Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ solve the differential equation.
$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$
The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.
Thanks in advance.
| This equation could have been solved easily in polar coordinates.
\begin{align}
x &= z \cos{s} \\
y &= z \sin{s}
\end{align}
Then
\begin{align}
dx = \cos(s) \, dz - z\,\sin(s)\, ds \\
dy = \sin(s) \, dz + z\,\cos(s)\, ds
\end{align}
as well as
$$x^2+y^2 = z^2\big(\cos^2(s) + \sin^2(s)\big) = z^2$$ which means that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$
Converging or Diverging? I guess I have to lower the fraction so that the roots will get away and I will have $\frac{1} {n}$ that diverges. But... | $$
0<\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}=
\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}\cdot
\frac{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}}
{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} \\
=\frac{(n+1)^2-n^2}
{
n(\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is... | A proof for $n=3$.
We'll prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$ for all positives $a$, $b$ and $c$.
Indeed, let $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$.
Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}=\frac{1}{1-\frac{ab+ac+bc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 1
} |
Basis vector find Let $\alpha_1=[ 2,1,3,0] $
$\alpha_2=[ 1,1,1,-1] $, $\alpha_3=[ 2,-1,5,4] $, $\alpha_4=[ 1,2,0,-3] $, $\alpha_5=[ 3,1,6,1] $
be vectors from $\mathbb{R}^4$ . From vectors system ($\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5 $) choose basis of vector space $V=lin(\alpha_1,\alpha_2, \alpha_3, \alpha... | Consider the matrix
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
and perform Gaussian elimination on it:
\begin{align}
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of the integral $\int \sqrt{t^4-t^2 + 1}\,dt$ My friend took his Calculus $2/3$ test yesterday.
One of the questions he had trouble with was this integral:
$$\int \sqrt{t^4-t^2 + 1}dt$$
My attempt
It seems rather clear that the only approach was trigonometric substitution.
First, completing the square:
$$\in... | The indefinite integral cannot be expressed in terms of elementary functions. Evaluating it requires knowledge of elliptic integrals. However, it is interesting to note that its definite counterpart, when evaluated over the entire real line, and subtracted from its quadratic or parabolic asymptote, yields
$$\int_{-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 +... | Without expanding, a simple substitution will help show that the result is true.
Put $r=n-k$, in which case when $k=0,n$ corresponds to $r=n,0$.
Hence
$$\sum_{k=0}^{n}(n-k)^2=\sum_{r=0}^n r^2=\frac16n(n+1)(2n+1)$$
NB: In fact, the original recurrence relationship telescopes quite readily as follows, without having ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ mu... | Continuing from what you have got,
Subtracting $13A + 4B + C = -2$ from $3A + 2B + C = 0$ yields $10A + 2B = -2$.
Now we have $3A + 2B = 0$ and $10A + 2B = -2$.
Hence $A = -2/7$ and $B$, $C$, $D$ can be calculated by putting back into $3A + 2B = 0$, $3A + 2B + C = 0$ and $A + B + C + D = 6$ in order.
Then $P'(0)$ can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}$? How do find the following sum
$$
\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}
$$
| Let's look at the more general case,
$$\sum_{\gcd (p,q) = d} \frac{1}{p^2q^2}.$$
$\gcd (p,q) = d$ means we have $p = d\cdot a$ and $q = d\cdot b$ with $\gcd(a,b) = 1$, so we obtain
$$\sum_{\gcd(p,q) = d} \frac{1}{p^2q^2} = \sum_{\gcd(a,b) = 1} \frac{1}{(da)^2(db)^2} = \frac{1}{d^4}\sum_{\gcd(a,b) = 1} \frac{1}{a^2b^2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Squeeze theorem: sequence of sums So I've been having some problems with finding the limit of this sequence:
$$\lim_{n \rightarrow \infty}\left(\frac{4}{ \sqrt[5]{1^{10}+1 }} + \frac{8}{\sqrt[5]{2^{10}+2}} + ... + \frac{4n}{\sqrt[5]{n^{10}+n}}\right)$$
I've tried to find it using the squeeze theorem, here's what I've ... | Consider $$s_n=\frac{4 n}{\sqrt[5]{n^{10}+n}}=\frac{4 n}{n^2\sqrt[5]{1+\frac{1}{n^9}}}=\frac{4}{n} \frac{1}{\sqrt[5]{1+\frac{1}{n^9}}}$$ So, for large values of $n$, it write $$s_n=\frac{4}{n}-\frac{4}{5 n^{10}}+\frac{12}{25
n^{19}}+O\left(\left(\frac{1}{n}\right)^{21}\right)$$ and, so, the sum $\sum_{n=1}^{\infty}s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Strategy for solving $7\vert2^{n+2}+3^{2n+1}$ by induction. So I have to show the following to be true using induction
$7\mid 2^{n+2}+3^{2n+1}$
This is easily checked with the case $n=0$ because $7 \mid 7$, but I assuming this holds for$n=k :$
$$7\mid 2^{k+2}+3^{2k+1}$$
I fail to see how I can use this for $n=(k+1):$... | You checked it for $n=0$, so now you have to prove implication:
$$ \left(\forall n \in \mathbb{N}\right)\left(\underbrace{7 \mid 2^{n+2} + 3^{2n+1}}_{\mathrm{Assumption}} \Longrightarrow
\underbrace{7 \mid 2^{(n+1)+2} + 3^{2(n+1)+1}}_{\mathrm{Thesis}}\right)$$
Simple thesis transformation:
$$\begin{align*}\begin{split}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\lim_{x\to-2}\frac{x+8}{x+3}=6 $ using $\epsilon-\delta$
Prove:
$$\lim_{x\to-2}\frac{x+8}{x+3}=6 $$
I started with:
$ |\frac{x+8}{x+3} - 6| < \epsilon => |\frac{x+8-6x-18}{x+3} | < \epsilon => |\frac{-5x-10}{x+3} | < \epsilon =>| \frac{-5(x+2)}{x+3} | < \epsilon $
From the definition of limit we know that:
$ ... | Hint:
Take $\delta=min(\frac{1}{2},\frac{\epsilon}{10})>0$
Then $\forall x, |x+2| <\delta $, $|x+3|>\frac{1}{2}$, hence $$| \frac{-5(x+2)}{x+3} |\le 10|x+2| < \epsilon $$
The key is to select $\delta$ such that $x$ is closed to $-2$ compared with $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The inequality $ abc\geq(a-b+c)(a+b-c)(b+c-a)$ holds for $a,b,c\geq 0$
What is the proof that:
$$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
I tried :
we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
then
$$a\ b\ c\geq(a-b+c)(a+b... | Write $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
Note that at most one of $x,y,z$ can be negative (consider the largest of $a,b,c$).
