Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Complex Modulus Thanks everyone that answered my last question! I had one more question for a different concept. If the modulus of a complex number $z = |z| = \sqrt{(a^2 + b^2)}$,where $a$ and $b$ are the real and imaginary constants of $z$, are you allowed to use the similar property for conjugates and say that the mo... | $|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{zz'+z+z'+1}=\sqrt{a^2+b^2+2a+1}$ if $z=a+bi$.
Alternatively, $|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{(a+1+bi)(a+1-bi)}=\sqrt{(a+1)^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trying to find the matrix by a given linear transformation and eigenvalue / eigenvector the question I'm trying to resolve is the following:
Given f: R3->R3 and f(1,1,0) = (1,0,0), f(0,1,1) = (1,1,0) and where 3 is an eigenvalue of the eigenvector (1,0,0).
a) Find the associated matrix to f on canonical base R3
Any he... | Denote $B= \{(1,1,0), (0,1,1), (1,0,0)\}$ a basis given. Let $C = \{e_1,e_2,e_3\}$ the canonical basis.
Note that $e_1 = 0\cdot (1,1,0) + 0\cdot(0,0,1) + 1(1,0,0), e_2 = 1 \cdot (1,1,0) + 0 \cdot(0,1,1)+ (-1)(1,0,0)$ and $e_3 = (-1)(1,1,0) + 1 \cdot (0,1,1) + 1 \cdot(1,0,0).$
Now, the matrix representation A of the li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ where $q+p = 1$ How to prove $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ ,where $q+p = 1$. And here $C_{m}^{n}=\frac{m !}{n !(m-n) !} $ for $m \geq n$,... | Here we seek to prove that
$$\sum_{M=k-1}^{2k-1}
\left({M\choose k-1} (1-x)^k x^{M-(k-1)}
+ {M\choose k} x^{k+1} (1-x)^{M-k}\right) = 1.$$
The LHS is
$$\sum_{q=0}^n {q+n-1\choose n-1} (1-x)^n x^q
+ \sum_{q=1}^n {q+n-1\choose n} x^{n+1} (1-x)^{q-1}
\\ = (1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q
+ x^{n+1} \sum_{q=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $ \prod\limits_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $ How would you prove that $$ \left(\forall n\in\mathbb{N}\right),\ \prod_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $$
Wolfram|Alpha says its limit would be exactly $ \frac{\cosh{\left(\frac{\sqrt{3}\pi}{2}\right)}}{\pi} $, but ... | You can also use the fact that
$$\ln(1+x)\leq x-\frac{x^2}{3}, x\in\left[0,\frac{1}{2}\right]$$
to conclude
$$\prod\limits_{k=1}^n\left(1+\frac{1}{k^3}\right)\leq2\prod\limits_{k=2}^n e^{\frac{1}{k^3}-\frac{1}{3k^6}}=
2\cdot e^{\sum\limits_{k=2}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)}$$
but
$$\sum\limits_{k=2}^n\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x+y^3,x^2+y^2,x^3+y$ are all integers, are $x,y$ both integers?
Let $x,y$ be both real numbers. If $x+y^3,x^2+y^2,x^3+y$ are integers, are $x,y$ both integers?
This question begins with two real numbers while usual number theory tricks are based on the precondition that the variables are integers. Showing $x,y$ a... | Here's a hint (or at least some useful steps that are too long for a comment): Let $t=xy$. Note that
$$\mathbb Z\ni (x^3+y)(x+y^3)-(x^2+y^2)^2=t^3+t-2t^2=t(t-1)^2$$
and
\begin{align*}
\mathbb Z
&\ni \big[(x^3+y)(x^2+y^2)-(x+y^3)\big]\big[(x+y^3)(x^2+y^2)-(x^3+y)\big]\\
&=t(t-1)^2+(x^2+y^2)^2t(t^2+t-1).
\end{align*}
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence. I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....
I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.
Assuming that $x_n \geq x_{n+1}$ , ... | Hint:
Actually, you only need to prove that all $x_n$ live in an interval $I$ (to be determined) such that
$$f(x)=\frac12\Bigl(x+\frac2x\Bigr)<x \quad\text{on }I.$$
So you have to determine the variations of the function $f$ and show by induction that $x_n\in I$ for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But ... | Note that $\{(x,y):16-x^2+y^2<0\}\cap\{(x,y):x^2+y^2-4<0\}=\emptyset$, and hence the two domains are the same in different expressions!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Limit of $\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with a > 0 and b > 0 Question:
$\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with $a>0$ and $b>0$.$a$ and $b$ are constant.
| Now that I have solved the problem, I think I should post the answers and thank everyone for their help.
