Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3... | Firstly, we can think about how multiples of $n$ appear in a sequence. For example, when $n=4$:
$$1,2,3,[4],5,6,7,[8],9,10, ...$$
Secondly, we are asked to determine the $k^{th}$ position in the sequence above, "skipping" multiples of $n$. For example when $k=5$, using the same sequence above:
$$1,2,3,[4],5, \langle 6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate... | $w_k$ are the roots of $$y^{15}-1=0$$
Now let $p_k=\dfrac1{(w_k-1)^3}, w_k-1=\sqrt[3]{\dfrac1{p_k}}$
Writing $p_k$ as $z$ and $w_k$ is a root of $y^{15}=1$
$$1=\left(1+\sqrt[3]{\dfrac1z}\right)^{15}$$
$$z^5=(1+\sqrt[3]z)^{15}$$
$$\iff - \sum_{r=0}^4z^r\binom{15}{3r}=z^{1/3}\sum_{r=0}^4 z^r\binom{15}{3r+1}+z^{2/3}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Compute the $\lim_{n\to\infty}\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}$ I know you should prove $f_n (x) = \frac1{1+x+\frac{x^n}n}$ converges uniformly and then the limit follows immediately however I’m struggling to prove the uniform convergence.
| $\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\lt\int_0^1\frac{\mathrm{d}x}{1+x}\\
&= \ln(2)\\
\end{array}
$
and
$\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\gt\int_0^1\frac{\mathrm{d}x}{1+x+\frac1{n}}\\
&=\ln(2+\frac1{n})-\ln(1+\frac1{n})\\
&=\ln(2)+\ln(1+\frac1{2n})-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum of Ranks of Two Complementary Matrices Suppose we have a binary matrix $A$, i.e, all the elements of $A$ are either $0$ or $1$. Let $B$ be the complementary matrix of $A$, i.e.,
$$
B_{ij}=
\begin{cases}
1,\textrm{ if }A_{ij}=0;\\
0,\textrm{ if }A_{ij}=1.
\end{cases}
$$
Then $A+B$ is a matrix with all entries being... | There is not enough information to tell.
For example, let $s=\operatorname{rank} B$. If $n=2$, we can take
*
*$A=\begin{bmatrix} 1&1\\0&0\end{bmatrix}$, then $r+s=2$
*$A=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, then $r+s=3$
*$A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$, then $r+s=4$
Similarly, with $n=3$:
*
*$A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Primes of the form $x^2 + 9 y^2$ and $x^2 + 12 y^2$ I've been studying primes of the form $x^2 + n y^2$ (where of course $x$, $y \in \mathbb{Z}$) and I noticed the following:
$$p = x^2 + 9 y^2 \iff p \equiv 1 \,\, (\textrm{mod} \,\, 12) \iff p = u^2 + 12 v^2$$
For example, $10^2 + 9 \cdot 7^2 = 541 = 23^2 + 12 \cdot 1^... | Not a concidence.
I will prove that if a prime p is of the form $x^2+9y^2$ then it must be of the form $a^2+12b^2$ and viceversa.
Case 1: $p=a^2+12b^2$ Prove $p=x^2+ 9y^2$ for some $x$ and $y$
$p=a^2+12b^2$ Note that $p$ is $\equiv 1\pmod{4}$ so -1 is also a quadratic residue, so $-1\times9=9$ also is one.
From Thue's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theor... | Integrand's concern about the case for $ \ xyz \ \ne \ 0 \ $ turns out to be unnecessary. If we solve the original Lagrange equations for $ \ \lambda \ $ , we obtain
$$ \left\{ \begin{array}{c}
2·(x^3-yz) \ = \ \lambda x \\ 2·(y^3-4xz) \ = \ \lambda y \\
2·(z^3- xy) \ = \ \lambda z
\end{array} \right. \ \ \Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that there's no ordered basis $E$ in which $T{x\choose y}={0\choose y}$ can be represented as $1\ 2\choose 2\ 4$ Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that:
$$T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix}0\\ y\end{pmatrix}$$
Show that there's no ordered-basis, $E$ such that:
$$[T]_E = \be... | With
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ y \end{pmatrix}, \tag 1$
we have
$T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \tag 2$
and
$T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 3$
which shows the eigenvalues of $T$ are $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
When converting a Riemann sum to an integral, how does one decide the bounds of the converted Riemann sum? Context problem: $$ \lim_{n\to \infty} \frac{1}{n} \sum_{r=1}^{r=2n} \frac{r}{\sqrt{r^2+n^2}}$$
So the thing messing me up is the '2n' in the upper of sum, I've already figured out that the function which this rie... | And now a synthesis of the other two answers that involves less (apparent) guessing than one and slightly more precision than the other.
Suppose what you are given is a Riemann sum similar to the ones you performed by hand when introduced to the concept of Riemann sums (left Riemann sum, right Riemann sum, midpoint ...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $N = y^2+1$ then the continued fraction of $\sqrt(N)$ has period 1. I am reading a book that says that if $N = y^2+1$ then the continued fraction of $\sqrt{N}$ has period 1, i.e $\sqrt{N} = [q_0;\overline{q_1}]$ or similarly
$$x=q_0+\cfrac{1}{q_1+\cfrac{1}{q_1+\cfrac{1}{ q_1+ \cdots}}}$$
What I've tried to do was t... | $\begin{array}\\
\sqrt{y^2+1}-y
&=(\sqrt{y^2+1}-y)\dfrac{\sqrt{y^2+1}+y}{\sqrt{y^2+1}+y}\\
&=\dfrac{1}{\sqrt{y^2+1}+y}\\
\end{array}
$
so
$\sqrt{y^2+1}
=y+\dfrac{1}{y+\sqrt{y^2+1}}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$.
Prove that for all triangles with angles $\alpha, \beta, \gamma$, $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma +... | For $1+\cos\alpha\ne0,$
$$\dfrac{\sin\alpha}{1+\cos\alpha}=\cdots=\tan\dfrac\alpha2$$
Now,
$$\tan\dfrac\alpha2+\tan\dfrac\beta2+\tan\dfrac\gamma2$$
$$=\dfrac{\sin\left(\dfrac\alpha2+\dfrac\beta2\right)}{\cos\dfrac\alpha2\cos\dfrac\beta2}+\dfrac{\sin\dfrac\gamma2}{\cos\dfrac\gamma2}$$
$$=\dfrac{\cos\dfrac\gamma2}{\cos\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\gcd\left(n^{a}+1, n^{b}+1\right)\mid (n^{\gcd(a, b)}+1)$
Let $a$ and $b$ be positive integers. Prove that $\operatorname{gcd}\left(n^{a}+1, n^{b}+1\right)$ divides $n^{\operatorname{gcd}(a, b)}+1$.
My work -
I proved this for $n=2$ but I am not able to prove this for all $n$ (if anyone wants I can give m... | Suppose that for some prime $p$ and positive integer $k$ we have $p^k$ divides both $n^a+1$ and $n^b+1$. Then, we need to prove that $p^k$ divides $n^{\gcd(a,b)}+1$. Denote $d=\gcd(a,b)$. Here, we will consider two cases:
Case 1. $p=2$. In this case, if $a$ or $b$ is even, then $k=1$ (because $m^2+1$ can't be divisible... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $g(x)$ if $f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$ $$f:[-2,2] \rightarrow \Bbb R$$
$$\text {and }f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$$
And, let $g(x)$ be equal to $|f(x)|+f(|x|)$
We need to find the value... | I used Desmos to draw the graphs:
$$f(x)$$
$$|f(x)|$$
$$f(|x|)$$
$$g(x)$$
While I love desmos, I think you solution for f(|x|) is more realistic.
And then, the graph of g(x) is:
Notice that because the graph f(|x|) is not defined at x<0, then g(x) is not defined there.
