}{3}$ and $y$ by $2x$ and obtain $$(1+2x)^{\frac{1}{3}}=1+\frac{2 x}{3}-\frac{4 x^2}{9}+\frac{40 x^3}{81}+O\left(x^4\right)$$ $$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Prove that $\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$ There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me :
If $a_1,a_2,a_3,...,a_n$ IS an arithmetic progression and ... | Here's sketch, not complete proof.
Since $A$ is an arithmetic progression, Let $a_{i+1}-a_i = d$
$$\begin{align}\dfrac{1}{a_ia_{i+1}} &= \dfrac{d}{d}\dfrac{1}{a_ia_{i+1}}
\\&= \dfrac{1}{d}\dfrac{a_{i+1}-a_i}{a_ia_{i+1}} \qquad \text{since } a_{i+1}-a_i = d
\\&= \dfrac{1}{d}\left(\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(n!)^ 2 \gt n^n$ Prove the above by
*
*by mathematical induction
*By any other method.
I was just asked to prove this so I thought of using mathematical induction.
My effort :
I started first by verification and the inequality was true for n=3, 4...
Then the assumption step $(k!)^ 2 \gt k^k$
But after... | Assuming that $(k!)^ 2 \gt k^k$ we write
$$((k+1)!)^ 2 = (k!)^2 \cdot (k+1)^2 \gt k^k \cdot (k+1)^2. \tag {*}$$
Now we compare the RHS (*) with $(k+1)^{k+1}\ $:
$$\dfrac{(k+1)^{k+1}}{ k^k \cdot (k+1)^2}=\dfrac{(k+1)^{k}}{ k^k \cdot (k+1) }=\dfrac{1}{k+1} \cdot {\left(1+\dfrac{1}{k}\right)^{k}}. $$
Now using the binomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integra... | Notice that applying the chain rule to $\frac{1}{3}\sin^3(x)$ yields $\sin^2(x)\cos(x)$. Indeed you have $\int\sin^2(x)\cos(x)\,dx=\int\sin^2(x)\,d(\sin(x))=\frac{1}{3}\sin^3(x)+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is... | Consider the series
\begin{align}
S = \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \ln\left(1 + \frac{1}{2k}\right) \right]
\end{align}
for which, by using the logarithm in series form, it becomes
\begin{align}
S &= \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \frac{1}{2k} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n \, (2k)^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$ I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
| Starting with the Beta function
\begin{align}
B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt
\end{align}
differentiate with respect to $x$ and $y$ twice. This leads to
\begin{align}
I &= \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \ln^{2}(t) \, \ln^{2}(1-t) \, dt \\
&= \partial_{x}^{2} \partial_{y}^{2} B(x,y).
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Remainder of a summation divided by $2^{12}$ For a positive integer $n$, let $f(n)$ be equal to $n$ if there is an integer $x$ such that $x^2-n$ is divisible by $2^{12}$, and let $f(n)$ be $0$ otherwise. Determine the remainder when $$\sum_{n=0}^{2^{12}-1} f(n)$$ is divided by $2^{12}$.
Source: Caltech Harvey Mudd Tou... | Basically, all we need is the fact that the squares in $(\mathbb{Z}/2^k\mathbb{Z})^\times$, for $k\geq 3$, are exactly those that are $1 \pmod{8}$. This follows from the standard fact that $(\mathbb{Z}/2^k\mathbb{Z})^\times \cong Z_2\times Z_{2^{k-2}}$, that all squares are $1 \pmod{8}$, and that there are exactly $2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/961692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of an identity involving binomial coefficients I have found numerically that the following identity holds:
\begin{equation}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{x^2+t^2-t}{2t-1},
\end{equation}
where $n$, $t$, and $x$ are positive integers ($x... | This is actually easier then I thought. All what we have to do in here is to use identities of the gamma function along with the Gauss' theorem for the hypergeometric function.
\begin{eqnarray}
&&\sum\limits_{n=0}^{\lfloor \frac{t-x}{2}\rfloor}
n 2^{t-x-2 n} \frac{\binom{t}{n+x} \binom{t-n-x}{t-2 n-x}}{\binom{2 t}{t+x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The integral of $x^3/(x^2+4x+3)$ I'm stumped in solving this problem. Every time I integrate by first dividing the $x^3$ by $x^2+4x+3$ and then integrating $x- \frac{4x^2-3}{x+3)(x+1)}$ using partial fractions, I keep getting the wrong answer.
| $$\frac{2x^3}{(x+1)(x+3)}=\frac{[(x+3)-(x+1)]x^3}{(x+1)(x+3)}=\frac{x^3}{x+1}-\frac{x^3}{x+3}$$
For $\displaystyle\int\dfrac{x^3}{x+a}dx,$ set $x+a=h$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to solve $f''(e^{x} \cdot \sin x)$ How do I find the derivative $f''(e^{x} \cdot \sin x)$.
I start to find $f'(x)$ by using the product rule $f'(e^{x} \cdot \sin x) = e^{x} \cdot \sin x + e^{x} \cdot \cos x = e^{x}(\sin x + \cos x)$
Now when I have $f'(x)$ I use it to find $f''(x)$.
I split upp the problem by findi... | The last line is wrong :
$f''(x) = (e^{x} \sin x + e^{x} \cos x) + (e^{x} *(-\sin x) + e^{x} \cos x)$
$f''(x) = 2 e^{x} (\cos x)$
The $2 \cos x$ just came out of inattention !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is known about $x^m + y^m = z^n$ over $\mathbb{N}$ when $m,n \geq 2$ and $m \neq n$? So Fermat's Last Theorem resolves the question of positive integer solutions to $x^m + y^m = z^n$ when $m = n \geq 3$. But what about if $m \neq n$ and $m,n \geq 2$? Is anything general known about when there are positive integer ... | I attempt to solve the equation $x^3+y^3=z^n, x,y,z, n\in N \textrm{ and }n>3.$
Noting that $(1,1,2,1)$ is a solution and multiplying it by $2^{3k}$, we have
$$1^3+1^3=2^1 \Rightarrow \left(2^k\right)^{3 }+\left(2^k\right)^3=2^{3 k+1}$$
Therefore there are infinitely many solutions
$$(2^k,2^k, 2,3 k+1),$$
where $k\in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Simplification of geometric series. Can someone please help simplify this series?
$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$
In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$
However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$.