Edit: Without loss of generality, assume that $a=\max(a,b,c)$, then $$2x=(a-c)+b\ge b\ge 0, 2z=(a-b)+c\ge c\ge 0.$$
So there are two ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence: ... | Using the following inequality
\begin{equation}
\frac{n}{\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}} } \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq \frac{(1-\frac{1}{2})+(1-\frac{1}{4})+...,(1-\frac{1}{2n})}{n} = 1-\frac{1}{2n}\sum_{k=1}^{n}\frac{1}{k}
\end{equation}
and another results
\begin{equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 11,
"answer_id": 9
} |
Use Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$ I am working on the following question from Lay's Introduction to Analysis with Proof:
Use the Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$
My work so far:
Define $f(t) = \sqrt t$ which we will wor... | Notice that
$$7\frac{1}{4}=7+\frac{1}{4}$$
and
$$7\frac{2}{7}=7+\frac{2}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute: $\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}$ Help me please to compute the limit of:
$
\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdot \left ( 1-\frac{2}{4} \right ) ^{\frac{4}{n}}\cdot \left ( 1-\frac{2}... | $(1-2/k)^k$ approaches $e^{-2}$, so most of the factors are near $e^{-2}$.
Then you take the $n$th root of them all, and there are $n$ factors, so the answer is $e^{-2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $\log_c{x}$ if $\log_a{x} = p$, $\log_b{x} = q$, and $\log_{abc} {x} = r$.
Given that $\log_a{x} = p$, $\log_b{x} = q$, and $\log_{abc} {x} = r$, find the value of $\log_c{x}$.
| Using the change of base and product rule for logs, we have:
$$
p = \frac{\log x}{\log a}
\qquad\text{and}\qquad
q = \frac{\log x}{\log b}
\qquad\text{and}\qquad
r = \frac{\log x}{\log abc} = \frac{\log x}{\log a + \log b + \log c}
$$
Taking reciprocals of each equation, we can combine them to obtain:
\begin{align*}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
| *
*By my answer here and with the condition given,
$$2\left(a^3+b^3+c^3\right)>2\left(a^2b+b^2c+c^2a\right)$$
$\left(a^2b+b^2c+c^2a\right)>\left(ab^2+bc^2+ca^2\right)\\\impliedby ab(a-b)+bc(b-c)-ca(a-b)-ca(b-c)>0\\ \impliedby a(b-c)(a-b)-c(a-b)(b-c)>0 \\\impliedby (a-c)(a-b)(b-c)>0$
$$\color{blue}{\boxed{\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Degree-4 Polynomial Solve the equation $x^4 - 14x^3 + 50x^2 -14x + 1 = 0$. I am not sure about how to best proceed, and would like a solution that does not involved the generalised quartic formula.
| A more detailed solution:
If we divide the equation by $x^2$:
$$\frac{x^4}{x^2} - \frac{14x^3}{x^2} + \frac{50x^2}{x^2} - \frac{14x}{x^2} + \frac{1}{x^2} = x^2 - 14x + 50 - \frac{14}{x} + \frac{1}{x^2}$$
Then, combining like terms, we notice that:
$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50$$
If we let $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Factorising quadratics - coefficient of $x^2$ is greater than $1$ In factoring quadratics where the coefficient of $x^2$ is greater than $1$, I use the grouping method where we multiply the coefficient and constant together and then factor.
My question is can someone explain the math behind that?
Example:
$5x^2+11x+2,\... | Sometimes, instead of using symbols, it can help to just use numbers to keep track of patterns (and then generalize with symbols later).
To understand this grouping method, it is instructive to work the problem backwards, and to expand, instead of factor.
For example
$$(2x+3)(5x+7)\text{ the constants are all distinct ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculating point 2P on an elliptic curve The equation for the curve is $$y^2=x^3+ax+b$$ and the point in question is $P(x,y)$. We have to verify that the $x$ coordinate of $2P$ is $(x^4-2ax^2-8bx+a^2)/4y^2$. However, the value I get is $(9x^4+6ax^2-8xy^2+a^2)/4y^2$. I derived the algebraic formula for $2P$ ($x(2P)=m^2... | We have:
$$\tag 1 y^2 = x^3 + ax + b$$
To add a point, we have:
*
*$\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}$
*$v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}x$
*$x_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}$
*$y_3 = -(\lambda x + v)$ (Calcula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equality involving Taylor coefficients Considering the following series expansion
$$
\frac{1}{{1 - 2x - x^2 }} = \sum\limits_{n = 0}^\infty {a_n } x^n
$$
prove that $
\forall n,\,\exists m
$ such that $
a_n ^2 + a_{n + 1} ^2 = a_m
$
I tried proving this using the coefficients computed by Wolfram but it didn't rea... | First find the zeros of $1-2x-x^2$:
$$x=\frac{-2\pm\sqrt{4+4}}2=1\pm\sqrt2\;.$$
Let $\alpha=1+\sqrt2$ and $\beta=1-\sqrt2$; then
$$(1-\alpha x)(1-\beta x)=1-(\alpha+\beta)x+\alpha\beta x^2=1-2x-x^2\;,$$
so you can decompose $\frac1{1-2x-x^2}$ into partial fractions with denominators $1-\alpha x$ and $1-\beta x$ as
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How many 2m-permutations, consisting only of cycles of even length? How many 2m-permutations, consisting only of cycles of even length?
I have found this formula:
$$Q_2(n) =((2n − 1)!!)^2$$
but how it can be proven?
| The combinatorial species of permutations consisting only of even cycles is given by
$$\mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{Z}) + \mathfrak{C}_{=4}(\mathcal{Z}) +
\mathfrak{C}_{=6}(\mathcal{Z})+ \mathfrak{C}_{=8}(\mathcal{Z}) + \cdots).$$
This translates to the generating function
$$G(z) = \exp\left(\frac{z^2}{2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$ In my textbook they asked me to show that
$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
but this is not true, I think.
I put down
$$\begin{align}
\frac12(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)&=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right)]\\
&=\frac12\left[-1+2\sqrt... | $$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
$$\frac{1}{2^3}(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right]$$
$$=\frac18\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Give a general formula for the following sequence I have a sequence where the terms are $1,0,-1,-1,0,1,1,0,-1,-1...$. Can someone help me find a general formula in terms of $n$ for this sequence?
| We write the $n^{th}$ term of the sequence as $a_n$, then
$\displaystyle a_n = \begin{cases} 1 &\textrm{ if } n \equiv 1 \pmod 6 \\ 0 &\textrm{ if } n \equiv 2 \pmod 6 \\ -1 &\textrm{ if } n \equiv 3 \pmod 6 \\ -1 &\textrm{ if } n \equiv 4 \pmod 6 \\ 0 &\textrm{ if } n \equiv 5 \pmod 6 \\ 1 &\textrm{ if } n \equiv 0 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0 ^{\pi}\left (\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$ How would you evaluate the integral
$$\int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$$
The answer from Wolfram is $0$.