So, in conclusion, $$g(x) = \begin{cases}
0&x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}... | Let $x=a/b$, $y=c/d$, you'll get $2+x/y+y/x$. Now use that for any positive number the sum of that number and its reciprocal is at least $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by T... | $\cos(\frac{1}{n}) = 1 - \frac{1}{2n^2} + o(\frac{1}{n^3})$. Let $a_n = \cos(\frac{1}{n})$ and $b_n = 1- \frac{1}{2n^2}$. We have $$\frac{a_n^{n^3}}{b_n^{n^3}} = (1 + \frac{o(\frac{1}{n^3})}{1-\frac{1}{2n^2}})^{n^3} = (1+c_n o(1))^{n^3} $$ where $c_n = \frac{1}{n^3} \cdot \frac{1}{1 - \frac{1}{2n^2}} = \frac{1}{n^3-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any... | We have the identity:
$a^3+b^3+c^3=3abc$
Condition is, $(a+b+c)=0$
In "OP" solution if we take,
$a=(r+s)(r-3s)$
$b=2r(2s-r)$
$c=(r^2-2rs+3s^2)$
We get, $(a+b+c)=0$
Hence, $a^3+b^3+c^3=3abc$ ----(1)
Therefore by cubing both sides of equation (1),
$(a^3+b^3+c^3)^3=(3abc)^3=27(abc)^3 =27(a^3)(b^3)(c^3)$
Let, $(x,y,z)=(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A parameterized log-sine integral equating to square of arctan I have a really convoluted proof of the following:
$$ (1) \quad \quad - \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } dy = \big(\arctan(\sqrt{r})\big)^2 $$
I proved it for $0<r<1.$ To my surprise it holds for complex $r$ if $-1... | From the generating function of the Chebyshev polynomials of the second kind:
\begin{equation}
\frac{1}{1-2xz+z^{2}}=\sum_{n=0}^{\infty}U_{n}\left(x\right)z^{n}
\end{equation}
valid for $\left|z\right|<1$, we chose $z=-r, x=\cos\left( 2y \right)$ to express
\begin{equation}
\frac{\sin(2 y) } {r + 1/r + 2 \cos{(2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I prove that $\max(x,\max(y,z)) = \max(\max(x,y),z))$ using an algebraic formula? The maximum of two numbers can be expressed by
$$\max(x,y) = \frac12\left(x+y+|x-y|\right)$$
Consequently, we can write
$$\max(x,\max(y,z))=\frac{1}{4}\left(2x+y+z+|y-z|+|2x-y-z-|y-z||\right)$$
$$\max(\max(x,y),z)=\frac{1}{4}\left(... | From
$$\max(x,\max(y,z))=\frac{1}{4}\left(2x+y+z+|y-z|+|2x-y-z-|y-z||\right)$$
$$\max(\max(x,y),z)=\frac{1}{4}\left(x+y+2z+|x-y|+|-x-y+2z-|x-y||\right)$$
Let $y-z=a, x-y=b$, so $z=y-a, x=y+b, x-z=a+b$.
Then
$$\frac{1}{4}\left(x+2y+z+b+|a|+|2x-y-z-|y-z||\right)$$
$$\frac{1}{4}\left(x+2y+z-a+|b|+|-x-y+2z-|x-y||\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then ... | We'll prove a more general result:
If $x,y,z$ are the sides of a triangle, then
$$2\left(xy+yz+zx\right)\geq4S\sqrt3+x^2+y^2+z^2.$$
Indeed, as the $RHS$ is a decreasing function of $xyz,$ then according to the $uvw$ principles for triangle sides, we only need to check in the latter $y=z=1$ and $x\in[0,1]$ and get
$$x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find an equivalent sequence as $n\to +\infty$ of $u_1>0, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}$ Let $u_1>0$ be a real number. Let us consider $(u_n)_{n\geq 1}$ the sequence such as:
$$
\forall n \geq 1, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}\quad (\star)
$$
Find an equivalent of $u_n$ as $n\to +\infty$.
So I found ... | 1. There is a heuristic argument which is useful for guessing the behavior of $u_n$: Rewrite the recurrence relation as
$$ u_{n+1} - u_n = \frac{u_n}{n} - u_n + \frac{1}{n^2}. $$
Its continuum analogue is the following differential equation:
$$ y' = \frac{y}{x} - y + \frac{1}{x^2}. $$
Using the standard method, this eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show $\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}$
I've been attempting to show that:
$$\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}\\
\sum_{k=2}^n {k+2\choose 4}{2n+1\choose n+k+1}={n\choose 2}4^{n-2}$$
Can anyone give some direction?
| Let's rearrange the sum as
$$
\eqalign{
& s(n) = \sum\limits_{k = 1}^n {\left( \matrix{
k + 1 \cr
2 \cr} \right)\left( \matrix{
2n + 1 \cr
n + k + 1 \cr} \right)} = \cr
& = \sum\limits_{k = \,0}^{n - 1} {\left( \matrix{
k + 2 \cr
2 \cr} \right)\left( \matrix{
2n + 1 \cr
n + k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Triangles and medians Let G be the ABC barycenter. A line intersects the medians AD, BE and CF in X, Y and Z, respectively. Prove that
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}=3$$
By areas relations I found that
$\frac{XG\cdot YG}{BG\cdot DG}=\frac{6SG_{\triangle XYG}}{S_{\triangle ABC}}\\
\frac{XG\cdot ZG}{CG\cdot ... | Without loss of generality, we can scale and place the triangle so that its center of mass is in the origin and one of its vertices is on the $y$-axis at a unitary distance from the center of mass. So the vertices are $A(0,1)$, B$(p,q)$, and $C(-p,r)$. By well known properties of the medians, the midpoints of $BC$, $AC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $\int \frac{dx}{(x^2+x+1)^2},$ without using any kind of substitutions. I know there exists a reduction formula for the integral: $$\int\frac1{(ax^2+bx+c)^n}\,dx.$$ But this uses substitutions. So for the integral in question, let:
\begin{align}
I&=\int\frac{dx}{(x^2+x+1)^2}\\\\
&=\int\frac{dx}{\{(x-\omega)(x... | \begin{align}
I=\int \frac{dx}{(x^2+x+1)^2}&=\frac{1}{B}\int \frac{1}{2x+1}d(A-\frac{B}{x^2+x+1}) \\
&=\frac{1}{B}\left[ \frac{1}{2x+1}(A-\frac{B}{x^2+x+1})+\int (A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2}dx \right] \\
\end{align}
Note
\begin{align}
&(A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2} \\
&=\frac{Ax^2+Ax+A-B}{x^2+x+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inverse trigonometric equation and finding value of given expression If $\arctan(4) = 4 \arctan(x)$ then value of $x^5-7x^3+5x^2+2x+9870$ is?
I used $2\arctan(x) = \arctan(2x/1-x^2)$ twice for RHS of the equation which gave me $x^4+x^3-6x^2-x+1=0$ and I am clueless how to proceed after that.
Also I am not sure about us... | Hint
From where you have left, use
$$x^5-7x^3+5x^2+2x=x(x^4+x^3-6x^2-x+1)-(x^4+x^3-6x^2-x+1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I ... | Another way for the first inequality.