I understand if the pr... | \begin{align*}
\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k &= \frac{1}{2} + 2 \frac{1}{2^2} + 3 \frac{1}{2^3} + 4 \frac{1}{2^4} \dots\\
&= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + 2 \frac{1}{2^3} + 3 \frac{1}{2^4} + \dots\right)\\
&= \left(\frac{1}{2} + \frac{1}{2^2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the limit of a Riemann Sum The function is $f(x) = 1-x^2$. I'm stuck as I can't factor the expression in the last line to find the limit.
| The expression from your last line can be simplified:
\begin{align}
\frac{1}{n} \bigg[n-\frac{1}{n^2}\bigg(\frac{n(n+2)(2n+1)}{6}\bigg)\bigg]
& = \bigg[\frac{n}{n}-\frac{n(n+2)(2n+1)}{6n^3}\bigg] \\[0.1in]
& = 1-\frac{1}{6} \bigg( \frac{n+2}{n} \bigg)\bigg(\frac{2n+1}{n}\bigg) \\[0.1in]
& = 1-\frac{1}{6} \bigg( 1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15}\right) = \pi$ How to prove the following identity
$$\sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15}\right) = \pi$$
I am totally clueless in this one... | Here is a summary of the proof given by Bailey, Borwein, Borwein, and Plouffe in The Quest for Pi in only a few lines of integration.
To begin they note the following definite integrals as summations, $n=1,\ldots,7$:
$$ \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^8} dx =
\int_0^{\frac{1}{\sqrt{2}}} \sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
How to find the value of $m$ which is a power of a polynomial, when divided by a linear polinomial gives some remainder?
Q) The value of $m$ if $2x^m+x^3-3x^2-26$ leaves remainder of 226 when divided by $x-2$.
(1) 0
(2) 7
(3) 10
(4) all of these
How i solved it
let $p(x)=2x^m+x^3-3x^2-26$ and $g(x)=x-2=0$
therefore $... | It is $2\cdot 2^m=2^ {m+1}$ $(\ne (2^2)^m,$ which is the wrong point in your solution). So your equation should read
$$2^{m+1}=2^8$$ from where you get $m=7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Continuous Functions such that $f(0) = 1$ and $f(3x) - f(x) = x$ Just had a midterm with the following problem:
Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.
I was just curious how this would end up.
During the exam I tried setting $f(1) = c$, and end up gettin... | We have: $f(x) - f\left(\frac{x}{3}\right) = \dfrac{x}{3}$
$f\left(\frac{x}{3}\right) - f\left(\frac{x}{9}\right) = \dfrac{x}{9}$
.....
$f\left(\frac{x}{3^{n-1}}\right) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3^n}$.
Add the above equations:
$f(x) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3}\cdot \left(1 + \frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrals: Partial Fractions $$ \int \frac{x^2-x+12}{x^3+3x} $$
I factored the denominator to get $ x(x^2+3) $. I then seperated the x and the $x^2+3$ into the partials $\frac{A}{x}$ and $\frac{Bx+C}{x^2+3}$.
After combining the two, I came up with $$ \frac{A(x^2+3)+ x(Bx+C)}{x(x^2+3)}$$
This is where I'm stuck. I'm as... | You are almots done. As you wrote $$ \frac{x^2-x+12}{x^3+3x}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$$ then $${x^2-x+12}=A(x^2+3)+(Bx+C)x$$Expand the rhs, so $${x^2-x+12}=(A+B)x^2+Cx+3A$$ Identify the coefficients for a given power of $x$; so $$A+B=1$$ $$C=-1$$ $$3A=12$$ So, $A=4$, $B=-3$,$C=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | We have $a+ib=2w$ where $w^3=1$
Taking modulus on either sides $\sqrt{a^2+b^2}=2|w|$
Directly taking modulus on either sides on $(a+ib)^3=8,$
we get $(\sqrt{a^2+b^2})^3=8\implies (a^2+b^2)^3=8^2=(2^3)^2=2^6$
As $a^2+b^2>0, a^2+b^2=\sqrt[3]{2^6}=(2^6)^{\frac13}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
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Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | If you want to use the power series way to do it(although I think it's overkill), by convex property of $f(x)=x^n$ we have $\frac{(a^n+b^n+c^n)}{3}\geq (\frac{a+b+c}{3})^n$ or equivalently $(a^n+b^n+c^n)\geq \frac{(a+b+c)^n}{3^{n-1}}$, where $n$ is positive integer.
Therefore your left hand side(series expansion) is gr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Solving the ODE $(x^2 - 1) y''- 2xy' + 2y = (x^2 - 1)^2$ I want to solve this ODE:
$$(x^2 - 1)y'' - 2xy' + 2y = (x^2 - 1)^2.$$
I found out that $y_1 = x$ and $y_2 = x^2+1$ are solutions of the associated homogeneous equation, $x$ by inspecting, and $x^2+1$ by multiplying $x$ by a function and solving. Now, I know ... | I have redone the calculations and solved the problem. I'll leave my solution here, in case it helps someone. This time, my mistake was not writing the equation as: $$y'' - \frac{2x}{x^2 - 1}y' + \frac{2}{x^2 - 1}y = (x^2 - 1)$$
before applying the method.Really, $y_1 = x$ and $y_2 = x^2+1$ are solutions for the homoge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inverse Trigonometric Integrals How to calculate the value of the integrals
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx
$$
and
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$
| The first integral is not that difficult to evaluate. Note that
$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_1^{\infty} dx \frac{\arctan^2{x}}{x^2}\\ &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Average order of Eulers totient function squared I was wondering if one has a nice asymptotic formula for the sum
$$\sum_{n\le x} \phi(n)^2$$
and if so, how does one calculate it. I know that one has $\sum_{n\le x} \phi(n) = \frac{3}{\pi^2}x^2+O(x\log x)$, but I can't seem to find similar results for other powers of $\... | We can get the first term of the asymptotics with the Wiener-Ikehara
theorem. For this purpose we need to evaluate
$$L(s) = \sum_{n\ge 1} \frac{\varphi(n)^2}{n^s}.$$
Recall that
$$ n = \prod_p p^v \quad\text{implies}\quad
\varphi(n) = \prod_{p|n} (p^v - p^{v-1})$$
which we may either quote or derive using the fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Limit calculation and discontinuity Having a function, which has a polynomial in the denominator like:
$$ \lim_{x \to 2}\,\dfrac{x+3}{x-2} $$
We see there is a discontinuity at x=2, because it sets the denominator to 0. But multiplying by its conjugate we get:
$$ \dfrac{x+3}{x-2}\cdot\dfrac{x-2}{x-2} = \dfrac{(x+3)(x-2... | The mistake you are making is that $\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to -2}\frac{x+2}{(x+2)(x^2-2x+4)} = \frac{0}{0}$ is not necessarily discontinuous. It is a so called indeterminant form. Typically, you have to do something algebraic to "get to" the actual limit, similar to what you did by factoring and canc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the coordinates of the expression $(\cos x + \sin x)^3$ in the basis {$1, \sin x, \cos x, \sin 2x, \cos 2x, \sin3x, \cos 3x$} I'm quite stumped here. I expanded out $(\cos x + \sin x)^3$ to $\sin^3x + \cos^3x + 3\sin x\cos^2x + 3\sin^2x\cos x$ And I've tried the trig identities for $\sin 3x = 3\sin x - 4\sin^3x$
I... | When you expand, you get $\cos^3 x+3\cos^2 x\sin x+3\cos x\sin^2 x+\sin^3 x$.