$\lim _{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(\frac{1}{2}\right)^{\frac{1}{x}}\cdot \left(a^x+b^x\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(a^x+b^x\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha... | Since by Vieta's formulas the sum of the roots is $-2$, there is not much to do here: given that two roots are $\alpha$ and $\alpha^2+2\alpha-4$, the third one has to be $-\alpha^2-3\alpha+2$.
The interesting question, however, is: how do we realize that there are two
roots $\alpha,\beta$ fulfilling $\beta=\alpha^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{al... | For the second one we get
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
\frac{k+1}{n} {n-1\choose k} {n\choose k+1}
x^k (1-x)^{n-1-k}
\\ = (-1)^{n-1} \sum_{k=0}^{n-1} {n-1\choose k} {n+k-1\choose k}
(-1)^k x^k.$$
The LHS is
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
x^k (1-x)^{n-1-k}.$$
The coefficient on $[x^q]$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
2-split of $n$ is $\left\{ \lfloor \frac{n}{2} \rfloor,\lceil \frac{n}{2} \rceil \right\}$. What about 3, 4, ...? Clarification: $k$-split of $n$ is an ordered integer sequence $\left\{ a_1,\cdots,a_k \right\}\quad \text{s.t.}$
*
*$0\le a_1\le\cdots\le a_k$
*$a_1+\cdots+a_k=n$
*${\left(a_k-a_1\right)}$ is minimize... | It is clear your question is satisfied by the following partition:
$$x = \sum_{k=0}^{n-1}\bigg\lfloor\frac{x+k}{n}\bigg\rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. I've been asked to prove that $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. Proving that $1+x+\frac{x^2}{2!}>0$ is simple as you just need to consider the discriminant and show that it's less that 0. Showing $1+x+\frac{x^2}{2!} + \frac{x^3}{3... | As you say, $1+x+x^2/2+x^3/6=0$ has a unique solution, (with $x<0$). For this $x$,
$$1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}=\frac{x^4}{4!}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In $\triangle ABC$ with integer sides, $BC$ is the average of the other sides, and $\cos C=AB/AC$. Find the smallest possible area.
In $\triangle ABC$, the length of side $\overline{BC}$ is equal to the average of the other two sides. Also, $\cos C = \frac{AB}{AC}$. Given that all the side lengths are integers, find t... | Let $BC=n$, $BA=n-k$ and $AC=n+k$. Then
$$\cos C = \frac{n-k}{n+k}= \frac{ (n+k)^2+n^2 -(n-k)^2}{2n(n+k)}$$
which yields $n=6k$. Thus, the smallest triangle has the integer side lengths 5, 6 and 7, and the corresponding area is $6\sqrt6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$ I have first written $x^4-2x^2+2=(x^2-1)^2+1$ then I choose $t=x^2-1$. From this I got $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$
From this how to proceed?
| Let
\begin{eqnarray}
I(a)=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx
\end{eqnarray}
Then,
\begin{eqnarray}
I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)}dx\\
&=&\frac{1}{1+a}\int_0^\infty\left(\frac{1}{1+x^2}+\frac{x^2-1}{1+a x^4}\right)dx
=\frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this first order ODE right Here is ODE of first order:
$$ y'=- \frac{x^2}{y^3} $$
with $y(0)=1 , y(0)=-1 $
Can my method be the separation method? If I use it, it doesn't work out:
$ \frac{dy}{dx}= -x^2 \frac{1}{y^3} $
$ \leftrightarrow y^3 dy = -x^2 dx $
By integration I get:
$$ \int y^3 dx = \int -x^2 dx... | You are correct. If $y\not=0$ then
$$y'=- \frac{x^2}{y^3}\leftrightarrow \int y^3 dy = \int -x^2 dx \leftrightarrow \frac{1}{4} y^4= - \frac{1}{3} x^3+c \leftrightarrow y= \pm \sqrt[4]{ - \frac{4}{3} x^3 +c }. $$
Now for $y(0)=1$, we get
$$ y(x)= \color{blue}{+}\sqrt[4]{1 - \frac{4}{3} x^3},$$
whereas for $y(0)=-1$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a^{2} + ab + b^{2} | ab(a+b)$ then $\left | a-b \right | > \sqrt[3]{ab}$ This is one of my university homework's problems this week:
Suppose that a and b are two positive integer and $a^{2} + ab + b^{2} | ab(a+b)$ then prove that $\left | a-b \right | > \sqrt[3]{ab}$
I tried to prove that $a^{2} + ab + ... | We need $a \neq b$.