For positives $x$, $y$ and $z$ we need to prove that:
$$\sum_{cyc}(x^3-x^2z)\geq0.$$
Indeed,
$$\sum_{cyc}(x^3-x^2z)=\sum_{cyc}(x^3-xy^2)=$$
$$=\sum_{cyc}x(x+y)(x-y)=\sum_{cyc}\left((x^2+x)(x-y)-\frac{2}{3}(x^3-y^3)\right)=$$
$$=\frac{1}{3}\sum_{cyc}(x-y)(3x^2+3xy-2x^2-2xy-2y^2)=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum of $P$ Given $a, b, c \geq 0$ and $a + b + c = 6$. Find the maximum of:
$$P = (a-b)(b-c)(c-a)$$
My guess is the maximum of $P$ would be just $0$ but I don't know whether this guess is true
And if it's true, how do I prove it?
| Suppose $c = \min\{a,b,c\},$ so
$$(a-c)^2 \leqslant a^2, \quad (b-c)^2 \leqslant b^2.$$
By the AM-GM inequality we have
$$\begin{aligned}P^2 = (a-b)^2(b-c)^2(c-a)^2 & \leqslant a^2b^2(a-b)^2 = \frac14 \cdot 2ab \cdot 2ab \cdot (a-b)^2 \\& \leqslant \frac14\left(\frac{2ab+2ab+(a-b)^2}{3}\right) ^3\\&=\frac{(a+b)^6}{108}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate this integral $\int \frac{dx}{16+5^{2-3x}}$? Evaluate the following integral
$$\int \frac{dx}{16+5^{2-3x}}=?$$
My attempt:
Assume that $2-3x=t$, $-3dx=dt$, $dx=-\frac{dt}{3}$
$$\int \frac{-dt/3}{16+5^{t}}=-\frac13\int \frac{dt}{16+5^{t}}$$
let $ 5^t=u$, $\implies 5^t\ln 5\ dt=du$, $dt=\frac{du}{u\ln 5}$... | Yes, you can use single substitution
Let $16+5^{2-3x}=t\implies 5^{2-3x}\ln 5\cdot (-3)=dt$, $dx=\frac{dt}{3(16-t)\ln 5}$
$$\int \frac{dx}{16+5^{2-3x}}=\int \frac{1}{t}\frac{dt}{3(16-t)\ln 5}$$
$$=\frac{1}{3\ln5}\int \frac{dt}{t(16-t)}$$
$$=\frac{1}{3\ln5}\frac{1}{16}\int\left( \frac{1}{t}-\frac{1}{t-16}\right)\ dt$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots$ My attempt
$$1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$$
By Leibniz alternative test for convergence. It is a convergent alternative series. How do I evaluate t... | $$
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)3^{n}}=\sqrt{3}\sum_{n=0}^\infty \frac{(-1)^{n}(1/\sqrt{3})^{2n+1}}{(2n+1)}
$$
$$
=\sqrt{3}\arctan(1/\sqrt{3}) = \frac{\sqrt{3}\pi}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$... | For $x\leq0$ it's obvious.
But for $x>0$ we obtain:
$$x^4+x^2+1-(x^3+x)=x^4-2x^3+3x^2-2x+1+x^3+2x^2+x=$$
$$=(x^2-x+1)^2+x(x+1)^2>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Consider $\dot{x}=4x^{2}-16$. I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.
Question
\begin{equation}
\dot{x}=4x^{2}-16
\end{equation}
Answer
\begin{equation}
\frac{\dot{x}}{4x^{2}-16} = 1\\
\frac{\dot{x}}{x^{2}-4} = 4\\
{{dx\over dt}\over x^2-4}=4 \... | The author has used the method of "partial fractions" to write
$$
\frac{1}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2},
$$
(solving for $A$ and $B$, which I'm not going to do), integrated each of the right hand items to get a "log" term, and then combined a difference-of-logs into a log-of-a-quotient. There's nothing subtle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How can I solve this definite integral: $\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}}$
Evaluate $$\int_{0}^{a}\dfrac{x^4dx}{\sqrt{a^2-x^2}}$$
I tried taking $t$ as
$$t = \sqrt{a^2-x^2}$$
Thus my final integral became
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt$$
but I couldn't go any further in solving this integral.
I also tried by... | You can proceed as follows
Let $t=a\sin\theta\implies dt=a\cos\theta d\theta$
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt=\int_{0}^{\pi/2}(a^2-a^2\sin^2\theta)^{3/2}a\cos\theta \ d\theta$$
$$=\int_{0}^{\pi/2}(a^{3}\cos^3\theta) a\cos\theta \ d\theta$$
$$=a^4\int_{0}^{\pi/2}\cos^4\theta d\theta$$
Using: $\color{blue}{\int_0^{\pi/2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2... | By the Cauchy-Schwarz inequality we have
$$(a^3+b^3+c^3+3abc)^2 = \left[\sum a(a^2+bc)\right] \leqslant \sum a^2 \sum (a^2+bc)^2.$$
Therefore we will show that
$$4(a^2+b^2+c^2)^2 \geqslant 3[(a^2+bc)^2+(b^2+ca)^2+(c^2+ab)^2],$$
equivalent to
$$a^4+b^4+c^4 + 5(a^2b^2+b^2c^2+c^2a^2) \geqslant 6abc(a+b+c). \quad (1)$$
Whi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wro... | Use the cosine of sums formula to decompose $\cos 3x$.
$$\cos 3x = \cos (2x + x) = \cos 2x \cdot \cos x - \sin x \cdot \sin 2x$$
The identities $\cos 2x = 2 \cos^2 x - 1$ and $\sin 2x = 2 \sin x \cos x$ are widely known. These are also derived from the cosine of sums and sine of sums formulas respectively. Plugging the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\operatorname{arcsinh}=\operatorname{arcsech}(x)$ analytically I am trying to obtain an analytical solution of the equation.
$$\operatorname{arcsinh}(x) = \operatorname{arcsech}(x)$$
Equating the logarithmic definitions leads to the rather unwieldy equation
$$x^4+x^3\sqrt{x^2+1} +x^2 -1.0 -\sqrt{1-x... | $$\log(x+\sqrt{x^2+1})=\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)$$
is equivalent to
$$x^2-1=\sqrt{1-x^2}-x\sqrt{x^2+1}.$$
Then with squaring,
$$x^4-2x^2+1=1-x^2-2x\sqrt{1-x^4}+x^2(x^2+1)$$
simplifies to
$$x=0\lor x=\sqrt{1-x^4}.$$
The last equation can be reduced to biquadratic.
$$x=\dfrac1{\sqrt\phi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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The quotient of a quotient group by another quotient group Let $G$, $M$, and $N$ be the cyclic groups given by
$$
G \colon= \left\langle a \colon a^{12} = e \right\rangle = \left\{ e = a^0, a, a^2, \ldots, a^{11} \right\},
$$
$$
M \colon= \left\langle a^2 \right\rangle = \left\{ e, a^2, a^4, a^6, a^8, a^{10} \ri... | Not only is your conclusion correct, you can probably find it stated as a theorem (possibly with general proof) in your favorite group-theory textbook [e.g it is in my copy of Fraleigh 3rd edition].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,d... | $I=\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx $
$2\cos x-\sin x=\sqrt5 \cos(x+a) $, where $\tan a=1/2$ and $ 3\sin x+5\cos x=\sqrt{34}\cos(x-b) $, where $\tan b=3/5$
$I=\int \frac{\sqrt 5 \cos(x+a) } {\sqrt{34}\cos(x-b)} \, dx$ Substitute $t=x-b$ so that
$I=\sqrt{5/34}\int \frac{\cos(t+a+b) } {\cos t}\, dt$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
| It is practical to set $x_n=a z_n$ and $r=\frac{b}{a}$ to get rid of a parameter:
$$ z_{n+1} = z_{n} + 1 + \sqrt{4z_n+r^2},\qquad z_0=0 $$
Obviously $z_n\geq n$, since $z_{n+1}\geq z_{n}+1$, but also
$$ z_{n+1} \geq (\sqrt{z_n}+1)^2, $$
hence by setting $z_{n}=w_n^2$ we get $w_{n+1}\geq w_n+1$ and $z_n\geq n^2$. On the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{... | Let $P(a, b, c)$ be any point on $(\pi): 3 x+2 y+z=7$.$\\\\$
$\\ OP^{2}=a^{2}+b^{2}+c^{2}$ is the minimum iff OP $\perp(\pi) \Leftrightarrow \displaystyle O P =\left|\frac{3(0)+2(0)+7}{\sqrt{3^{2}+2^{2}+1^{2}}}\right| =\frac{7}{\sqrt{14}} =\sqrt{\frac{7}{2}}$
$\therefore$ the minimun value of $\displaystyle a^{2}+b^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate this integral? $\int^1_0\frac{(\frac{1}{2}-x)\ln (1-x)}{x^2-x+1}\mathrm{d}x$ $$\int^1_0\frac{\left(\frac{1}{2}-x\right)\ln (1-x)}{x^2-x+1}\mathrm{d}x$$
The indefinite integral didn't seem to be helpful...which include the $\operatorname{Li}(x)$.