Note that $3\cos^2 x\sin x=3(1-\sin^2 x)\sin x=3\sin x-3\sin^3 x$, and similarly
$3\cos x\sin^2 x=3\cos x-3\cos^3 x$.
So our expression is equal to $3\cos x+3\sin x-2\cos^3 x-2\sin^3 x$.
You know how to express $\sin^3 x$ in terms of $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A combinatorics and algebra problem I'm given two numbers $n$ and $m$. I have to make $n$ with $m$ numbers (only taking their sum). For example, if $n=6$, $m=3$, $6$ is formed with $3$ numbers in the following
way:
*
*$1+1+4=6$
*$2+2+2=6$
*$1+2+3=6$
For every element of the sets above, if I use $C(r,2)$, $0<r<n... | It suffices to consider two summands $a,b$ and verify that $C(a,2)+C(b,2)<C(1,2)+C(a+b-1,2)$ whenever $a,b>1$. Indeed,
$$ C(a,2)+C(b,2) = \frac12(a^2-a+b^2-b)$$
$$ C(1,2)+C(a+b-1,2) = \frac12(a+b-1)(a+b-2)$$
and the difference is after simplification
$$C(1,2)+C(a+b-1,2) -(C(a,2)+C(b,2)) = \frac12(ab-a-b+1)=\frac12(a-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form of $I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx$ Does the integral below have a closed-form:
$$I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx,$$
where $\tan^{-1} (\cdot)$ is inverse tangent function.
Neither Wol... | Step 1: Introducing an extra parameter
Define
$$\phi(\alpha)=\int^\frac{\pi}{2}_0\arctan\left(\frac{\sin{x}}{\cos{x}-\frac{1}{\alpha}}\right)\cot{x}\ {\rm d}x$$
Differentiating yields
\begin{align}
\phi'(\alpha)
=&-\int^\frac{\pi}{2}_0\frac{\cos{x}}{1-2\alpha\cos{x}+\alpha^2}{\rm d}x\\
=&\frac{\pi}{4\alpha}-\frac{1+\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.
| We have these identities. (Do you see why these are true? Hint: FOIL. Or, if you learned $``{(a+b)^2=a^2+2ab+b^2}
``$, use that.)
$$\left(a+\frac1a\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
$$\left(a^2+\frac1{a^2}\right)^2=\left(a^4+\frac1{a^4}\right)+2$$
Now, plug in $(a+1/a)=\sqrt3$.
$$\left(\sqrt3\right)^2=\left(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How can I show that $n! \leqslant \left(\frac{n+1}{2}\right)^n$?
Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$
I know that it can be done by induction but I always find line where I do not know what to do next.
| Hint:
$$
(n!)^2 = 1\times n \times 2\times (n-1) \times \dots
= \prod_{k=1}^n k(n+1-k)
$$
then use
$$
k\times (n+1 -k) =
\left(\frac {n+1}2 + \frac {n+1}2 - k\right)\left(\frac {n+1}2 - \frac {n+1}2 + k\right)
\\= \left(\frac {n+1}2\right)^2 - \left( \frac {n+1}2 - k\right)^2
\le \left(\frac {n+1}2\right)^2
$$
to get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions Have been working on this for years. Need a system which proves that there exists a number $C$ which has certain properties. I will give a specific example, but am looking for a system which could possibly be generalized.
Find (or prove th... | Well if $ax^2+bx+c$ has integer roots, then $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ is an integer. Since in this case $a=1$ and $b=-47$ and we want to find $C$, $$\frac{47\pm \sqrt{(-47)^2+4C}}{2}=\frac{47\pm \sqrt{2209+4C}}{2}$$ is an integer. $47$ is odd. Since the fraction is an integer, the numerator must be divisible by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
} |
Finding the range of equation. Any tricks? I m working on the following problem
For real numbers $a,b$, if $a+ab+b=3$, then find the range of $m=a-ab+b$. Is there any inequalities here to use?
| You are given $a+ab+b=3$, which you can make $(a+1)(b+1)=4$. Let $x=a+1, y=b+1$, so $xy=4$ $m=a-ab+b=x-1-(x-1)(y-1)+y-1=-xy+2x+2y-3=2x+2y-7=2x+\frac 8x-7$
Can you find the range of that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary proof of an inequality with $e^x$ when $|x|<1$. Assume $|x| <1 $ and we already know $0 \leq e^x - 1 - x$. Note that last inequality was god given, we know it is true, but we do not know how it was proved. Can we deduce from here that $e^x - 1 - x \leq x^2$ without using derivatives etc., just only with elem... | If $f$ is convex and differentiable, we have $f(x)-f(y) \ge f'(y) (x-y)$ for all $x,y$.
The function $\exp$ is convex and differentiable, so we have $e^x-1 \ge x$ for all $x$.
Let $\phi(x) =e^x-1-x$, then $\phi(x) \ge 0$, $\phi(0)=\phi'(0) = 0$ and $\phi''(x) = e^x$.
The mean value theorem gives $\phi(x) = \phi(0) + \p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the sum. Evalute the following sum: $ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$.
I tried doing it but I keep getting the wrong answer. I've used known sums to solve it.
| Reverse the sequence and add term-wise to the original:
$$ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$ n + n + (n-1) + . . . . + 6 + 6 + 5 + 4 + 4 + 3 + 2 + 2 + 1$$
All triples will sum to $(n+1)+(n+2)+(n+1)=3n+4$, and each number is counted twice. But, beware, $n$ must be even and there are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Limit proof for rational function $\frac{1}{x}$ A while ago I posted another one like this with a incorrect approach, please see this one!