Would you use a substitu... | Here are the steps
$$ \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\frac{\sin\left(\frac{3x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Area of Triangle The position vectors of $A$ $B$ and $C$ relative to an origin $O$ are given by $OA=(2,1,3)$
$OB=(0,-1,7)$ and $OC=(2,4,7)$
Part i) Show that angle $BAC= \cos^{-1}(\frac{1}{3})$
Part ii) Using the previous part calculate the exact area of the triangle $ABC$
I was able to do both parts easily. For par... | Due to the Cosine Theorem:
$$\cos A = \frac{b^2+c^2-a^2}{2bc}$$
and since, through the Pythagorean Theorem:
$$c^2 = \|OA-OB\|^2 = 2^2+2^2+4^2 = 24,$$
$$b^2 = \|OA-OC\|^2 = 0^2+3^2+4^2 = 25,$$
$$a^2 = \|OB-OC\|^2 = 2^2+5^2+0^2 = 29,$$
we have:
$$\cos A = \frac{25+24-29}{10\sqrt{24}}=\frac{1}{\sqrt{6}}$$
so:
$$\sin A = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral with sector of ellipse An ellipse is given by the following equation:
$$
152 x^2 - 300 x y + 150 y^2 - 42 x + 40 y + 3 = 0
$$
After solving for the midpoint we have:
$$
152 (x-1/2)^2 - 300 (x-1/2) (y-11/30) + 150 (y-11/30)^2 = 1/6
$$
Introducing polar coordinates:
$$
x = 1/2 + r \cos(\theta) \quad ;... | $$\int\frac{dx}{a\cos^2x-b\cos x\sin x+c\sin^2x}=\frac2{\sqrt\Delta}\cdot\tanh^{-1}\bigg(\frac{b-2c\tan x}{\sqrt\Delta}\bigg)$$ where $\Delta=b^2-4ac$. If $\Delta<0$, just use Euler's formula to transform the hyperbolic arctangent of complex argument into a trigonometric one of real argument. For $\Delta=0\iff b=\pm2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I prove this seemingly simple trigonometric identity $$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... | I'd use the tangent half-angle substitution:
\begin{align*}
t&=\tan\frac\theta2 &
\sin\theta&=\frac{2t}{1+t^2} &
\tan\theta&=\frac{2t}{1-t^2} &
\sec\theta&=\frac{1+t^2}{1-t^2}
\\
u&=\tan\frac\phi2 &
\sin\phi&=\frac{2u}{1+u^2} &
\tan\phi&=\frac{2u}{1-u^2} &
\sec\phi&=\frac{1+u^2}{1-u^2}
\end{align*}
Then you have
\begin... | {
"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": 3
} |
Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
| Here are a set of complete solutions, with all cases accounted for.
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}
\overset{x=\frac{bt}{\sqrt{1-t^2}} }=\int_0^{\frac1{\sqrt{1+b^2}} }\frac{dt}{a^2+(b^2-a^2)t^2}
$$
which leads to the solutions for the three distinctive cases below
\begin{align}
& \Phi(b<a) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Equations with variable powers Find the roots of the equation $3^{x+2}$+$3^{-x}$=10 . By inspection the roots are $x=0$ and $x=-2$. But how can I solve this equation otherwise?
| Hint: Multiply both sides by the positive quantity $3^x$ to clear the fraction implied by the negative exponent, and rearrange a bit:
$$3^{x+2} + 3^{-x}=10$$
$$3^x\cdot3^{x+2} + 3^x\cdot 3^{-x} = 3^x\cdot 10$$
$$3^{2x+2} + 1 = 3^x\cdot 10$$
$$3^{2x}\cdot 3^2 + 1 = 3^x\cdot 10$$
$$(3^x)^2\cdot 9 + 1 = 3^x\cdot 10$$
$$9\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$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$$
Along my proof I am stuck at the above section where it w... | The base case is trivial, now we follow to the inductive step by asuming the induction hypothesis and proving for $n + 1$.
So:
\begin{align*}
1 + 4 + 7 + ... + 3n-2 + 3(n+1)-2 & = \frac{n(3n-1)}{2} + 3(n+1)-2\\
& = \frac{n(3n-1)}{2} + \frac{2(3(n+1)-2)}{2}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 10
} |
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ M... | This is indeed possible. Starting from your fourth line: $$\frac{n^2(n+1)^2}{4} + \frac{4(n+1)^3}{4} = \frac{n^2(n+1)^2 + 4(n+1)(n+1)^2}{4} = \frac{(n^2 + 4(n+1))(n+1)^2}{4}$$ $$=\frac{(n^2 + 4n + 4)(n+1)^2}{4} = \frac{(n+2)^2(n+1)^2}{4}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Writing an alternating series as two non-alternating series? How does one calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
if given this information: $$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6}.$$
How does one account for the $(-1)^{n+1}$ in the first series?
Note: in this earlier post, user David Hodde... | There are two kind of terms in series $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$: terms with sign $+$ and terms with sign $-$. Note that:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots$$
So $+$ occurs with terms for odd $n$ and $-$: for even t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence test for the series $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$
Determine convergence of the series
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$$
My proof: using comparison test I have
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} = \sqrt{n+\frac{1}{n}} - \sqrt{n} \to 0$$ for large $n$.
Th... | Hint: $\dfrac{\sqrt{n^2+1} - n}{\sqrt{n}} = \dfrac{1}{\sqrt{n}\cdot \left(\sqrt{n^2+1} + n\right)} < \dfrac{1}{n^{3/2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}... | Note
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}\color{#C00000}{\cdot 2x}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$
And Not:
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$
This is because:
$$\frac{d}{dx} {f(x)}^n = n (f(x))^{n-1}\color{gree... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
| $\bf{My\; solution::}$ Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
Let $$(\sec x+\tan x)= t\;,$$ Then $$\left(\sec x\cdot \tan x+\sec^2 x\right)dx = dt $$
So $$\displaystyle \left(\sec x+\tan x\right)dx = \frac{dt}{\sec x}\Rightarrow \displaystyle dx = \frac{dt}{\sec x}$$
Now Using $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 4
} |
Why is the extension $k(x,\sqrt{1-x^2})/k$ purely transcendental? Consider the function field $k(x,\sqrt{1-x^2})$ of the circle over an algebraically closed field $k$. Is $k(x,\sqrt{1-x^2})$ a purely transcendental extension over $k$?
I'm curious because I was reading the answer here. The proof shows $k(x+\sqrt{1-x^2})... | You are correct that there was a mistake in my original proof, but the result is still true. Instead, in the case that $\mathrm{char}(k) \neq 2$, consider the purely transcendental extension $k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$. This field contains
$$
\frac{1 - \left(\frac{\sqrt{1-x^2}}{1+x}\right)^2}{1 + \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$ For any natural number $n$ ,let $S(n)$ denote the sum of the digits of $n$.Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$
| Write $n=abc=100a+10b+c.$ Then, $S(n)=a+b+c.$ Note that $S(n)$ has two digits. Then,
if $a+b+c<10$ then $a+b+c=2$ (since $S(S(n))=a+b+c);$
if $10\le a+b+c<20$ then $a+b+c=11$ (since $S(S(n))=a+b+c-9);$
if $20\le a+b+c$ then $a+b+c=20$ (since $S(S(n))=a+b+c-18).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
The minimum range of a $n \times n$ grid Let $n > 1$ be a integer, we put $1, 2, \cdots n^2$ into the cells of a $n \times n$ grid.
Let the range of the grid be the maximal difference between two cells that are in the same row or in the same column. Find the minimum range of all possible grids.
| For $n$ even let $n=2k$.