Also, I can't set up a parameter to differentiate it.
Any ... | First, substitute $y=1-x$:
$$
\Rightarrow \int _0^1 \frac{(y-1/2)\log(y)}{y^2-y+1}\,dy
$$Then use integration by parts with $u=\log(y)$:
$$
=\left. \frac{1}{2}\log(y^2-y+1)\log(y)\right|_{0}^{1} -\frac{1}{2}\int_0^1\frac{\log(y^2-y+1)}{y}\,dy
$$The boundary terms vanish (check this yourself). Note that $y^3+1=(y^2-y+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$ $$\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$$
Insane integral! So far I have tried to complete the square for the denominator then substitute and use taylor series for the n... | I assume that $\phi$ is the golden ratio. Consider $u=x-\frac{1}{x}$ so that $\frac{x^2}{x^2+1} du=dx$:
$$I=\int_0^1 \frac{x^2}{x^2+1} \cdot \frac{x^2+1}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{x^2}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{1}{x^2-1+\frac{1}{x^2}} \ln{\left(1+u\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Sum of geometric series when exponent is $2n$, not $n$? I have a probability below which denotes the chance of catching a fish.
$$P = \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + \dots $$
I can find a generalized form of $P$ by assu... | $$\sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}=\frac{1}{4}\sum_{n=0}^\infty\left(\frac{9}{16}\right)^{n}=\frac{1}{4} \frac{1}{1- \frac{9}{16}}=\frac{4}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find coordinates of point in $\mathbb{R^4}$. There is a problem here in Q. $5$ on the last page. It states to find coordinates of point $p$.
Taking point $a=(3,2,5,1), \ b=(3,4,7,1), \ c= (5,8,9,3)$.
Also, $b$ has two coordinates in common with $a$, and $p$ lies on the same line as $a,b$.
So, those two coordinates of... | Having $p=(3,x,y,1)$ does not reflect that $p$ is on the line $ab$. However, parametric equation of an arbitrary line passing through arbitrary points $a,\,b$ in Euclidean space is $x=a+t(b-a)$ with some parameter $t\in \mathbb{R}$, hence
$$p=a+t(b-a)=(3,2,5,1)+t\,(0,2,2,0)$$
or, more compactly $$p=(3,2+2t,5+2t,1)$$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with Summation of series Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?
Answer: $\frac{13}{36}$
My Approach:
I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\... | You have solved it already. Recognise, that you can factor out $\frac 13$ from all terms. You have an infinite terms of +/- fractions, from which the first positive three ($\frac 12$, $\frac 13$, $\frac 14$) remains in the sum, every other is cancelled out by a negative counterpart with a 3 step gap. Therefore the end ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the equation $\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$ Problem
Solve the equation $$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$$
What I've tried
First I tried factoring the denominators but only the second one can be factored as $(x+4)(x-2)$.
Then I tried substit... | You started very well. To make things easier, set $A=x^2+7x-8$ (*) and the equation rewrites
$$\frac{1}{A+4x} + \frac{1}{A-5x} + \frac{1}{A-20x} = 0.$$
Denominators are not allowed to be zeros. We solve
$$(A-5x)(A-20x)+(A+4x)(A-20x)+(A+4x)(A-5x)=0$$
or equivalently
$$3A^2-42Ax=0,$$ or even
$$3(x^2+7x-8)(x^2-7x-8)=0,$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Given the equation $\alpha \mathbf{v} + \mathbf{v}\times\mathbf{a} = \mathbf{b}$, solve for $\mathbf{v}$. I'm reading a textbook at the moment that provides the following linear equation,
$$
\alpha \mathbf{v} + \mathbf{v}\times\mathbf{a} = \mathbf{b},
$$
and asks to solve for $\mathbf{v}$. The form of $\mathbf{v}$ is g... | Taking cross product with $\mathbf{a}$ on both sides, we get,
\begin{align*}
&\alpha \mathbf{v} + \mathbf{v}\times \mathbf{a} = \mathbf{b}\\
\implies &\alpha(\mathbf{v}\times \mathbf{a})+(\mathbf{v}\times \mathbf{a})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}\\
\implies &\alpha(\mathbf{b}-\alpha \mathbf{v})+(\mathbf{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$ Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$
I know $x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$ but does it help? I think $u=\ln{x}$ might be necessary some point.
| Write this as:
$$\int_{0}^1\frac{(x^2-1)(x-1)\,dx}{(x^5-1)\ln x}$$
Note that $$\int_0^1 x^y\,dy = (x-1)/\ln x$$
Replacing this and converting the problem into a double integral gives
$$\int_{0}^1\int_0^1\frac{(x^2-1)}{(x^5-1)}x^y\,dx\, dy$$
Writing $\frac{1}{1-x^5} = \sum_{k=0}^\infty x^{5k}$
This becomes
$$\int_{0}^1\... | {
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Find value of $\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$ My attempt :
$$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$
Change into variable form
$$\dfrac{(1+a^2)(1-b^2)}{ab}$$
$$\df... | $$\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$$
$$=\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{5\pi}{12}}$$
$$=\dfrac{4}{\left(\dfrac{2\tan\frac{5\pi}{12}}{(1+\tan^2\frac{5\pi}{12})}\right)\left(\dfrac{2\tan\frac{\pi}{... | {
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Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and ... | $\left|3x-5\right|$ is negative for $x<-\frac{3}{2}$ and $-\frac{3}{2}\le x<\frac{5}{3}$. Positive for $x\ge \frac{5}{3}$.
For $\left|2x+3\right|$ you have in the same intervals the signs: $-$, $+$ and $+$.
Hence for $x<-\frac{3}{2}$ you have $5-3x-\:2x-3\:>\:0 \iff x<\frac{2}{5}$. and so on and so removing the absolut... | {
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Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}... | Your second last equality is not justified: You cannot take limit of enumerator and denominator separately as the limit of the denominator is (trivially) equal to 0. Try the appropriate version of l'Hospitals rules (which helps, because enumerator and denominator (as you have formally shown) converge to $0$.
| {
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Finding real $(x,y)$ solutions that satisfies a system of equation. I was given:
$x + y^2 = y^3 ...(i) \\ y + x^2 = x^3...(ii)$
And was asked to find real $(x,y)$ solutions that satisfy the equation.
I substracted $(i)$ by $(ii)$:
$x^3 - y^3 + y^2 - x^2 + x - y = 0$
Then factored it out so I have:
$(x-y)(x^2 + xy + y^2... | You can begin by noting that the $2$ functions are inverses of each other (and only involve odd non-zero exponents). Using the fact that inverse functions are reflections in the line $y=x$, we can now see that the intersection points must be along the line $y=x$. Substituting $y$ into $x$ or vice-versa, we obtain the e... | {
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"source": "stackexchange",
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Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{... | A Geometric Proof:
Take a point $O$ in the plane and consider three line segments (this can be done since $x,y,z$ are positive real numbers) $OA, OB, OC$ with $$|OA|=x,|OB|=z,|OC|=y$$ and $$\angle AOB=60^{\circ},\angle BOC=60^{\circ}$$ therefore $\angle AOC=120^{\circ}$. In the triangle $\Delta ABC$ we have $$AB+BC\geq... | {
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Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's ... | When $n=2$, it directly works.