Is this an accurate proof for limits for the function $\frac{1}{x}$
$\displaystyle \lim_{x\to1} \frac{1}{x} = \frac{1}{1} = 1$
Using $\epsilon-\delta$ So,
$\displaystyle \frac{|x-1... | I think you have the right idea, but it is a bit unorganized, separate the scratch work from the proof. This is how I would approach it.
Let $\epsilon>0$ be arbitrary, let $\delta=\min(1/2,\epsilon/2)$ such that $\vert x−1\vert<\delta$, then $$\left\vert\frac{1}{x}−1\right\vert= \frac{\vert x−1\vert}{\vert x\vert}< ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to find the solution to this system of equations Given the system of equations:
$
xy+xz=54+x^2 \\
yx+yz=64+y^2 \\
zx+zy=70+z^2 $
Need to find all of the solutions of $ x,y$ and $z$.
Tried to sum up all three equations but got stuck with nothing to factorize.
| Adding, $x^{2}$ ,$y^{2}$, & $z^{2}$ to 1st,2nd & 3rd equations respectively we get,
$x(x+y+z)=2x^{2}+54$
$y(x+y+z)=2y^{2}+64$
$z(x+y+z)=2z^{2}+70$
So, $\dfrac{2x^{2}+54}{x}=\dfrac{2y^{2}+64}{y}=\dfrac{2z^{2}+70}{z}=x+y+z=k(say)$
From these, $2x^{2}-kx+54=0,2y^{2}-ky+64=0,2z^{2}-kz+70=0$.
So, $x=\dfrac{k\pm \sqrt{k^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the matrix $A$ with this condition.... If $\theta \in\mathbb{R}\setminus\{k\pi, k\in\mathbb{Z}\}$ and $A\in M_{2\times 2}(\mathbb{C})$ such that $$A^{-1} \begin{pmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta \\
\end{pmatrix} A= \begin{pmatrix}
e^{i\theta} & 0 \\
... | This is not the 'right' way to answer the question, but if you use your idea, and note that you can scale $A$ so its determinant $ad-bc$ is 1 (giving a nice expression for the inverse), and multiply everything out, you get
\begin{align*}
\begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2 Prove that:
If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$
From Barnard & Child's "Higher Algebra".
I know that the highest power of a prime $p$ contained in $N$! is
$$ \left... | It is an easy exercise to show that for all real numbers $x$ we have
$$
\lfloor 3x\rfloor=\lfloor x\rfloor+\lfloor x+\frac13\rfloor+\lfloor x+\frac23\rfloor.
$$
Thus for all $n$ and all prime powers $p^t\ge3$ we have
$$
\begin{aligned}
\lfloor \frac{3n}{p^t}\rfloor&=\lfloor \frac{n}{p^t}\rfloor+
\lfloor \frac{n}{p^t}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Proving the following formula of $\ln(2)$ Proving the following formula of $\ln2$
$$\ln2=\frac{1}{2}\left(\frac{3}{2}\right)-\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{6}\left(\frac{9}{8}\right)-\frac{1}{8}\left(\frac{15}{16}\right)+\frac{1}{10}\left(\frac{33}{32}\right)-\frac{1}{12}\left(\frac{63}{64}\right)\cdots$... | Your formula is :
\begin{align}
S&:=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}\frac{2^n-(-1)^n}{2^n}\\
&=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}+\sum_{n=1}^\infty \frac{(-1)^{2n}}{2\,n}\frac 1{2^n}\\
&=-\frac 12\sum_{n=1}^\infty \frac{(-1)^n}{n}+\frac 12\sum_{n=1}^\infty \frac 1{n} \left(\frac 12\right)^n\\
&=\frac 12\log(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Question on modulo Find the last two digits of $3^{2002}$.
How should I approach this question using modulo? I obtained 09 as my answer however the given answer was 43.
My method was as follows:
$2002\:=\:8\cdot 250+2$
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
The last two digits is the remainder when ... | Observe that $\text{gcd(3,100)} = 1$, by Euler totient formula: $3^{\phi(100)} \cong 1(\mod 100)$. But $\phi (100) = 100\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{5}\right) = 40$. Thus: $3^{40} \cong 1(\mod 100)$. So $3^{2002} = (3^{40})^{50}\cdot 3^2 \cong 1\cdot 9(\mod 100) \cong 9(\mod 100)$. Thus the remainder i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
computation on hyper surface $z=x^2+y^2$ I have problem with following exercise
Consider the hypersurface $M$ parametrized by $z=x^2+y^2$.
Endow this with the Riemannian metric induced from the $\mathbb{R}^3$.
Compute the sectional curvature.
| This method was using Monge surface parametrization, which was introduced by @THW. in my former question.
Endowment of $\mathbb{R}^3$, we can do Gauss approach as follows.
\begin{align}
\mathbf{x}(r,\theta)=(f(r),r\cos(\theta), r\sin(\theta))
\end{align}
where $f(r)=r^2=z$.
\begin{align}
&\mathbf{x}(r,\theta)=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
| $$3\left(2x-1\right)^{2}=147\iff\left(2x-1\right)^{2}-49=0\iff$$$$\left[\left(2x-1\right)-7\right]\left[\left(2x-1\right)+7\right]=0\iff\left(2x-8\right)\left(2x+6\right)=0$$
A product $p\times q$ equals $0$ if and only if $p=0$ or $q=0$.
Applying that here gives $2x-8=0$ or $2x+6=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Express this sum of radicals as an integer? I have read somewhere that the radical $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ and I don't understand it. How do you solve this(when the RHS is unknown)?
| Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Write $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$. Then
$$
x^3=
a+b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2}
=
a+b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})
=4-3x
$$
Now, the derivative of $x^3+3x-4$ is always positive, which means there is only one real root. By inspection, $x=1$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Discriminant of quadratic formula I noticed something that i find interesting about the quadratic formula. I hope someone can explain it to me.
$f(x) = Ax^2 + Bx + c$
$f'(x) = 2Ax + B$
If I say that $f'(x) = 0$ I get that $x = -\frac{B}{2A}$. From this I get that $f(-\frac{B}{2A})$ is a max/min point.
The quadratic fo... | I follow you up to
$$f\left(-\frac{B}{2A}\right) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} + C = \frac{4AC - B^2}{4A}.$$
But the square root of this gives you a $2\sqrt{A}$ in the denominator, so it's not quite what you have.
Here's the connection I see. In order to get real roots, $B^2 - 4AC$ needs to be non-negative.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find this sum $S$ using Real-analysis methods only $$S = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$
I have tried a lot and failed, any help is appreciated. $H_k$ is the harmonic number.