$\left(\begin{array}{cccccc}
1&\cdots & k^2-k + 1&k^2+1&\cdots&2k^2-k+1\\
\vdots & \ & \vdots& \vdots&\ &\vdots\\
k &\cdots & k^2 & k^2+k& \cdots \ & 2k^2\\
n^2-2k^2+1&\cdots&n^2-k^2-k+1&n^2-k^2+1&\cdots&n^2-k+1\\
\vdots&\ &\vdots&\vdots&\ &\vdots\\
n^2-2k^2+k&\cdots&n^2-k^2&n^2-k^2+k&n^2-k&n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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. Let $A_k = 2^{3k-1}+5\cdot 3^k = \frac{1}{2}\cdot 8^k+5\cdot 3^k$.
Once you prove that the sequence $\{A_k\}_{k\geq 1}$ fulfills the recurrence relation
$$ A_{k+2} = 11\cdot A_{k+1} -24\cdot A_k $$
you trivially have that $A_{k+2}\equiv -2\cdot A_k\pmod{11}$ holds.
$A_2=77\equiv 0\pmod{11}$ and the conclusion is ... | {
"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": 8
} |
Limit of function using Taylor's Formula To find:
$$\lim_{x \to 0} \left(\frac{ \sin(x)}{x}\right)^\frac{1}{x}$$
by using Taylor's formula.
So I used the Taylor's formula for $\sin(x)$ and got::
$\sin(x) = x - \frac{x^3}{6} + O(x^4)$
And then my function becomes:
$$\lim_{x \to 0} \left(\frac{ x - \frac{x^3}{6} + O(x^2)... | $$
\lim_{x \to 0} \left( 1 + \frac {x^2}6\right )^{\frac 1x} = \lim_{x \to 0} \left[\left ( 1 + \frac {x^2}6\right )^{\frac 6{x^2}} \right ]^{\frac x6} = \lim_{x \to 0}e^{\frac x6} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac... | Try using this:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Let $a^3= \cos{\frac{2\pi}{7}},b^3= \cos{\frac{4\pi}{7}},c^3= \cos{\frac{6\pi}{7}} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Congruence properties of $a^5+b^5+c^5+d^5+e^5=0$? It is known that given a solution to,
$$a^4+b^4+c^4 = d^4\tag1$$
then either $-c+d,\;c+d$ is always divisible by $2^{10}$. For example,
$$95800^4+414560^4+217519^4=422481^4$$
then $217519+422481=2^{10}\cdot5^4$.
Duncan Moore noticed, but could not prove, a similar cong... | It is helpful to consider separately the cases with exactly two and exactly four odd terms (these are the only possibilities since a solution with zero odd terms would not be primitive, while a sum with an odd number of odd terms could not equal zero).
The case with exactly four odd terms looks difficult, but that with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Differentiating $ \left( 1 - \frac {1}{x} \right)^x $ I have a calculus question. How does one differentiate $\left(1-\frac{1}{x}\right)^x$, for x>1? It should be positive right?
| Notice that the power is $x$, which is not a constant, so if we let $ y = \left( 1 - \frac {1}{x} \right)^x $, Then we can take the natural logarithm to get $$ \ln y = \ln \left( 1 - \frac {1}{x} \right)^x = x \cdot \ln \left( 1 - \frac {1}{x} \right). $$
Differentiating each side (implicit differentiation), we get $$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find all integers that make this expression rational I came up with this difficult problem a while ago while solving another relatively easy problem.
Find all integers m and n, such that $m^2 + n^2$ is a square, and such that $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational.
I've already tried the pythagorean substitution ... | Too long for a comment. As requested by OP as a sidenote, if we relax the requirements on $m,n$ and allow rationals, then the system,
$$m^2+n^2 = x^2,\quad \frac{2(m^2+1)}{n^2+1}=y^2\tag1$$
does have an infinite number of rational solutions. Other than the obvious case $n=1$, we have two quadratic polynomials in $m$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$ Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$
Please guide me to solve this problem. I have differentiated it with respect to $x$ and make equal to zero, but couldn't get any point.
| HINT: Here we can use the trigonometric substitution $x=\cos 2\theta$ where $\theta\in[0,\frac{\pi}{4}]$
Now $$f(x)=(x+1)^{1/3}-(x-1)^{1/3}=(2\cos^2 2\theta)^{1/3}-(2\sin^2 2\theta)^{1/3}.$$ Then we can use the parametric differentiation for find the minimum and maximums.
OR:
Using the identity $a^3-b^3=(a-b)(a^2+ab+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding the residue of a function with a surd variable $$f(z) = \frac{1}{z-2\sqrt{z}+2}$$
Is this the correct way of doing this, please advice - Thanks, what i did was try to rationalise the expression first as follows:
$$f(z) = \frac{1}{z-2\sqrt{z}+2} = \frac{1}{(z+2)-2\sqrt{z}}*\frac{(z+2)+2\sqrt{z}}{(z+2)+2\sqrt{z}}... | Why are calculated The function {f\left( x \right)e^x }
\begin{array}{l}
{\mathop{\rm Re}\nolimits} s\left( {f\left( x \right); - 2i} \right) = \mathop {\lim }\limits_{z \to - 2i} \left( {z + 2i} \right)\left( {f(z)} \right) \\
= \mathop {\lim }\limits_{z \to - 2i} \frac{{z + 2 + 2\sqrt z }}{{z - 2i}} = \frac{{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $ \lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1} $ How can you calculate this limit?
$$
\lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1}
$$
I really don't have a clue what to do with the $e^{x-1}$
| The form of the expression in the limit is something like a difference quotient, and in fact, applying the difference-of-squares factorization $a^2 - b^2 = (a + b)(a - b)$ to the bottom shows that we can write the expression as the product of a difference quotient (whose limit is, by definition, a derivative) and a fun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Trigonometic Substitution VS Hyperbolic substitution The following tables were taken from University of Pennsylvania's page about Calculus:
Trigonometric Substitution
Hyperbolic Substitution
As you can see, the forms $1+x^2$ and $x^2-1$ are repeated in the tables. How does one know when a trigonometric substitution i... | In my opinion, the following substitutions are enough:
appears in integrand
substitution
convention
1
$\sqrt{x^2+1}$
$x=\sinh\theta$
$\theta\in\mathbb R$
2
$\sqrt{x^2-1}$
$x=\pm\cosh\theta$
$\theta\geq0$
3
$\sqrt{1-x^2}$
$x=\sin\theta$
$-\frac{\pi}2\leq\theta\leq\frac{\pi}2$
$\sinh$ and $\cosh$ are bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Series expansion of $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$? How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again??
| I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
An integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively Find an integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively. (Without Modular Arithmetic).
I don't know if it is right, but I got this
$$n=5x+4=13y+7$$
$$n=5(x-11)+59=13(y-4)+59$$
The $\operatorname{lcd}(5,13)=65$... | $5x - 13y = 7-4 = 3 \Rightarrow 5x = 15y + 3 - 2y \Rightarrow x = 3y + \dfrac{3-2y}{5} \Rightarrow 5\mid 3-2y \Rightarrow 3-2y = 5k \Rightarrow y = \dfrac{3-5k}{2} = \dfrac{3-k}{2} - 2k \Rightarrow 2\mid 3-k \Rightarrow 3-k = 2h \Rightarrow k = 3-2h \Rightarrow 3 - 5k = 3 - 5(3-2h) = -12 + 10h \Rightarrow y = \dfrac{3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Integration without complex analysis on rational-improper integral Evaluate:
$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$
Without the use of complex-analysis.