Assume it works for $n=k\ge 2$, and try to show that for $n=k+1$.
$1+...+\frac{1}{k^2}>\frac{3k}{2k+1}$ is given, so it suffices to show $\frac{1}{(k+1)^2}\ge\frac{3(k+1)}{2(k+1)+1}-\frac{3k}{2k+1}=\frac{3}{4(k+1)^2-1}\Leftrightarrow (k+1)^2\ge 1$, which is obvious.
| {
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$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$ I was reading a solution in which author used this identity without giving any hint that how it comes ...{maybe it is obvious}
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3... | For those of you who would like to see the full version of this:
Applying the factorization $x^3+1=(x+1)(x^2-x+1)$ to $x=a^{3^k}$ gives
$$ a^{2 \cdot 3^k}-a^{3^k}+1 = \frac{a^{3^{k+1}}+1}{a^{3^k}+1}. $$
Therefore,
$$ \prod_{k=0}^{n-1} \left(a^{2 \cdot 3^k}-a^{3^k}+1\right) = \prod_{k=0}^{n-1} \frac{a^{3^{k+1}}+1}{a^{3^... | {
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Find $\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$ Original question is $$\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$$
I got it down to $$\sum_{n=1}^{\infty} \frac{(n-1)!}{(k+n)!}$$
Here I am confused. Possible fraction decomposition but its ugly! Maybe this approach is not good... | \begin{align*}
\sum_{n=1}^\infty \frac{1}{n(n+1)...(n+k)} &= \frac{1}{k} \sum_{n=1}^\infty \frac{k}{n(n+1)...(n+k)} \\
&= \frac{1}{k} \sum_{n=1}^\infty \left[ \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)...(n+k)} \right], \\
\end{align*}
and this series telescopes so that every minus cancels with a plus, and we are lef... | {
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Closed form of $\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}}$ How to prove the following formula,
$$\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}} =
\begin{cases}
2 \cos\Big[\frac{1}{2^{n}}\Big(\pi + \arctan{\frac{\sqrt{x(4-x)}}{x-2}}\Big)\Big] & 0 < x < 2 \\
2 \cos\Big[\frac{1}{2^{n}}\arc... | Hint If $a=2\cos(y)$ then
$$2+a= 2 + 2\cos(y)= 4 \cos^2(\frac{y}{2}) \,.$$
So write $\sqrt{x}=2 \cos(\alpha)$ and use the above formula:
$$2 +\sqrt{x}= (2 \cos(\frac{\alpha}{2}))^2 \\
2+\sqrt{2+\sqrt{x}}=2+2\cos(\alpha/2)= (2\cos(\alpha/4))^2 \\
...$$
| {
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"source": "stackexchange",
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How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdo... | Think it through: to find the lowest common divisor of $a$ and $b$ you take the prime factorization of $a$ and $b$. And find the number whose prime factorization consists of the prime factors of $a$ and of $b$ and raising them to the higher power they are raised to in $a$ and $b$. In other words:
If $\{p_i\}$ are t... | {
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How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the... | $$I=\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, dx$$
Get rid of the denominator
$$u=\sqrt {2x} - \sqrt {x+4}\implies x=3 u^2+2 \sqrt{2}u \sqrt{u^2+4 }+4\implies \frac{dx}{du}=\frac{\sqrt{2} \left(4 u^2+8 \right)}{\sqrt{u^2+4 }}+6 u$$
$$I=\int6 du+\int\frac{4 \sqrt{2} u}{\sqrt{u^2+4}}du+\int\frac{8 \sqrt{2}}{u\sqrt{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761762",
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How to find indefinte integral $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using substitution of $u=\cos(\frac{x^2}{4})$?
Can someone help me find $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using the substitution $u=\cos(\frac{x^2}{4})$ step-by-step?
All I've done so far is:
$$\frac{du}{dx}=-\frac{1}{2}x\sin(\fr... | From your steps
\begin{align*}
&\int_{1}^{\frac{1}{\sqrt{2}}} \frac{-4u}{\sqrt{1 - u^2}}du\\
=&~2 \int_{0}^{\frac{1}{2}} \frac{dt}{\sqrt{t}} \quad \text{on the substituion } (1 - u^2) = t\\
=&~2 \int_{0}^{\frac{1}{\sqrt{2}}}\frac{w}{w}d\omega\quad \text{on the substituition } t = \omega^2\\
=&~4 \bigg[\frac{1}{\sqrt{2}... | {
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Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or... | Your expression is $\displaystyle\sum_{k=0}^n \frac{x^{k+1}}{k+1} \binom{n}{k}$ evaluated at $x=2$.
Hint. Can you think of where else $\displaystyle\frac{x^{k+1}}{k+1}$ shows up (specifically, in calculus)?
| {
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"source": "stackexchange",
"question_score": "2",
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Find the recursive formula for $I_m:=\int \frac{dx}{\sqrt{x^2+1}^m},\enspace \,m\in\mathbb{N}$ Find the recursive formula for:
$$I_m:=\int \frac{dx}{\sqrt{x^2+1}^m}\,\,,\,\,\,\,m\in\mathbb{N}$$
My attempt:
$$I_m=\int\frac{dx}{\sqrt{x^2+1}^m}=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\int \frac{x^2}{\sqrt{x^2+1}^m(x^2+1)}... | Another way:
$$\dfrac{d\dfrac x{(x^2+1)^{n/2}}}{dx}=\dfrac1{(x^2+1)^{n/2}}-\dfrac{nx}{2(x^2+1)^{n/2+1}}\cdot2x$$
$$=\dfrac1{(x^2+1)^{n/2}}-n\cdot\dfrac{1+x^2-1}{(x^2+1)^{n/2+1}}$$
Integrate both sides w.r.t $x,$
$$\implies nI_{n+2}-(n-1)I_n=\dfrac x{(x^2+1)^{n/2}}+K$$
| {
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What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i d... | For any natural $n$ we obtain:
$$\begin{aligned}5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}&=2^{n+1}\left(10\cdot5^{2n}+9\cdot6^n\right) \\
&= 2^{n+1}\left(10\cdot(6+19)^{n}+9\cdot6^n\right) \\
& \equiv2^{n+1}\left(10\cdot6^n+9\cdot6^n\right) \pmod{19} \\ &\equiv 0 \pmod{19}.\end{aligned}$$
| {
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Contradiction on an Inequality The problem on which I have a problem is this-
Let a, b, c be non-negative real numbers. Prove that $$ \sum_{cyc} {a^2-bc \over 2a^2+b^2+c^2} \ge 0 $$
While solving and after some resolution, we get $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le 3 $$
And by C-S, we have,
$$ \sum_{cyc} {... | Another solution.
By your work we need to prove that:
$$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq3,$$ which is true by C-S:
$$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{c^2}{c^2+a^2}\right)=3.$$
| {
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A trick system of equations \begin{cases}
x+y+\dfrac{x^2}{y^2}=7\\
\dfrac{(x-y)x^2}{y^2}=12
\end{cases}
I don't have any idea to solve this. I tried to subtract, add and multiply the given equations, but nothing help me to find $x$ or $y$.
Can someone help me to solve this system?
Thanks for attention.
| The second equation tells you that $$\frac{x^2}{y^2}=\frac{12}{x-y}$$
(The possibility $x-y=0$ is ruled out, because this would violate the second equation.)