Thanks (real method only please)
| We have:
$$\frac{1}{2}\log^2(1-x)=\sum_{k=1}^{+\infty}\frac{H_k}{k+1}x^{k+1}$$
but since $\int_{0}^{1} x^n\log^2 x\,dx = \frac{2}{(n+1)^3}$ it follows that:
$$\begin{eqnarray*} S &=& \frac{1}{2}\int_{0}^{1} \log^2(1-x)\log^2 x\,dx=\frac{1}{2}\left.\frac{\partial^4}{\partial^2 a\,\partial^2 b}B(a,b)\,\right|_{a,b=1}\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A limit coming from high school I'm sure you guys can do this in many ways, using integrals, Taylor series, but I need a
high school way, like the use of a simple squeeze theorem. Can we get such a way?
$$\lim_{n\to\infty}\left(1+\log\left(1+\frac{1}{n^2}\right)\right)\left(1+\log\left(1+\frac{2}{n^2}\right)\right)\cd... | Approach 1
We can use the inequality
$$
x-\frac12x^2\le\log(1+x)\le x\tag{1}
$$
for $x\ge0$, to good result.
First, note that
$$
\begin{align}
\color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}
&=\frac{n^2+n}{2n^2}\\
&=\frac12+\frac1{2n}\tag{2}
\end{align}
$$
Next, since $\frac{k}{n^2}\le\frac1n$, we get
$$
\begin{align}
\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
How find this limits $\lim_{N\to\infty}\sum_{n=1}^{N}\frac{1}{(n+1)}\sum_{i=1}^{n}\frac{1}{i(n+1-i)}$ Find this limits
$$\lim_{N\to\infty}\sum_{n=1}^{N}\left(\dfrac{1}{(n+1)}\left(\dfrac{1}{1\cdot n}+\dfrac{1}{2\cdot(n-1)}+\cdots+\dfrac{1}{(n-1)\cdot 2}+\dfrac{1}{n\cdot 1}\right)\right)$$
I know this
$$\dfrac{1}{1\cd... | $\displaystyle \begin{align} \sum_{n=1}^{\infty}\dfrac{H_n}{(n+1)^2} &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n+1}}{(n+1)^2} - \frac{1}{(n+1)^3}\right) \\ &= \sum_{n=2}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= 2\zeta(3) - \zeta(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
| It's quite easy to make that polynomial a square. Let,
$$P = (4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2$$
Then,
$$P = (a d)^2,\quad \text{if}\, a = 2b\tag{1}$$
and all other variables are free. Similarly,
$$P = (b c)^2,\quad \text{if}\, c = 2d\tag{2}$$
$$P = (2b d)^2,\quad \text{if}\, a = 2b,\, \text{or}\, c = 2d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Number of length $8$ binary strings with no consecutive $0$'s How many $8$ bit strings are there with no consecutive $0$'s?
I just sat an exam, and the only question I think I got wrong was the above(The decider for a high-distinction or a distinction :SSS)
I took Number with consecutive $0$'s
1 zero $0$
2 zeros $\frac... | A bit string with no consecutive zeros is composed of 1 and 10 units, but may also begin with one 0 unit. (That is, every zero is either preceeded by a one or is at the start of the string).
Let $N_{x,y}$ count the permutations of a bit string composed of $x$ 1 and $y$ 10 units.
$$N_{x,y} = \frac{(x+y)!}{x!\, y!}$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
convergence of this series $a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$?? The sequence with nth term is given as
$a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$
this sequence will converge to??
what I did is:
for this we can use sandwich theorem i.e $f(x)\leq g(x)\leq h(x)$ then $lim h(x)=lim g(x)=lim f(x)... | Simply note that,
$\displaystyle \begin{align} a_n = \sum\limits_{k=n}^{2n}\frac{1}{k} = \sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n-1}\frac{1}{k} &= \sum\limits_{k=1}^{2n}\frac{1}{k} - 2\sum\limits_{k=1}^{n-1}\frac{1}{2k} \\ &= \left(\sum\limits_{k=1}^{n}\frac{1}{2k-1} + \sum\limits_{k=1}^{n}\frac{1}{2k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the maximum area of polygon? Firstly , I divided the polygon to three triangles and I used the Heron's formula to find the area of triangles which formed the polygon $$A=\sqrt{p(p-a)(p-b)(p-c)}$$.I couldn't find easily the relationship between $d$ and area for the three triangles. Now I need the distance $d$... |
$$ h1=\sqrt{5^2-(\frac{d}{2})^2} \\$$upper triangle area is $$ s1=\frac{1}{2}d\sqrt{5^2-(\frac{d}{2})^2}$$
$$h2=\sqrt{4^2-(\frac{d-5}{2})^2}\\$$so 2 lower triangle area =$$s2=2*\frac{1}{2}\frac{d-5}{2}\sqrt{4^2-(\frac{d-5}{2})^2} $$ rectangle area is =$$s3=5*\sqrt{4^2-(\frac{d-5}{2})^2} $$the problem reduce to max s=s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Disproving existence of real root in some interval for a quintic equation Disprove the statement: There is a real root of equation $\frac{1}{5}x^5+\frac{2}{3}x^3+2x=0$ on the interval (1,2).
I am not sure whether to prove by counter-example or by assuming the statement is true and then proving by contradiction.
| It's sufficient to show that:
*
*f(1) > 0
*For all n > 0: f(1 + n) > f(1)
Step #2 means that the function is increasing at all points to the right of x=1. From this you can say that f(x) doesn't equal zero (has a real root) in the range (1,2) because f(x) is always above 1 in that range.
Working out the algebra:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3. Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3.
How do I start with this question? Rationalising the denominator? But how would I expand it?
| $$\frac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}=\frac{(x-3)^2\sqrt{(x-3)^2+4}+2(x-3)^2}{(x-3)^2}$$ $$\frac{(x-3)^2\sqrt{(x-3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers) Question is to Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers) .
I took rectangle with sides $x$ and $y$ .
Since Perimeter is constant so i took $2(x+y) = 2k$ , ... | A simpler approach:
If the perimeter is
$2a$,
the sides are
$x$ and $a-x$.