With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?
| Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx
$
$$\begin{align}
I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\
&=
\frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int \frac{dx}{x^2+x+1}$ I am trying to evaluate the following integral:
$$I=\int \frac{dx}{x^2+x+1}$$
I am not supposed to do it with complex numbers so it's kind of hard.
I checked the answer on WolframAlpha.
It gives $$I=\frac{\sqrt{3}}{2}\arctan{\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\right)}+C $$
Having... | Answering your second question.
Since $ax^2 + bx+c = a\Bigg[x^2 + \frac{b}{a}x + \frac{c}{a}\Bigg] = a\Bigg[(x+\frac{b}{2a})^2 - \frac{b^2 - 4ac}{4a^2} \Bigg]$ then
$$\int \frac{1}{ax^2+bx+c}dx = \frac{1}{a}\int \frac{1}{(x+\frac{b}{2a})^2 - \frac{b^2 - 4ac}{4a^2}}dx$$
And the next step will depend on the sign of $b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Cube roots escape $ \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c}, $
where $ a, b $ and $ c $ are positive integers. What is the value of $ a+b+c $?
This question appeared in one of my exams
| Let
$ x^3=5$
$ y^3=4$
$ 9\sqrt{\strut x-y}$
$ =(x^3+y^3)\sqrt{\strut x-y}$
$ =(x^2-xy+y^2)\sqrt{\strut(x-y)(x+y)^2}$
$ =(x^2-xy+y^2)\sqrt{\strut x^3+x^2y-xy^2-y^3}$
Now we can substitute the value already
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut 5+\sqrt[3]{\strut 100}-\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?
| Note that $$ \sqrt{x^2+6x}-\sqrt{x^2-2x} = \frac{8x}{\sqrt{x^2+6x}+\sqrt{x^2-2x}}, $$so our answer is the limit of the above as $ x \to -\infty $. But note that $ \sqrt {x^2 + 6x} + \sqrt {x^2 - 2x} \sim \mathcal{O}(2x) $, so the answer is $\frac{8}{2}=\boxed{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| You already know that $\sqrt{x^2 + 1} \sim x$. The problem is that that approximation isn't good enough for computing this limit.
However, you've done exercises to do this exact thing: use a differential approximation and the value of $\sqrt{x^2}$ to estimate the value of $\sqrt{x^2 + 1}$. e.g. maybe you've been asked ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
How to solve following limit I've been struggeling a bit with the following limit:
$\lim\limits_{x \to 0} \frac{a- \sqrt{a^2 - x^2}}{x^2}$
The solution is:
If a < 0 then -$\infty$ .
If a > 0 then $\frac{1}{2a}$
But I don't know how to get there.
Thank you.
| $$\begin{align*}
\lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}
&= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\
&= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\
&= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\
&= \frac{1}{a+\sqrt{a^2}}\\
\end{align*}$$
Now the tricky part i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $\sqrt {1+\sqrt 5}$ I considered simplifying $\sqrt {1+\sqrt 5}$.
So I started $(a+b \sqrt 5)^2 = 1 + \sqrt 5$.
This gave me $a\color{blue}{^2} + 5b^2 =1 , 2ab = 1$ so the result was
$ \sqrt{1+\sqrt 5} = \sqrt{1/2 - i} + \dfrac{\sqrt 5}{\sqrt{2 - 4i}}.$
I continued in the same spirit : $(a_2+b_2 i)^2 = 1/2... | (Too long for a comment.)
I think what he means is that since from $(1)$,
$$ \sqrt{1+\sqrt 5} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5}) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}\tag1$$
Or,
$$ \dfrac{\sqrt{1+\sqrt 5}}{2} = -\frac{\sqrt 5-1}{4}\Bigr(\sqrt{1+\sqrt 5}\Bigl) i +\dfrac{\sqrt 5}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \ldots $ $$
\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} \times\ldots$$
I already know a way to calculat... | (Solution by OP in question: converted to a community wiki answer)
I already know a way to calculate it:
$$\cos{\frac{\pi}{4}} = \sqrt{\frac{1}{2}}, \frac{\pi}{4} = x$$
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} = \sqrt{\frac{1+\cos{x}}{2}} = \cos{\frac{x}{2}}$$
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 0
} |
What is wrong with the sum of these two series? Could anyone help me to find the mistake in the following problem? Based on the formula of the sum of a geometric series:
\begin{equation}
1 + x + x^{2} + \cdots + x^{n} + \cdots = \frac{1}{1 - x}
\end{equation}
\begin{equation}
1 + \frac{1}{x} + \frac{1}{x^{2}} + \cdots ... | The domains of convergence of these two sequences don't coincide. One converges for $|x|>1$ and the other for $|x| < 1$. Therefore, the sum is meaningless.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Maximum of $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0How Can I Maximise $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0<x < 12$
Using Inequality
$\bf{My\; Try::}$ In $0<x<12\;,$ The value of $(45-2x)\;,(24-2x)\;,2x>0$
and we can write $\displaystyle f(x) = \frac{1}{2}\bigg[(45-2x)\cdot (24-2x)\cdot (4x)\bi... | We can also use AM-GM as follows:
$$f(x) = \frac{1}{2\cdot 5\cdot 7} [2(45-2x)]\cdot [5(24-2x)] \cdot [7(2x)]$$
$$\leq \frac{1}{2\cdot 5\cdot 7\cdot 27}\left[2(45-2x) + 5(24-2x) + 7(2x)\right]^3 = \frac{210^3}{2\cdot 5\cdot 7\cdot 27} = 70^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
| One thing we tend to like to do when dealing with this stuff is have the denominator of fractionish things as a plain number, as much as possible. To do this, we'll multiply both top and bottom by something that will cancel out any radicals in the bottom.
$$\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Find Min Values Of $P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$ Given $x,y,z>0$ and $y+z=x(y^2+z^2)$
Find Min Values Of
$P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$
Could Someone give me an idea ?
| by the Lagrange Multiplier Method we get
$$P\geq \frac{91}{108}$$ the equal sign holds if $$x=\frac{1}{5},y=z=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Polynomial annihilator method $y''+8y=5x+2e^{-x}$ $y''+8y=5x+2e^{-x}$
It is asked to solve the equation by this method.
My result was $c_1\sin(\sqrt8x)+c_2\cos(\sqrt8x)+5x+\frac{2}{9}e^{-x}$
Then it is aked a particular solution so that $y(0)=0$ and $y'(0)=-{1/15}$
I arrived to $c_2=-\frac{2}{9}$ and to $c_1=-\frac{218... | We are given
$$\tag 1 y'' + 8y = 5x + 2e^{-x}, y(0)=0, y'(0)=-\dfrac 1{15}$$
We are asked to use the Annihilator Method, but here are some clearer notes that I will follow.