WLOG assume $a>b$. Let $d=(a,b)$. Then there exists two coprime positive integers $x> y$ such that $a=dx,b=dy$.
$a^{2} + ab + b^{2} | ab(a+b)$ equivalent to $x^2+xy+y^2 \mid dxy(x+y)$
Since $(x,y)=1$, $x^2+xy+y^2$ and $xy(x+y)$ are coprime. So we must have $x^2+y^2+xy \mid d$. Then $d\geq x^2+y^2+xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Schur inequality Show that for all positive real numbers $a$, $b$ and $c$ such that $abc=1$, the inequality $a+b+c+2a^4+2b^4+2c^4\ge \dfrac{3}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right)$ is true
I tried using Schur inequal... | We need to prove that
$$\sum_{cyc}(2a^4+a^2bc)\geq\frac{3}{2}\sum_{cyc}a^3(b+c)$$ or
$$\sum_{cyc}(4a^4-3a^3b-3a^3c+2a^2bc)\geq0$$ or
$$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^4-a^2bc)\geq0,$$ which is true by Schur and Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\r... | The derivative of $\sqrt{1-x}$ follows a pattern
$$f'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}}$$
$$f''(x) = \frac{1\cdot 3}{2\cdot 2}(1-x)^{-\frac{5}{2}}$$
$$\vdots$$
$$f^{(n)}(x) = \frac{(2n)!}{4^n n!}(1-x)^{-\frac{(2n+1)}{2}}$$
a product of odds divided by a power of $2$ which makes your derivative
$$\frac{100\cdot 198!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
show that limit $\lim_{n\to+\infty}f_{n}(x)=x+3$,if $f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$ let $x$ is give postive real number,if $f_{0}(x)=0,0<x\le\dfrac{1}{2}$, and such $$f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$$
show that
$$\lim_{n\to+\infty}f_{n}(x)=x+3$$
This problem is from AMM 11967(2017),this solution
| Let $g_n(x) = f_n(x) - x - 3$, then $g_0(x) = -(x + 3)$ and it suffices to prove that
$$
\lim_{n\to \infty} |g_n(x)| = 0.
$$
Note that since
$$
\begin{align}
g_{n+1}(x) = f_{n+1}(x) - (x + 3) &= \sqrt{6(1+x) + x^2 + 3 + g_n(x^2)} - (x + 3)\\
&=\frac{g_n(x^2)}{\sqrt{(x+3)^2 + g_n(x^2)} + (x + 3)},
\end{align}
$$
we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose tha... | It turns out that I solved the equations for the wrong variables. If I rewrite $(1)-(3)$ as
$$\begin{bmatrix}0&c&b\\ c&0&a\\ b&a&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}a\\ b\\ c\end{bmatrix}$$
and solve for $x,y,z$ instead, I will get the law of cosines, i.e.
$$x=\frac{b^2+c^2-a^2}{2bc}$$
et... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
} |
Proving a general reasult in polynomials. If $f(x) = (x-a)(x-b)(x-c)(x-d) - 1$ where $a, b, c, d$ are distinct integers. Prove that $f(x)$ can't be factorized into integer polynomial with $deg ≥1$
In the above question I proved for three degree polynomial using contradiction. But couldn't use the same for the even degr... | By rational root test or Gauss’s Lemma (see the proof there for the lemma over the integers), if $f(x)$ has linear factor over $\mathbb Q$, then $f(x)$ has a factor of the form $x-t,t\in {\mathbb Z}$. But $f(t)=0$ implies $$(t-a)(t-b)(t-c)(t-d)=1,$$ which is impossible, as $t-a,t-b,t-c,t-d$ are nonzero distinct integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $ Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$
where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.