If $d$ is the diagonal,
$\begin{array}\\
d^2
&=x^2+(a-x)^2\\
&=x^2+a^2-2ax+x^2\\
&=2x^2-2ax+a^2\\
&=2(x^2-ax+a^2/2)\\
&=2(x^2-ax+a^2/4+a^2/4)
\text{ (completing the square)}\\
&=2((x-a/2)^2+a^2/4)\\
\end{array}
$
This is maximized when
$(x-a/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx$ How do I integrate
\begin{equation}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx,
\end{equation}
which has arisen from a problem I'm working on? I've noticed I can do the following:
\begin{align}
\int_0^1\sqrt{-x^6+x^4-x^2+1}\:dx & =\int_0^1\sqrt{\left(-x^4\right)\left(x^2-1\... | $\int_0^1\sqrt{-x^6+x^4-x^2+1}~dx$
$=\int_0^1\sqrt{x^4(1-x^2)+1-x^2}~dx$
$=\int_0^1\sqrt{(1-x^2)(x^4+1)}~dx$
$=\int_0^\frac{\pi}{2}\sqrt{(1-\sin^2x)(\sin^4x+1)}~d(\sin x)$
$=\int_0^\frac{\pi}{2}(\cos^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}(1-\sin^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $|\alpha|\leq 1$ and $|\beta|\leq 1$, prove that $|\alpha+\beta|\leq |1+\overline{\alpha}\beta|$ Note $\alpha$ and $\beta$ are complex numbers and $\overline{\alpha}$ is the conjugate of $\alpha$. I've tried using variations of the triangle inequality and I couldn't find anything to work.
| $|\alpha + \beta|^2 = |(x_1+x_2) + (y_1+y_2)i|^2 = (x_1+x_2)^2+(y_1+y_2)^2$
$|1+\bar{\alpha}\beta|^2 = |1+(x_1-y_1i)(x_2+y_2i)|^2 = |1+x_1x_2+y_1y_2 + (x_1y_2-x_2y_1)i|^2 = (1+x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2$. Thus we prove:
$x_1^2+x_2^2+2x_1x_2 + y_1^2+y_2^2+2y_1y_2 \leq 1+x_1^2x_2^2+y_1^2y_2^2 + 2x_1x_2 + 2y_1y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving binomial coefficients identity: $\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that:
$$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.$$
Tried proving it by induction but got stuck. Any help with p... | Let $E = \{1,2, \dots , n+1\}$. The number $\binom{n+1}{r+1}$ is the number of subsets $A$ of $E$ with $r + 1$ elements.
Classify these subsets $A$ according to their largest element $b$, which can be any number among $r + 1$, $r + 2$, ..., $n + 1$. The number of $(r+1)$-element subsets of $E$ with largest element $b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve $x^4 - 2x^3 + x = y^4 + 3y^2 + y \wedge (x,y) \in \mathbb{Z}^2$ I want to solve equation $x^4 - 2x^3 + x = y^4 + 3y^2 + y$ in integers. The task comes from the LXVI Polish Mathematical Olympiad. Series with this task ended twenty days ago, so it is legal to talk about it.
It is of course equal to $x(x-1)(x^2-x-1... | Hint: $x^4-2x^3+x=y^4+3y^2+y\let\leq\leqslant\let\geq\geqslant$ is equivalent to $(2x^2-2x-1)^2-(2y^2+3)^2=4y-8$.
We have $2x^2-2x-1+2y^2+3\mid4y-8$, hence $|2x^2-2x-1+2y^2+3|\leq|4y-8|$. As $x^2\geq x$ for $x\in\mathbb Z$ this gives $y^2+1\leq|2y-4|$. For $y\geq2$ this is $(y-1)^2\leq-5$; for $y\leq2$ this is $(y+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
| Divide the numerator & the denominator by $\cos^4x$ to find
$$I=\int\frac{\sec^4x}{\tan^4x+1+\tan^2x}dx$$
Putting $\tan x=y,$
$$I=\int\frac{1+y^2}{1+y^2+y^4}dy=\int\frac{1+\dfrac1{y^2}}{\dfrac1{y^2}+1+y^2}dy$$
$$=\int\frac{1+\dfrac1{y^2}}{\left(y-\dfrac1y\right)^2+3}dy $$
Finally set, $u=y-\dfrac1y=\dfrac{y^2-1}y=-2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
| It's Quite Simple..
You know, $a^2 - b^2 = (a-b)(a+b)$
Now, $(x-5)^2 - 4 = (x-5)^2-2^2$
Let $a=x-5 $ and $b=2$
So, $a^2-b^2 = (x-5)^2-2^2 = (a-b)(a+b)=(x-5-2)(x-5+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$ This induction problem is giving me a pretty hard time:
$$\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
I am struggling because my math teacher explained us that in this case ($n^2$) when... | Hint: What is $\frac{4(n+1)}{2(n+1)+1}-\frac{4n}{2n+1}$? (What is the corresponding difference on the other side of the inequality when you move from $n$ to $n+1$?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| Do you know that $(\cos^{-1}t)'=-\frac{1}{1+t^2}$?
So $$f'(x)=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot (\frac{1-x}{1+x})'
=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot \frac{-(1+x)-(1-x)}{(1+x)^2} $$
Is it okay?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ How can we prove that:
$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$
| Different approach:
Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$
By using the identity $4ab=(a+b)^2-(a-b)^2$ and setting $a=\ln(1-x)$ and $b=\ln(1+x)$ we have
$$4I=\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx-\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx$$
The fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
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Limit of a difference equation Given $y_{k+1} = 1 + \sqrt{y_k}$ for $k \geq 0$ and $y_0 = 0$, we have a limit of the form $L = \lim_{k \rightarrow \infty} y_k = 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ... \sqrt{1 + \sqrt{2}}}}}$.
Apparently this can be rewritten as $L = \frac{3 + \sqrt{5}}{2}$, but I am at a loss for how t... | To show that the limit actually exists, we can proceed as follows:
Let
$$
\textstyle a_n=\overbrace{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots+\sqrt{1+\sqrt2}}}}}^{n\ 1\text{s}}\tag{1}
$$
so that
$$
(a_n-1)^2=a_{n-1}\tag{2}
$$
Subtracting $(2)$ substituting $n\mapsto n-1$ from $(2)$ and dividing by $(a_n-1)+(a_{n-1}-1)$, gives
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to calculate $\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$? Question :
$$\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$$
Here I tried $\tan x$ and $\sin x$ expansion in numerator and denominator which are as follows :
$$\tan x =x +\frac{x^3}{3}+\frac{2x^5}{15} \cdots$$
$$\sin ... | It looks like a bad guy. I used Wolfram Mathematica to compute
$$
\tan^3 x -x^3 = x^5+\frac{11}{15}x^7 + O(x^8)
$$
and
$$
e^{\tan^3 x -x^3}=1+x^5+\frac{11}{15}x^7 + O(x^8).