If we substitute this in the first equation, we get
$$x+y+\frac{12}{x-y}=7$$
which implies
$$(x+y)(x-y)+12=7(x-y)$$
$$12=7(x-y)-(x-y)(x+y)$$
$$12=(x-y)[7-(x+y)]$$... | {
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How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\t... | A variation of other answers (that more closely parallels a common pattern when the numerator and denominator are polynomials) is "big part factoring". \begin{align*}
\lim_{x \rightarrow 0} \frac{12^x - 4^x}{9^x-3^x}
&= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{4}{12} \right) ^x \right)}{9^x \left( ... | {
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Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that s... | Try some examples. I suggest researching Simon's Favorite Factoring trick, it helps with these types of questions. Asking on artofproblemsolving.com is a great idea, Simon is a real person who invented this factoring trick, and it's on that website.
I'll give you one example. Let's take $\frac{1}{a}+\frac{1}{b}=\frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{... | Consider using differentiation under the integral sign. Parameterize the integral as the following:
$$I(a)=\int_0^{\infty} \frac{\ln{\left(1+ax^2\right)}}{9+x^2} + \frac{\ln{\left(a+x^2\right)}}{1+9x^2}+ \frac{\ln{\left(a+x^2\right)}}{9+x^2} \; \mathrm{d}x$$
The integral in question is $I(4)$. First, differentiate $I... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that a non-decreasing function $F$ with the properties $F(0) = 0$, $F(x/3)=F(x)/2$ and $F(1-x) = 1 - F(x)$ is the Cantor function $\begingroup$
I'm trying to prove by induction an assertion of Chalice (1991) article (http://people.math.sc.edu/girardi/m555/current/hw/FunkyFunctions/Cantor.pdf). I know that if $a_{... | I think I solved the step I was stucked on. Since $a_{n_{k-\left(2^{n+1}\right)}} \in E_{n} \cup \{0,\ 1\} \subset K $, we have $a_{(n+1)_{k}} = \frac{2}{3}+\frac{1}{3}a_{n_{k-\left(2^{n+1}\right)}} \Rightarrow a_{(n+1)_{k}} = \frac{2}{3}+\frac{1}{3}\sum_{i=1}^{\infty} \frac{a_{n_{k-\left(2^{n+1}\right)_{i}}}}{3^{i}} ... | {
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Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)... | One way is using differentiation. Let $$f(x) = \sin^{-1}(x) + \frac{\pi}{4}$$ And $$g(x) = {\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) $$
Edit:
Note that domain of $\sin^{-1} x$ is $-1\le x \le 1$. The domain of $f(x) - g(x)$ is $A = \{x | x \in \mathbb{R}, x \in D_f \cap D_g \}$. So we have $$D_f ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}... | The integral test requires an upper bound which is not immediate. Instead we can relate the series to the Taylor expansion of the exponential.
$$n^5=n(n-1)(n-2)(n-3)(n-4)+10(n-1)(n-2)(n-3)(n-4)\\+65(n-2)(n-3)(n-4)+285(n-3)(n-4)+781(n-4)+1024$$ so that
$$\sum_{n=1}^\infty\frac{2^nn^5}{n!}=
\\\sum_{n=1}^\infty2^n\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conditional Probability Denominator doubt Let X be a rv such that $P(X = 2) = 1/4$ and its CDF is given by
$$F_{X}(x) =
\begin{cases}
0,& x< -3\\
\frac{3}{4}(x+3),& -3\leq x <2 \\
3/4,& 2 \leq x <4\\
\frac{3}{64} x^2,& 4 \leq x < \frac{8}{\sqrt{3}} \\
1,& x \geq \frac{8}{\sqrt{3}}
\end{cases}
$$
$x=2... | $P(X>2)=\int_4^{8/\sqrt3 } \frac 3 {32} xdx=\frac 1 4$. Add $P(X=2)=\frac 1 4$ to this to get $P(X \geq 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)... | A general way:
By elimination of $z$ we get
$$(x+y)(x^2+y^2-xy+x+y-2)=0$$
Case 1: So two branches one: $x+y=0 \implies z=1,x=n, y=-n$, where $n\in I$
Case 2: The other set of solutions are given by
$$(x^2+y^2+xy+x+y-2)=0$$, write this as a quadratic of $x$ and treat $y$ as constant then
$$x=\frac{-(1-y)\pm \sqrt{(1-y)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Compute the matrix of norms of $A=\begin{bmatrix}3&4\\1&-3\end{bmatrix}$ My work so far
Using the following
$\hspace{30px} L^1\ =\displaystyle \max_{\small 1\le j\le m}(\displaystyle \sum_{i=1}^n |a_{ij}|)\\ \hspace{30px} L^2\ =\sigma_{max}(A)\\ \hspace{30px} L^F\ =\sqrt{\displaystyle \sum_{i} \displaystyle \sum_{j} |a... | Alternatively, recall that
$$\|A\|_2 = \sup_{\|(x,y)\|_2=1} \|A(x,y)\|_2 = \sup \left\{\left\|\begin{bmatrix}3&4\\1&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\right\| : x^2+y^2=1\right\}$$
so subject to $x^2+y^2=1$ we wish to maximize $$f(x,y) = (3x+4y)^2+(x-3y)^2 = 10x^2+25y^2+18xy.$$
Plugging in $y=\sqrt{1-x^2}$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | the pqr method:
Let $p = a+b+c, q = ab+bc+ca = 3, r = abc$.
Since $(a+b+c)^2 - 3(ab+bc+ca) = a^2+b^2+c^2 - ab-bc-ca \ge 0$, we have $p \ge 3$.
Since $ab+bc+ca \ge 3\sqrt[3]{ab \cdot bc \cdot ca} = 3\sqrt[3]{(abc)^2}$, we have $r\le 1$.
We have
\begin{align}
(1+a^2)(1+b^2)(1+c^2) - 8 &= a^2b^2c^2+a^2b^2+b^2c^2+c^2a^2+a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that... | Let $\sqrt[4]{x}=a$ and $\sqrt[4]{x+1}=b$.
Thus, $a\geq0,$ $b\geq1$, $b^4-a^4=1$ and $$a+b=\sqrt[4]{a^4+b^4}$$ or
$$(a+b)^4=a^4+b^4$$ or $$2ab(2a^2+3ab+2b^2)=0,$$ which gives $$ab=0.$$
Can you end it now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\newcommand{\dd}{\mathrm{d}}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\... | Your series can be written:
$$\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1} x^{n} \ln{\left(2 \right)}^{n + 1}}{\left(n + 1\right)!}=\sum_{n=0}^{\infty} \frac{x^{n} \left(- \ln{\left(2 \right)}\right)^{n + 1}}{\left(n + 1\right)!}$$
being $(-1)^{n+1}=-1$ forall $n\in\Bbb N$.
And rewriting
$$\sum_{n=0}^{\infty} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Bernoulli's Inequality for $-1 \leq x\leq 0$ My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower ... | As regards your original goal, there is a shorter way. We have that
$$e^x-1-x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n-1-x=\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}.$$
Hence, for $x\in [-1,1]$, $$|e^x-1-x|=\left|\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\right|\leq x^2\lim_{n\to\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit num... | Write $0$ at the front of four digits number so we always have five digits number. The fifth number is one of $(0,1,6,7,8,9)$.
$6\times 5! =720$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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(BAMO $2013/3$) $ABH$, $BCH$ and $CAH$ is congruent to $ABC$.
Let $H$ be the orthocenter of an acute triangle $ABC$. Prove that the triangle formed by the circumcenters of triangles $ABH$, $BCH$ and $CAH$ is congruent to $ABC$.
I have already seen many answers on MSe But my doubt is diffferent,
In this solution (firs... | Because $AO_B=CO_B$, $\measuredangle AO_BC=2\beta$ and from here:
$$\measuredangle CAO_B=90^{\circ}-\beta.$$
Another way.