For the homogeneous part of $(1)$, we have:
$$(D^2+8)y = 5x + 2e^{-x} \implies D^2 + 8 = 0 \implies D = \pm 2 \sqrt{2}~i$$
This gives us a homogene... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double integral $ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy$, $D$=$\{(x,y): x^2+y^2 \le1 , y\ge0\}$
Solve $$ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy \ \ \ \ . . . \ (*)$$
where $D$=$\{$$(x,y): x^2+y^2 \le1 , y\ge0 $$\}$
$$
$$
Here is my attempt.
$$\begin{align}
&(1).\ \ \ (*)=\int_{-1}^1 \int_{0}^{\sqrt{1... | The integrand function $\frac{y}{x^2+(y+1)^2}$ suggest to put
$$\left\{
\begin{align}
x&=r\cos\theta\\
y+1&=r\sin\theta
\end{align}\right.
$$
so that $x^2+(y+1)^2=r^2$ and the Jacobian is $r$.
From $y\ge 0$ we have $y=r\sin\theta-1\ge 0$ that is $r\ge\frac{1}{\sin\theta}$ and from $x^2+y^2\le 1$ we have $$x^2+y^2=r^2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
| $$\left(\frac{\sin x}{x}\right)^2 + \left(\frac{\tan x}{x}\right) = \\ \left(\frac{x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots}{x}\right)^2 + \left(\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots}{x}\right) = \\ \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \ldots\right)^2 + \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
What is the appropriate substitution for this equation? I need to solve this equation:
$x^\frac{1}{3}+x^\frac{1}{7}=2(x^5)^\frac{1}{21}.$
I can't find the substitution which reduces it to a system or equation without roots.
| $$\text{The equation can be rewritten as follows:}$$
$$x^{\frac{7}{21}} + x^{\frac{3}{21}} = 2(x^{\frac{5}{21}})$$
$$$$
$$\text{Now, let } y = x^{\frac{1}{21}}:$$
$$y^7 + y^3 = 2y^5 \implies y^7 - 2y^5 + y^3 = 0$$
$$$$
$$\text{If } y = 0, \text{ then } x = 0, \text{ else we have: }$$
$$y^4 - 2y^2 + 1 = 0 \implies (y^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting
$$\frac{d^2y}{dt^2}=\frac{-... | $$
\dfrac{d}{dt}g(x)\dot{x}= \dfrac{dg}{dt}\dot{x} + g(x)\ddot{x}
$$
Now let's have $g(x) = -1/x^2$
We have
$$
\dfrac{dg}{dt} = \dfrac{dg}{dx}\dot{x} = \frac{2}{x^3}\dot{x}
$$
So hopefully it is clear to see that you have compute the derivative of $-1/x^2$ wrong?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove convergence of series Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$
so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since
both series ... | Hint. Here is an approach.
Recall that, for $x$ near $0$, you have by the Taylor series expansion or simply by the finite sum of a geometric sequence:
$$
\frac{1}{1+x}=1-x+x^2+\mathcal{O}(x^3),
$$
You may then write, as $n$ is great:
$$
\frac{(-1)^n}{n^{1/3}+(-1)^{\frac{n(n+1)}{2}}}=\frac{(-1)^n}{n^{1/3}\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me ... | This gif is a nice visualization of the following algebraic manipulations (color to correspond to the colors in the gif):
\begin{align*}
(1 + 2 + 3 + \cdots + n)^2
&= \sum_{i=1}^n \sum_{j=1}^n ij \\
&= \color{green}{\sum_{1 \le i < j \le n} ij} + \color{red}{\sum_{1 \le j < i \le n} ji} + \color{blue}{\sum_{1 \le j \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
If $0 \leq a, b, c \leq 1$, the inequality $| a - b + c| \cdot |(a - c)^2 + b^2| \leq 1$ holds Please, help me to prove the inequality or find a counter-example
For $0 \leq a, b, c \leq 1$, prove that
$|a - b + c| \cdot |(a - c)^2 + b^2| \leq 1$.
| *
*$|a-c|\le 1 \implies(a-c)^2\le |a-c|$
*$b\le 1 \implies b^2\le b$
$\therefore (a-c)^2+b^2\le |a-c|+b=a-c+b=a+(b-c)$$\hspace{10mm}$(if $a\ge c$)
Now
\begin{align}
| a - b + c| \cdot |(a - c)^2 + b^2 | & = |[a - b + c] \cdot [(a - c)^2 + b^2] | \\[6pt]
& \le |[a - b + c] \cdot [a+(b-c)]| \\[6pt]
& = |[a - (b - c)] \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\frac{\cos^2 x}{1-\sin x }dx$ $\int\frac{\cos^2 x}{1-\sin x} dx $ can someone explain me how to solve this one and please show your complete solution? So am I supposed to make the numerator $1+sinx$? but I think that doesn't help. Should I do long division?
| We know that $\displaystyle \cos^2 x = 1 - \sin^2 x = (1-\sin x)(1+ \sin x)$
Hence,
$$
\require{cancel}
\begin{align}
\int\frac{\cos^2x}{1-\sin x}\mathrm dx &= \int \frac{\cancel{(1-\sin x)}(1+\sin x)}{\cancel{1- \sin x}}\mathrm dx\\
&= \int \mathrm dx + \int \sin x\\
&= x - \cos x + \color{gray}{\mathcal C}
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\bigl(1-\frac{1}{2k}\bigr)$
i cant see why we have :
*
*$$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$
*$$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$
Even i see the notion of Double factorial
this question is relat... | $\dfrac{(2n)!}{4^nn!^2}=\dfrac{(2n)!!(2n-1)!!}{(2^nn!)^2}=\dfrac{(2n)!!(2n-1)!!}{((2n)!!)^2}=\dfrac{(2n-1)!!}{(2n)!!}$
$\prod_{k=1}^n(1-\frac{1}{2k})=\prod_{k=1}^n\frac{2k-1}{2k}=\dfrac{\prod_{k=1}^n(2k-1)}{\prod_{k=1}^n(2k)}=\frac{(2n-1)!!}{(2n)!!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire... | Both sides are $n (n+1)(2n+1)/2$ which also happens to be
$3 (1^2 + 2^2 + \ldots + n^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 2
} |
How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\c... | I think it is better to count directly.
We have $5+10+15+...+5[n/5]=5\frac{[n/5]([n/5]+1)}{2}$ counts one factor $5$ from each multiple of $5$ raised to the power the number has.
Then $25+50+75+...+25[n/25]=25\frac{[n/25]([n/25]+1)}{2}$ counts one factor $5$ from each multiple of $25$ raised to the power it appears. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{... | $$f(x) = x e^{x} = \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} $$
Integrate to get
$$\int_0^x dt \, t e^t = \sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)(n-1)!} $$
Integrate by parts...
$$\int_0^x dt \, t e^t = x e^x - \int_0^x dt \, e^t = (x-1) e^x +1$$
Plug in $x=1$ on both sides to get
$$\sum_{n=1}^{\infty} \frac{1}{(n+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 3
} |
Closed form solution for $\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}$. Let
$$
\text{S}_k = \sum_{n=1}^\infty\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{\ddots1+\cfrac{1}{1+n^2}}}},\quad\text{$k$ rows in the continued fraction}
$$
So for example, the terms of the sum $\text{S}_6$ a... | This is not a full answer but a partial result.