I was thinking on partial fra... | $$\dfrac{ 1 }{(x^4+c)(x^2+1) }=\frac{1}{\left(x^2+1\right) \left(x^2-i \sqrt{c}\right) \left(x^2+i \sqrt{c}\right)}$$ Using partial fractions, it is
$$-\frac{1}{2 \left(\sqrt{c}-i\right) \sqrt{c} \left(x^2-i \sqrt{c}\right)}-\frac{1}{2
\left(\sqrt{c}+i\right) \sqrt{c} \left(x^2+i \sqrt{c}\right)}-\frac{i}{2
\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that the value of $\Delta$ is an integer for the given determinant
$\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$
taking $\sqrt 6$ out from the first column and performing the following operations ... | From here
$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$
Pull out a factor of $\sqrt{3}$ from the second row and a factor of $\sqrt{2}$ from the third to get
$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$ I'm in a little struggle with this limit, can anyone help me, please?
$$\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$$
I tried to use the logarithm to then use L'Hospital's rule but I got stuck here:
$\ln(L)=\lim_{x \to \... | Let $f(x) = (1-2\sin x)^{\tan(\frac{\pi}{6}-x)}$, then $f(x) = e^{g(x)}$ with $g(x) = \tan(\frac{\pi}{6}-x) \log (1-2\sin x)$.
$$\begin{align}
\lim\limits_{x \to \frac{\pi}{6}^- } g(x)
&= \lim\limits_{x \to \frac{\pi}{6}^- } \frac{\tan\left(\frac{\pi}{6}-x\right)}{\frac{\pi}{6}-x} \left(\frac{\pi}{6}-x\right)\log \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$ For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4\,{p}^{4}q+8\,{p}^{3}r+27\... | If $a+b-c<0$ and $a+c-b<0$, we obtain $a<0,$ which is a contradiction.
Thus, it's enough to prove our inequality for $a+b-c>0$, $a+c-b>0$ and $b+c-a>0.$
Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2\geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)\geq\frac{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we ge... |
Just some work, that was too big for a comment.
Well, when we have a $\triangle\text{ABC}$:
We know that:
$$
\begin{cases}
\angle\alpha+\angle\beta+\angle\gamma=\pi\\
\\
\frac{\text{a}}{\sin\left(\angle\alpha\right)}=\frac{\text{b}}{\sin\left(\angle\beta\right)}=\frac{\text{c}}{\sin\left(\angle\gamma\right)}\\
\\
\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistak... | The product formula for sine says
$$
\frac{\sin(x)}{x}=\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)\tag1
$$
Thus, for $0\lt x\le\pi$
$$
\begin{align}
\frac{\sin(x)}x
&\le1-\frac{x^2}{\pi^2}\tag2\\[3pt]
&\le1-\frac{x^2}{(x+\pi)^2}\tag3\\
x-\sin(x)
&\ge\frac{x^3}{(x+\pi)^2}\tag4
\end{align}
$$
Explanation:
$(2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x... | You don't have to make polynomials and solve it by generating functions,
let $a=3k$, and $c=2j$ for $k,j \in \mathbb{N_0}$
$$3k + 2j + b = 15 \iff 3k+2j=15-b$$
This is a classic linear diophantine equation, and since $\gcd(2,3)=1$, it has solutions. So, all we need to do is find initial solutions to $k,j$ in terms of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the stronger inequality of $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geq \frac{1}{\sum ab}$ For $a,b,c>0$ and $a+b+c=1.$ Prove$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}$$
This inequality is easy and there are two nice proof by AM-G... | An improvement of your last inequality is the following: For $a, b, c > 0$ with $a+b+c=1$,
\begin{align}
&\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\\
\ge\ & \frac{1}{ab+bc+ca} + 4\cdot \frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)}.\tag{1}
\end{align}
$4$ is the best constant in the sen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modular calculation high exponent? I want to show, that $5^{96}\equiv -1 \pmod{193}$, without using the formula for quadratic residue.
So far I have :
$5^{96}\equiv 5^{4\cdot24} \equiv 625^{24}\equiv 46^{24}\equiv 186^{12}\equiv -7^{12}\equiv 7^{12}\equiv 7^{3\cdot4}\equiv 150^4\equiv -43^4\equiv 43^4\equiv 112^2\equi... | Let $x=5^{96}$. Then in the field $\Bbb F_{193}$ we have
$$
1=5^{\phi(193)}=(5^{96})^2=x^2,
$$
so that $(x-1)(x+1)=0$ in $\Bbb F_{193}$. Since a field has no zero divisors, we must have either $x=1$ or $x=-1$.