$$
Therefore
$$
e^{\tan^3 x}-e^{x^3}=x^5+\frac{11}{15}x^7 + O(x^8).
$$
Since
$$
\log(1+x^3 \sin^2 x) = x^5 +O(x^6),
$$
the limit exists and is eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluting $\tan15°$ using difference formula
Evalute $\tan15°$ using difference formula
Steps I took:
$$\begin{align}
\tan(45-30)&=\frac { \tan(45)-\tan(30) }{ 1+\tan(45)\tan(30) }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \... | $$\frac{1-\dfrac1{\sqrt3}}{1+\dfrac1{\sqrt3}}=\frac{\dfrac{\sqrt3-1}{\sqrt3}}{\dfrac{\sqrt3+1}{\sqrt3}}$$
$$=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\ph... | Hint: Divide both sides of $8bc=a[4b^2 + (b^2-c^2)^2]$ by $bc$. You will end up with this:
$$8 = a[4\frac{b}{c}+bc(\frac{b}{c}-\frac{c}{b})^2]$$
Take the RHS and you can prove that it is equal to 8.
Process: Calculate $\frac{b}{c}$ from the given equations.
$$b = \frac{\sin (\theta + \phi)}{\cos \theta. \cos \phi}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
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Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2... | HINT:
If the $r$th term
$$T_r=\frac1{2r}-\frac1{r+1}+\frac1{2(r+2)}$$
$$2T_r=\frac1r-\frac2{r+1}+\frac1{(r+2)}=\left(\underbrace{\frac1r-\frac1{r+1}}\right)-\left(\underbrace{\frac1{r+1}-\frac1{r+2}}\right)$$
Recognize the two Telescoping series
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Number of solutions to equation, range restrictions per variable Find the number of solutions of the equation $x_1+x_2+x_3+x_4=15$ where variables are constrained as follows:
(a) Each $x_i \geq 2.$
(b) $1 \leq x_1 \leq 3$ , $0 \leq x_2$ , $3 \leq x_3 \leq 5$, $2 \leq x_4 \leq 6$
I believe I understand part A. I can fix... | By similar reasoning to (a) we can simplify the problem to:
\begin{eqnarray*}
x_1+x_2+x_3+x_4 &=& 9 \qquad\qquad\text{(*)} \\
\text{with } && 0 \leq x_1 \leq 2 \\
&& 0 \leq x_2 \\
&& 0 \leq x_3 \leq 2 \\
&& 0 \leq x_4 \leq 4.
\end{eqnarray*}
Define sets
\begin{eqnarray*}
S\;\, &=& \{\text{all solutions to (*) without u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent How do I prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent for every x in real numbers except for $x=0$?
I tried using the ratio test, but it doesn't seem to be conclusi... | You could use the limit comparison test.
Use $\displaystyle b_n = \frac{1}{n^2}$
Let $\displaystyle a_n = \frac{1}{n^2x^2 + 1}$
$\displaystyle \frac{a_n}{b_n} = \frac{n^2}{n^2x^2 + 1}$
$x$ is a constant $x \ne 0$
$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{1}{x^2}$
Provided $x \ne 0, x^2 > 0, \frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding $\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$ The WolfarmAlpha couldn't give me the sum of $$\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$$ therefore I thought that this problem is difficult so I used my calculator to get $(1/24)$
Is this value right or not?
If this right, why the Wol... | Consider the sum:
$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} $$
This sum is a bit easier than it looks, if you know the residue theorem. The main observation is that the numerator either takes the value $\sqrt{3}/2$, $-\sqrt{3}/2$, or $0$. Then you may rewrite the sum as
$$\frac{\sqrt{3}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ... | Another way using Taylor series $$A=\sqrt{n^2 + 1} - n=n\Big(\sqrt{1+\frac{1}{n^2}}-1\Big)$$ Remembert that, for samm values of $x$, $$\sqrt{1+x} =1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Replace $x$ by $\frac{1}{n^2}$ so $$\sqrt{1+\frac{1}{n^2}}-1=1+\frac{1}{2 n^2}-\frac{1}{8}
\left(\frac{1}{n^2}\right)^2+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | let f(n)=3n-2
let s(n)=[f(1)+f(2)+f(3)+...+f(n-1)+f(n)]
s(n) =(3(1)-2)+(3(2)-2)+(3(3)-2)+...+(3(n-1)-2)+(3(n)-2)
s(n) =(3(1) + 3(2) + 3(3) +...+ 3(n-1) + 3(n) )-2n
s(n) = 3(1 + 2 + 3 +...+ (n-1) + n )-2n
s(n) =3(sum of all #s from 1 to n) -2n
s(n) =3(sum of all #s from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 7
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Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Find the number of elements of order $1$, order $2$, order $3$, and order $5$. But this scenario can't happen. Why not?
Let $G$ be a group... | My way: you deduced that for such a $G$ it must be $n_5=6$. These six subgroups are cyclic of order $5$ (every group of prime order is necessarely cyclic). Let $K_1,\dots,K_6$ be these $5$-Sylow. It must be $K_i\cap K_j=1$ for $i\neq j$ and every element different from the identity must has order 5. Hence $G$ contains ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
| Hint
Let $$A=\left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}$$ and take logarithms of both sides $$\log(A)=9x^2 \log\Big(\frac{x^2+2}{x^2-4}\Big)=9x^2\log\Big(\frac{1+\frac{2}{x^2}}{1-\frac{4}{x^2}}\Big)$$ $$\log(A)=9x^2 \Big[\log \left(1+\frac{2}{x^2}\right)-\log \left(1-\frac{4}{x^2}\right) \Big]$$ Now, use the fact that,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| The hard way, by Taylor:
First establish the derivatives,
$$\begin{align}
f(x)(x-1)&=x^6-1\\
f'(x)(x-1)+f(x)&=6x^5\\
f''(x)(x-1)+2f'(x)&=30x^4\\
f'''(x)(x-1)+3f''(x)&=120x^3\\
f''''(x)(x-1)+4f'''(x)&=360x^2\\
f'''''(x)(x-1)+5f''''(x)&=720x\\
f''''''(x)(x-1)+6f'''''(x)&=720\\
f'''''''(x)(x-1)+7f''''''(x)&=0\\
\dots\\
f^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 9
} |
Weak principle of induction for $5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$
Show that
$$5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$$
Proving the base case $n(1)$:
$5(1)= \frac{5(1)(1+1)}{2}$
$5 = \frac{5(2)}{2}$
$5 = 5$
Induction hypothesis:
$n = k$
$5+10+15+\ldots+5k = \frac{5k(k+1)}{2}$
Induction step (adding $k+1$):
$5+10+15... | When you plug in $n=k+1$ you have that
$$5+10+\cdots+5k+5(k+1)\not=5+10+\cdots+5k+5k+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What is the smallest possible value of their sum? The product of two positive numbers is 36. What is the smallest possible value of their sum?