You can prove by induction that the described fraction of yours with $k$ horizontal lines is equal to $\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}$ with $k\ge3$. Perhaps with partial fractioning, you can compute the series using the well known result:
$$
\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 2,
"answer_id": 0
} |
How is the integral $\int \frac{1}{ae^{mx}+be^{-mx}} \, dx$ solved? I am trying to determine how the following integral was solved, or at least the name of article deriving this solution:
$$
\int \frac{1}{ae^{mx}+be^{-mx}} \, dx
= \frac{1}{m\sqrt{ab}}\,\tan^{-1} \left(e^{mx}\sqrt{\frac{a}{b}}\,\right)$$
| I don't know of the article where this result comes from, but it is definitely doable without it! First, notice that equality will fail if $a$ or $b$ are equal to zero. You cannot calculate the integral at all if both $a$ and $b$ equal zero. But the integral is very easy to solve if either $a$ or $b$ is zero. If both a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the value of this summation equation? The question is:
$$\sum_{i=1}^n (i^2+3i+4)$$
I get that
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(n+2)}{6}$$ and $$3\sum_{i=1}^n i = \frac{3n(n+1)}{2}$$ so one would get
I'll call this form1: $$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n$$
However, the textbook that I usin... | $$\sum_{k\le n} c=cn\neq c$$
In your example, $\displaystyle\sum 4=4n,$ yet you wrote $4$.
EDIT:
$$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n = n(n^2+6n+17)/3$$
$$\frac{n(n+1)(n+2)}{2} + \frac{9n(n+1)}{2} + 12n = n^3+6n^2+17n$$
$$\frac{n^3+3n^2+2n}{2} + \frac{9n^2+9n}{2} = n^3+6n^2+5n$$
$${n^3+3n^2+2n} + {9n^2+9n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\ta... | This is not a fully sketched answer.I enjoyed while going through it and hence thought others might like it.
Since $\displaystyle\sin\frac{\pi}{10}=\frac{1}{4}(\sqrt{5}-1)$ we obtain $\displaystyle\sin\frac{\pi}{30}=\frac{1}{8}(\sqrt{30-6\sqrt{5}}-1-\sqrt{5})$ and consequently $$\tan\frac{\pi}{30}=\sqrt{7-2\sqrt{5}-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Show that x is independent of this trig expression For some reason I cannot get the solution.
Show that $x$ is independent of: $\sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z)$
I have used all the identities but seem to be missing something.
| $\sin^2(x+y)+\sin^2(x+z) =1-[\cos^2(x+y)-\sin^2(x+z)]$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$\cos^2(x+y)-\sin^2(x+z)=\cos(2x+y+z)\cos(y-z)$
and then use Werner formula, for $2\sin(x+y)\sin(x+z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can we say two algebraic expressions are "equal" if one is undefined at certain points and the other isn't? I'm trying to understand why it is that we can say $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{(x-1)} = x+1$ but then have it also be the case that the two functions $f(x) = \frac{x^2-1}{x-1}$ and $g(x)=x+1$ are n... | we have $$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1$$ if $x \ne 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Convert the Polar Equation to Cartesian Coordinates $$
r^2=\sec 4\theta
$$
I graphed this equations using Wolfram Alpha and found it to be 2 hyperbolas. I'm having difficulty showing this using the standard equations
$$
x=r\cos\theta \;, \; y=r\sin\theta \;, and \; x^2 +y^2 =r^2
$$
My work so far:
$$
r^2 = \sec4\theta... | i can take $$r^2 =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}$$ and turn it into a cartesian equation.
$\begin{align}
1 &=\frac{r^2}{r^4-8r^4\sin^2\theta + 8r^4\sin^4\theta}\\
&=\dfrac{(x^2+y^2)}{(x^2+y^2)^2 - 8(x^2+y^2)y^2 +8y^4}\\
&= \dfrac{(x^2+y^2)}{x^4 + y^4-6x^2y^2}\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
find limit by rationalizing? I'm extremely new to calculus so please excuse my lack of lingo/formatting.!
Here is the problem:
$$\lim_{n\to\infty} \sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n - 1}$$
| To rationalize:
*
*Multiply by $$\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$
$$\left(\sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n -
1}\right)\cdot\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$ $$= \frac{6n+3}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve for $y$ in $x=\sqrt{(y-1)/(y+1)}$ I always struggle with this:
Express $y$ in terms of $x$ where
$$x = \sqrt\frac{y-1}{y+1}$$
I know to square both sides and get $x^2 = \frac{y-1}{y+1}$
Then I'm thinking multiply both sides by $y+1?$ So we get $(y+1)x^2 = y-1$
But where do I go from here? If there was only o... | Here is what I would do:
\begin{align}
x=\sqrt{\frac{y-1}{y+1}} &\Longleftrightarrow x^2 = \frac{y-1}{y+1}\\[1em]
&\Longleftrightarrow (y+1)x^2 = y-1\\[1em]
&\Longleftrightarrow yx^2+x^2=y-1\\[1em]
&\Longleftrightarrow yx^2-y=-(x^2+1)\\[1em]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Closed form of a recursive relation A sequence $\langle a_n\rangle$ is defined recursively by $a_1=0$, $a_2=1$ and for $n\ge 3$,
$$a_n=\frac 12 na_{n-1}+\frac 12n(n-1)a_{n-2}+(-1)^n\left(1-\frac n2\right).$$
Find a closed form expression for
$$f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\b... | With Abhishek Bakshi hint ,Now I have complete this answer,
since
$$\dfrac{a_{n}}{n!}=\dfrac{1}{2}\dfrac{a_{n-1}}{(n-1)!}+\dfrac{1}{2}\dfrac{a_{n-2}}{(n-2)!}+\dfrac{(-)^n(1-\dfrac{n}{2})}{n!}$$
let
$b_{n}=\dfrac{a_{n}}{n!},b_{1}=0,,b_{2}=\dfrac{1}{2} $then we have
$$b_{n}=\dfrac{1}{2}b_{n-1}+\dfrac{1}{2}b_{n-2}+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
the general solution of $y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$ I have some trouble finding the correct solution for the difference equation
$$y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$$
I've found that the characteristic equation of the difference equation is $\lambda^3-\frac{2}{3}\lambda+\frac{1}{3}$.
By co... | How did you input the equation into Wolfram Alpha?
You have as solutions
$$
\begin{align}
a_n&=(-1)^n\tag{1}\\[9pt]
b_n
&=\frac12\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n+\left(\frac{1-\frac i{\sqrt{3}}}{2}\right)^n\,\right]\\
&=\mathrm{Re}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n\,\right]\tag{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1... | We have
$$2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}=\sqrt n\left(2-\sqrt{1-\frac1n}-\sqrt{1+\frac1n}\right)\\\sim_\infty \sqrt n\left(2-1+\frac1{2n}+\frac1{4n^2}-1-\frac1{2n}+\frac1{4n^2}\right)=\frac1{2n^{3/2}}$$
so the given series is convergent by comparison with a convergent Riemann series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Overdetermined system with parameters $$\begin{cases}
ax & +y &=2a\\
x &+by &= b \\
ax &+ (5-b^2)y &= 1 \tag{A,B,C (in order)}
\end{cases}$$
Where $(x,y)$ are variables and $a,b$ are constants.
*
*What combination of $a$ and $b$ yields infinitely many solutions to our system?