Let $A'$, $B'$ and $C'$ be circumcenters of $\Delta BHC$, $\Delta CHA$ and $\Delta AHB$ respectively.
Thus, since $B'HA'C$ is a rhombus and $HC=c\cot\gamma$ in the standard notation, we obtain:
$$A'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find an approximation to $\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor (j-1)$ I want to find an approximate (ideally asymptotic) function $f_1:\mathbb{R}\to\mathbb{R}$ in order to approximate a function $f_0:\mathbb{R}\to\mathbb{N}$ with $f_0$ defined by
$$\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloo... | I suppose that $f_0(x) = 0$ for $x < 1$. Otherwise it's not clear how the sum should be interpreted, but with reasonable interpretations the behaviour can be determined much like the behaviour for $x \geqslant 1$.
First we split the terms $\bigl\lfloor \frac{x}{j} - 1\bigr\rfloor (j-1) = \bigl\lfloor \frac{x}{j}\bigr\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the side of the square ,by using trigonometry There is a square with diagonal length of 'a", the question is to find the length of the sides. it can be found by Pythagorean theorem. but I tried to do it with trigonometry, considering the properties of a square but it doesn't resemble the first answer, so I ... | Let the side length of the square equal $b$, and the hypotenuse $a$. By Pythagoras' Theorem, we have
\begin{align}
b^2+b^2&=a^2 \\
2b^2&=a^2 \\
b^2&=\frac{a^2}{2}\\
b&=\sqrt{\frac{a^2}{2}}=\frac{\sqrt{a^2}}{\sqrt 2}=\frac{a}{\sqrt 2}=\frac{\sqrt 2}{2}a
\end{align}
Alternatively, the angle between $a$ and $b$ is $45$ de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The value of $\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $ is $$\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $$
My working:
$$\int_0^1\ [(x+1)(x+2)(x+3)...(x+n)]\times[\frac{1}{x+1}\ + \frac{1}{x+2}\ + \frac{1}{x+3}\ +\ ...\ + \frac{1}{x+n}].dx \\$$
$\int_0^1[(x+1)(x+2)(x+3)...(x+n)]... | The integrand function is the derivative of $(x+1)(x+2)\cdots(x+n)=(x+1)_n$, hence the outcome simply is
$$\left[(x+1)_n\right]_{0}^{1}=(n+1)!-n!=n\cdot n!.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} ... | By the definition of hyperbolic functions and $e^x=1+x+\frac{x^2}2+o(x^2)$ we have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}=\frac{e^{2x}+1}{e^{2x}-1} - \frac{1}{x}=\\=\frac{2+2x+2x^2+o(x^2)}{2x+2x^2+o(x^2)} - \frac{1}{x}=\frac{2x+2x^2-2x-2x^2+o(x^2)}{2x^2+o(x^2)}=\frac{o(1)}{2+o(1)} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Product/division of exponencials with factorials as the base I put this formula on WolframAlpha $$\frac{(26!)^{n+2}}{13!}$$ and it simplified to $$2^{23n+36}\cdot175^{3n+5}\cdot7429^{n+2}\cdot34749^{2n+3}$$
I tried solving it by hand
\begin{align}
\frac{(26!)^{n+2}}{13!} & = \frac{(26\cdot25\cdot\ldots\cdot3\cdot2)^{n+... | For more systematic approach, let $v_p(x)$ be a $p$-adic valuation of $x$, then by some basic rules such as $v_p(x^n)=n\cdot v_p(x)$ and $v_p(x/y)=v_p(x)-v_p(y)$. Also we can use Legendre's formula $v_p(n!)=\sum_{i=1}^{\infty}\lfloor \frac{n}{p^i} \rfloor$, so
\begin{align}
v_2\left(\frac{(26!)^{n+2}}{13!}\right)&=(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The c... | We normally do exponents before multiplication and the minus sign here is a multiplication. I think the problem wants
$$-(8^{1/3}) \mbox{ non } (-8)^{1/3}.$$
So work out $8^{1/3}$ to get
$$2, 2 \mbox{ cis } \left(\frac{2\pi}{3}\right), 2 \mbox{ cis } \left(\frac{4\pi}{3}\right).$$
Now multiply by $-1$
$$-2, -2 \mbox{... | {
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"timestamp": "2023-03-29T00:00:00",
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How $\int_{0}^{\infty} \frac{\arctan(x)}{1+x}\frac{dx}{\sqrt[4]{x}}=\frac{\pi}{\sqrt2}\big(\pi/2+\ln{\beta}\big)$ $$
\int_{0}^{\infty}\frac{\arctan\left(x\right)}{1 + x}
\,\frac{\mathrm{d}x}{\sqrt[{\large 4}]{x}} =
\frac{\pi}{\,\sqrt{\,{2}\,}\,}
\left[{\pi \over 2} + \ln\left(\,{\beta}\,\right)\right]
$$
$$
\mbox{Find ... | Write
\begin{align*}
I
:= \int_{0}^{\infty}\frac{\arctan x}{(x+1)x^{1/4}} \, \mathrm{d}x
= \int_{0}^{1}\int_{0}^{\infty}\frac{x^{3/4}}{(x+1)(t^2x^2+1)} \, \mathrm{d}x\mathrm{d}t.
\end{align*}
The inner integral can be computed via a standard technique involving the contour integration along the keyhole contour. Indeed,... | {
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"url": "https://math.stackexchange.com/questions/3792655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$ In what ways can I evaluate $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
I tried a few methods but none work or simplify things and i cant think of substitutions that could turn things better, while the substitution ... | Split the integration range
\begin{align}
I& =\int _0^{\infty }\frac{\ln (1+x)}{1-x^2+x^4}dx= \int _0^1\frac{(1+x^2)\ln (1+x)-x^2\ln x}{1-x^2+x^4}dx\\
\end{align}
Then, integrate by parts via
$$d\left( \cot^{-1}\frac x{x^2-1}\right)=\frac{1+x^2}{1-x^2+x^4}dx$$
$$d\left( \frac12\tan^{-1}\frac x{1-x^2} - \frac1{2\sqrt3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Verifying the closed form of $\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}$ My question is at the end of the solution.
We have
$$\int_0^1 x^{n-1}\ln(1-x)=-\frac{H_n}{n}$$
Differentiate both sides with respect to $n$
$$\int_0^1 x^{n-1}\ln x\ln(1-x)=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$
Next ... | @User 628759 gave the solution. It is quite surprising to see how sensitive it the result to this parameter.
On my side, I computed exactly
$$\sum_{n=1}^{10000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.23740$$
$$\sum_{n=1}^{20000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.25448$$
To be slow, it is !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$ Question:
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$
My working:
$D=a^2-8(b+3)\geqq0$
$a^2\geqq8(b+3)=8b+24$
Add $b^2$ to both sides
$a^2+b^2\geqq b^2+8b+24=(b+4)^2+8\geqq 8$
$\therefore {(a^2+b^2)... | By your work: $$a^2+b^2=a^2+b^2-\frac{3}{4}(a^2-8(b+3))+\frac{3}{4}(a^2-8(b+3))\geq$$
$$\geq a^2+b^2-\frac{3}{4}(a^2-8(b+3))=\frac{1}{4}(a^2+4(b+3)^2)+9\geq9.$$
The equality occurs for $a=0$ and $b=-3$, which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794277",
"timestamp": "2023-03-29T00:00:00",
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Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displa... | Consider $A = (\frac{x+1}{x-1})^\frac{1}{3} + (\frac{x-1}{x+1})^\frac{1}{3}$ and $x = \tan u$. Since $x^4+x^2 = \frac{11}{5}$ then $\frac{\sin^2u}{(1-\sin^2u)^2}=\frac{11}{5}$ by solving a simple quadric equation we get that $\sin u = \pm \sqrt{\frac{27-7\sqrt5}{22}}$.