so far I got
$$xy=36$$
$$y=\frac{36}{x}$$
| First note that
$$f(x)=x+\frac{36}{x}=x+36x^{-1}$$
$$\frac{d}{dx}[f(x)]=1-36x^{-2}=1-\frac{36}{x^2}$$
So now let's find the critical points
$$\frac{d}{dx}[f(x)]=0 $$
$$ 1-\frac{36}{x^2}=0 $$
$$ 1=\frac{36}{x^2} $$
$$ x^2=36= (\pm 6)^2 $$
$$ x=\sqrt{(\pm 6)^2}=\pm 6 $$
Since we're only considering positive values, we'll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3... | Hint:-
$5\cdot2^n=x^2+y^2 \implies 5\cdot 2^{n+2}=(2x)^2+(2y)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} ... | Since $t=\tan \frac{x}{2}, \tan^{-1}t=\frac{x}{2}, x=2\tan^{-1}t$, and $dx=\frac{2}{1+t^2}dt$.
Then $\displaystyle\int\frac{1}{\sin x} dx=\int\frac{1+t^2}{2t}\cdot\frac{2}{1+t^2}dt=\int\frac{1}{t}dt=\ln|t|+C=\ln\left|\tan\frac{x}{2}\right|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | Hint $\ $ Times $2$ it is equivalent to $\,{\rm mod}\ 11\!:\ 8^n - 3^n\equiv 0,\,$ i.e.$\ 3^n\equiv (-3)^n\equiv \color{#c00}{(-1)^n} 3^n.\,$ Thus it suffices to prove $\ n$ even $\,\Rightarrow\, \color{#c00}{(-1)^n}\equiv 1,\,$ which is straightforward (by induction or not).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 5
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If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $... | Like Sameer Kailasa,
$\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \dfrac{5}{4}\ \ \ \ (1)$
and let $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) =x \ \ \ \ (2)$
$(1)-(2)\implies2(\sin \phi+\cos\phi)=\dfrac{5}{4}-x$
$\implies \sin \phi+\cos\phi=\dfrac{5-4x}8 $
Squaring we get $1+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of
$$
\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}
$$
My idea was to use ratio test:
$$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$
$... | How about computing the partial sums? For any $N>2$ we have:
$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$
hence:
$$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$
ensures convergence (towards $\frac{3}{2}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Simplify a quick sum of sines Simplify $\sin 2+\sin 4+\sin 6+\cdots+\sin 88$
I tried using the sum-to-product formulae, but it was messy, and I didn't know what else to do. Could I get a bit of help? Thanks.
| $$\sum_{n=1}^{N} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(N+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$
$$\sum_{n=1}^{44} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(44+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$
$$\sum_{n=1}^{44} {\sin(2 n)} = \frac{1}{2} \cot{\frac{2}{2}} - \frac{\cos{[(44+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
| Consider the contour integral
$$\oint_C dz \frac{(z^2-1) \log^2{z}}{1+z^4} $$
where $C$ is a keyhole contour about the positive real axis having an outer radius $R$ and an inner radius $\epsilon$. As $R \to \infty$ and $\epsilon \to 0$, the integral may be shown to be equal to
$$-i 4 \pi \int_0^{\infty} dx \frac{(x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
... | Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides.
$$ \begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 4
} |
Understanding Generating Function I have been looking at This Problem and Answer about generating functions. The problem asked for the generating function of: $$a_n=4a_{n-1}-4a_{n-2}+{n\choose 2}2^n+1$$ I understand how Ron Gordon simplified it to: $$\sum_{n=2}^{\infty} (a_n-4 a_{n-1}+4 a_{n-2}) x^n = \sum_{n=2}^{\inft... | Easiest is to define:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
then shift indices so there aren't subtractions there:
$$
a_{n + 2} = 4 a_{n + 1} - 4a_n + \binom{n + 2}{2} 2^{n + 2} + 1
$$
Multiply by $z^n$ and sum over $n \ge 0$:
$$
\sum_{n \ge 0} a_{n + 2} z^n
= 4 \sum_{n \ge 0} a_{n + 1} z^n - 4 \sum_{n \ge 0} a_n z^n
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables I have come as far as deducing that the denominator can be written as a geometric series. Hence, for $m=2$,
\begin{align*}
\int_{-\infty}^\infty \frac{1-x}{1-x^5} dx &= 2 \pi i ( B_1 + B_2 ) - \int_{C_R} \frac{1-z}{1-z^5} d... | There seems to be a tiny error in the last step, otherwise your solution is perfect.
You have that
\begin{align}
2\pi i \left(\frac{1-e^{i2\pi/5}}{-5e^{i8\pi/5}}+ \frac{1-e^{i4\pi/5}}{-5e^{i16\pi/5}} \right)
& = -\frac{2\pi i}{5}\left(e^{-i8\pi/5} - e^{-i6\pi/5} + e^{-i16\pi/5} - e^{-i12\pi/5}\right)\\
& = -\frac{2\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a solution to $a^4+(a+d)^4+(a+2d)^4+(a+3d)^4+\dots = z^4$? Is there a,
$$a^4+(a+d)^4+(a+2d)^4+(a+3d)^4+\dots = z^4\tag1$$
in non-zero integers? One can be familiar with,
$$31^3+33^3+35^3+37^3+39^3+41^3 = 66^3\tag2$$
I found that,
$$29^4+31^4+33^4+35^4+\dots+155^4 = 96104^2\tag3$$
which has $m=64$ addends. The ... | Let,
$$S(n)=a^4+(a+d)^4+\dots+(a+nd)^4\tag1$$
Then we can write
$$S(n)
=a^4+(a^4+4a^3d+6a^2d^2+4ad^3+d^4)+\dots+(a^4+4a^3dn+6a^2d^2n^2+4ad^3n^3+d^4n^4)\\
=(n+1)a^4+4a^3d(1+2+\dots+n)+6a^2d^2(1^2+2^2+\dots+n^2)+4ad^3(1^3+2^3+\dots+n^3)+d^4(1^4+2^4+\dots+n^4)$$
Using Faulhaber's formula for $F(x) = 1^x+2^x+\dots+n^x$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"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.