By subtracting $A$ from $C$ we get:... | My advice is to avoid dividing by variable expressions (since we might lose solutions or otherwise be implicitly making certain assumptions). We can row reduce the corresponding augmented matrix to get:
$$
\left[\begin{array}{cc|c}
a & 1 & 2a \\
1 & b & b \\
a & 5 - b^2 & 1
\end{array}\right]
\sim
\left[\begin{array}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\in... | By definition, $\arcsin(x)$ is the angle $\alpha$ such that $\sin(\alpha) = x$ and $-\pi/2 \le \alpha \le \pi/2$, while $\arccos(x)$ is the angle $\beta$ such that $\cos(\beta) = x$ and $0 \le \beta \le \pi$. Since $-\pi/2 \le \alpha \le \pi/2$, $\cos(\alpha) \ge 0$, so we have $\cos(\alpha) = \sqrt{1 - x^2}$. Simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
How to divide trigonometric ratios using identities? $$\frac{1-\tan^2x}{1+\tan^2x}$$
We know:
$$\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}$$
Now what? Flip denominator and times numerator?
Which equals ???
Please help - Thanks
| $$
\tan^2 x + 1 = \sec^2 x
$$
Thus
$$
\frac{1-\tan^2 x}{\sec^2 x} = \cos^2x - \sin^2x
$$
But going from your point of view
$$
\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}= \frac{\frac{1}{\cos^2x}}{\frac{1}{\cos^2x}}\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}=\\
\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}
$$
then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that distinct Fermat Numbers are relatively prime The Fermat numbers are defined by $F_m = 2^{2^m} + 1$.
Prove that for $m \ne n$ we have $(F_m, F_n) = 1$.
I have to first prove that $F_{m+1} = F_0F_1 \cdots F_m + 2$ by representing $F_{m+1}$ in terms of $F_m$.
| To see that $F_{m+1} = F_0 F_1\cdots F_m + 2$, proceed as follows.
\begin{align}F_{m+1} &= (2^{2^m} - 1) + 2\\
&= [(2^{2^{m-1}})^2 - 1] + 2\\
&= (2^{2^{m-1}} - 1)F_m + 2\\
&= [(2^{2^{m-2}})^2 - 1]F_m + 2\\
&= (2^{2^{m-2}} - 1)F_{m-1}F_m + 2\\
&\ldots\\
&= (2^{2^0} - 1)F_1F_2\cdots F_m + 2\\
&= F_0F_1\cdots F_m + 2.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}$ for $S=a_2+\dots +a_n.$ Let $a_2,\dots,a_n>0$ and $S=a_2+\dots +a_n.$ How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
| Because $\sum_{i=1}^na_i^{1-\frac{1}{i}}=\sum_{i=2}^na_i^{1-\frac{1}{i}}+1 < S+2\sqrt{S}+1=(\sqrt S +1)^2 \Leftrightarrow \sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
$ \ $
It's sufficient to show $(\sqrt S +1)^2 > {\sum_{i=1}^na_i^{1-\frac{1}{i}}}\tag 1 $
Suppose $S_n=a_2+\dots +a_n.$
Prove by induction:
When $n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Combinatorics, surjective functions with conditions Question: $A=\left\{ 1,2,3,4,5\right\} $
, $B=\left\{ 1,2,3\right\} $
. How many surjective functions are there such that $ f(1)\neq1$
,$f(2)\neq2$
,$ f(3)\neq2$
.
Solution: Overall we have $3^{5}-{3 \choose 1}2^{5}+{3 \choose 2}1^{5}=150$
functions that is s... | As an alternate method, let $A_i$ be the preimage of $i$ for $1\le i\le3$.
$\textbf{1)}$ If $1\in A_2$,
then there are $3\cdot3=9$ possible distributions of 4 and 5 if $2\in A_i$ and $3\in A_j$ with $i\ne j$
and there are $3^2-2^2=5$ possible distributions of 4 and 5 if $2, 3 \in A_i$ for $i=1$ or $i=3$.
$\textbf{2)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that there are no positive integers $a, b$ and $n >1$ such that $a^n – b^n$ divides $ a^n + b^n$. Prove that there are no positive integers $a$ , $b$ and $n>1$ such that $a^{n}–b^{n}$ divides $a^{n}+b^{n}$.
Can someone provide me a proof of this and explain it to me please.
| This is a problem that appears in this number theory book by Nivens et al (1.2.46, p 19). The following is a solution readily attained.
Problem: Prove that there are no positive integers $a,b,n > 1$ such that $(a^n-b^n)\mid (a^n+b^n).$
Proof. We may first prove the following fact:
$$
\text{For the integers $x,y$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
existence of solution to congruence $x^4 \equiv -4 \pmod p$ I stuck with the following question:
For which $p$ (prime numbers) there is a solution for the following congruence:
$x^4 \equiv -4 \pmod p$
I would greatly appreciate any help
| $$x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)$$
If $p$ divides both $(x^2+2x+2),(x^2-2x+2);$
$p$ must divide $(x^2+2x+2)-(x^2-2x+2)=4x$
If $p$ divides $4,p=2$
Else if $p|x\implies p$ must divide $-4\implies p=2$
So, for odd prime $p,$ it must divide exactly one of $x^2\pm2x+2=(x\pm1)^2+1$
So, we need $(x\pm1)^2\equiv-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Principal value of Fourier Integral I have tried to find the principal value of
$$\int_{-\infty}^\infty {\sin(2x)\over x^3}\,dx.$$
As $ {\sin(2x)\over x^3}$ is an even function, its integral may not be zero in the given limits. I cannot calculate its principal value as I met with a third order pole at $z=0$. The expect... | Unfortunately $$PV\int_{-\infty}^\infty {\sin 2x\over x^3}\,dx$$ diverges.
The integrals $\int_\frac{\pi}{4}^{\infty} {\sin 2x\over x^3}\,dx=\int_{-\infty}^{-\frac{\pi}{4}}{\sin 2x\over x^3}\,dx$ exist.
Since $$\sin 2x\ge \frac{4}{\pi}x$$ for $0<x<\frac{\pi}{4}$, we have $$\frac{\sin 2x}{x^3}\ge \frac{4}{\pi}\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$ Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$
I tried to redo this problem and got to this point
$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$
Let $z=x+iy$ then
$$2\sqrt{x^2+y^2}>1+(x... | let $D = \{z \colon 2|z| < |1 + z^2|\}.$ observe that the region $D$ is symmetric with respect to the transformations $z \to -z, z \to \bar z$ and $z \to \dfrac{1}{z}$ therefore it is enough to worry about the portion of $D$ in the first quadrant.
Let $o = 0, a = 1$ and $z = re^{it}, p = 1 + r^2e^{2it}$ so that $z$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{... | Calculation of $$I = \int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx\;,$$ Now Put $(1+x^n) = t^n\;,$ Then $nx^{n-1}=nt^{n-1}dt\Rightarrow x^{n-1}dx = t^{n-1}dt$
So we get $$I=\int\frac{(t^n-x^n)}{t^{n+1}}dx = \int \left[\frac{dx}{t}-\frac{dt}{t^2}\right] = \int d\left(\frac{x}{t}\right)=\frac{x}{t}+\mathcal{C}$$
So we get $$\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle ... |
An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
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.