Therefore $A = (\frac{\tan u+1}{\tan u-1})^\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Show that $f(x) = x|x|$ is continuous and differentiable - solution verification? Another exercise I did without any solutions.
I highly doubt this is correct so pls correct me :)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be given by $f(x):=x|x| .$ Show that $f$ is continuous and differentiable on $\mathrm{R}$
$$
\b... | For $x\neq 0$, $$\begin{align*} \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}&= \lim_{h\to 0} \frac{(x+h)|x+h|-x|x|}{h}\\&=\lim_{h\to 0} \frac{x|x+h| + h|x+h|-x|x|}{h}\\&=\lim_{h\to 0} \frac{x(|x+h|-|x|)+h|x+h|}{h}\\&= \lim_{h\to 0}\frac{x(|x+h|-|x|)}{h} + \frac{h|x+h|}{h}\\&=\bigg[x\lim_{h\to 0}\frac{|x+h|-|x|}{h}\bigg]+|x|\\&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$ L(P)= (1+X^2)P''(X)-2XP'(X)$
*
*$E=\mathbb{R}[X]$
*$L$ is an endomorphism of $E$ and $L(P)= (1+X^2)P''(X)-2XP'(X)$
What are the possible eigenvalues of $L$ and the dimension of the eigenspace ?
Let's $p(x)=\sum_{k=0}^{n} a_k X^k$ be an eigenvector associated to the eigenvalue $\lambda$.
$a_{n+1} = a_{n+2} = .. =0... | Let $\lambda$ be an eigenvalue of $L$ and $p(x)=\sum_{k=0}^n a_kx^k=\sum_{k=0}^\infty a_kx^k$ with $a_n=1$ and $0=a_{n+1}=a_{n+2}=\cdots$ be a corresponding eigenvector. From $(1+x^2)p''(x)-2xp'(x)=\lambda p(x)$, we obtain the recurrence relation
$$
\left[(k+2)(k+1)a_{k+2}+k(k-1)a_k\right]-2ka_k=\lambda a_k
$$
or equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does $(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) $ hold? Does there exist a positive constant $c>0$ such that
$$(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) \tag{1}$$
holds for any nonnegative $x,y$?
Let me add some context for this question:
The motivation comes from the case were $x,y$ are interpreted as sing... | Let $x=y=0,$ we get $c \geqslant 2.$
For $c =2,$ inequality become
$$(x-1)^2+(y-1)^2 \leqslant 2\big((x-y)^2+(xy-1)^2\big),$$
equivalent to
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8xy.$$
Using the AM-GM inequality, we have
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8\sqrt[8]{(x^2y^2)^2 \cdot x^2 \cdot y^2 \cdot (xy)^2}=8xy.$$
So, y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
w... | $$9-x^2>x^4+2x^2+1$$
$$x^4+3x^2-8<0$$
$$\left(x^2+\frac32 \right)^2 < 8+\frac94$$
$$\left(x^2+\frac32 \right)^2< \frac{41}4$$
$$0\le x^2<\color{red}-\frac32 + \frac{\sqrt{41}}2$$
Now, your answer should coincide.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$ without L'Hôpital's rule How would you solve the limit
$$\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$$
without using L'Hôpital's rule?
| $$\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)=\frac{1}{x} \log \frac{1}{x}+\frac{\log \left(1+\frac{1}{x} \right)}{\frac{1}{x}\cdot x^2}-\frac{\log \left(1+\frac{1}{x^2} \right)}{\frac{1}{x^2}\cdot x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=... | Let $a=\min\{a,b,c\}$, $b=a+u,$ $c=a+v$ and $u^2+v^2=2tuv$.
Thus, by AM-GM $t\geq1$ and we need to prove that:
$$2(a+b+c)\sum_{cyc}(a-b)^2\geq9\sqrt{\prod_{cyc}(a-b)^2}$$ or
$$2(3a+u+v)(u^2+v^2+(u-v)^2)\geq9\sqrt{u^2v^2(u-v)^2},$$ for which it's enough to prove that
$$4(u+v)(u^2-uv+v^2)\geq9\sqrt{u^2v^2(u-v)^2}$$ or
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\tan(x)\tan(x+\frac{\pi}{3})+\tan(x)\tan(\frac{\pi}{3}-x)+\tan(x+\frac{\pi}{3})\tan(x-\frac{\pi}{3}) = -3$ Let's assume that $\tan(x) = y$.
So, $\tan\Big(x+\dfrac{\pi}{3}\Big) = \dfrac{\tan(x) + \tan\Big(\dfrac{\pi}{3}\Big)}{1-\tan(x)\tan\Big(\dfrac{\pi}{3}\Big)} = \dfrac{y+\sqrt{3}}{1-\sqrt{3}y}$
Similarly... | Like Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$
$$t_0=\tan y,t_1=\tan\left(y-\dfrac\pi3\right),t_2=\tan\left(y+\dfrac\pi3\right)$$ are the roots of
$$t^3-(3\tan3y)t^2-3t+3\tan3y=0$$
Using Vieta's formula $$t_0t_1+t_1t_2+t_2t_0=\dfrac{-3}1$$
$$t_0+t_1+t_2=\dfrac{3\tan3y}1$$
$$t_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving the recurrence relation $T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$ I have been trying to solve the following recurrence relation
$T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$
$T(0) = 0$
I've tried to use substitution which wasn't very useful as I couldn't figure out a way to simplify the resu... | The calculation should be continued: We get
$ \begin{array}{rcl}
T_n&=&T_0+10n- 5\sum\limits_{k=0}^{n-1}\frac{1}{k+1} \\
&=&T_0+10n- 5 H_n
\end{array} $
Here the $H_n=1+\frac{1}{2}+\cdots +\frac{1}{n}$ are the so called harmonic numbers. It is known that
$H_n= \ln n+\gamma + \varepsilon_n$ where $(\varep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$... | Since the two posted Answers so far (that provide a concrete value of the indicated sum) disagree, I'll pursue the approach I'd suggested in a Comment.
First form an equation whose three roots are $y = (x+1)$ instead of $x$. That is:
$$ (y-1)^3 - 2007(y-1) + 2002 = 0 $$
$$ y^3 - 3y^2 - 2004y + 4008 = 0 $$
An equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $a = \frac{1+\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$? My question is this:
If $a = \frac{1+\sqrt5}2,\frac{1-\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$?
It is an AMC style question and is timed, so I will not be able to use solutions with a lot of case work.
I defined $a$ from part a of a 2 part ... | Okay, actually we can do better than what I suggested in the comments. The basic move is to repeatedly use the identity $a^2 = a + 1$ together with binary exponentiation. We have
$$a^3 = a(a^2) = a(a + 1) = a^2 + a = 2a + 1$$
$$a^6 = (a^3)^2 = (2a + 1)^2 = 4a^2 + 4a + 1 = 8a + 5$$
$$a^{12} = (a^6)^2 = (8a + 5)^2 = 64a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is my proof that $\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n)$ correct? Prove that: $$\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n) $$
$$\frac{n^3-3n^2+2n}{n^2-4n}≥\frac{n^3-3n^2+2n-17}{n^2-4n+22} $$
now we choose$$ N≥6$$
thus
$$\frac{n^3-3n^2+3n^2-18n+2n}{n^2-4n}≥\frac{n^3-3n^2+2n}{n^2-4n}$$
and
$$\frac{n^3-16n}{n^2-4n}=\frac{n^2-1... | First a small notation error, you define $N$ but don't use it. I think by $N\ge6$, you mean $n\ge N$ for $N=6$.
To prove $a_n = O(n)$ you need to prove $|a_n| \le Cn$ for $n>N$, but you have proven $a_n\le Cn$. You should also have a separate check that for $n\ge6$ (or any larger number) that $a_n\ge 0$. Then $a_n=|a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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