But $5^4=625=46$, so that $5^8=46^2=186$ and $5^{16}=49$,
$5^{32}=85$, so that
$$
5^{96}=(5^{32})^3=85^3=-1.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc +... | Also, we can use SOS here:
$$\sum_{cyc}\sqrt{a^2+ab+b^2}-\sqrt3(a+b+c)=\sum_{cyc}\left(\sqrt{a^2+ab+b^2}-\frac{\sqrt3(a+b)}{2}\right)=$$
$$=\sum_{cyc}\frac{4(a^2+ab+b^2)-3(a+b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}=\sum_{cyc}\frac{(a-b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$ Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$. I remember that I need to use the extended Euclidean algorithm, but it has been some time, so I am a bit rusty. Thanks in advance!
Edit: I trie... | Extended Euclidean algorithm is exactly right. Maybe
$x^3 -2 = (x^2 -x +1) (x+1) - 3 $
$ x+1 \, \, = (-x/3)(-3) + 1$. This, we may write as $(-x/3)(-3) = (x+1) -1$
So that finally, $(x^3 - 2) (-x/3) = (x^2 -x +1) (x+1) (-x/3) - 3 (-x/3) $
The right side further becomes $(x^2 -x +1) (x+1) (-x/3) + (x+1) - 1 = (x+1) \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$, find $|a_1|$
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$. Find $|a_1|$
I could conclude that $\displaystyle \lim_{n \to \infty}a_n$ must be $\infty$. Bu... | By the Wallis product
\begin{align*}
\frac{2}{\pi } & = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\frac{{(2n - 1) \cdot (2n + 1)}}{{2n \cdot 2n}}} = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\frac{{a_{2n - 1} a_{2n} a_{2n + 1} a_{2n + 2} }}{{a_{2n} a_{2n + 1} a_{2n} a_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Surface integral of piecewise volume boundary? How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with findi... | The outward surface normals for the two surfaces are given by the gradients of the expressions:
$$z+x^2+y^2-\frac{1}{2}=0 \implies \vec{n} = \frac{1}{\sqrt{3-4z}}\langle 2x, 2y, 1\rangle$$
$$x^2+y^2-z^2 = 0 \implies \vec{n} = \frac{1}{2\sqrt{2}|z|}\langle 2x, 2y, -2z \rangle$$
and consider the two vector fields
$$F_1 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$ While trying to prove it, I proved the following two inequalities:
$a^4+b^4+c^4\ge abc(a+b+c)$ and
$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$
The later one, on some simplification gives
$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$
But we can't claim... | Using “Titu's Lemma” (also called “Engel's form” of the Cauchy-Schwarz inequality) you have for $a, b, c > 0$
$$
\frac{a^3b + b^3c + c^3a}{abc} = \frac{a^2}{c} + \frac{b^2}{a} +\frac{c^2}{b}
\ge \frac{(a+b+c)^2}{c+a+b} = a+b+c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Given $x\cdot x =x$ Prove that it is a commutative ring I have seen some threads related to this, but I want to know whether my approach is correct.
Given $R$ a ring and that $\forall$ $x \in R, x\cdot x=x$. Then prove
that it is a commutative ring.
My try:
Considering $$(x+y)\cdot(x+y)=x\cdot x+x\cdot y+y\cdot x+y... | Your proof looks fine! Just for fun, here is a another one using the same ideas.
Let $a,b \in R$. We have that \begin{align}
(a+b)(a+b) &= a\cdot a + a \cdot b + b \cdot a + b \cdot b\\
&= a + a \cdot b + b \cdot a + b\\
&\overset{*}{=} a + b
\end{align}
and using the Cancellation Law in $*$ gives us that $a\cdot b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in t... | I am confused by the suggestion to use trigonometric substitution, since $$\frac{\sqrt{16x^2 - 9}}{x} = 16 x \frac{\sqrt{16x^2 - 9}}{16x^2},$$ and the substitution $$u^2 = 16x^2 - 9, \quad 2u \, du = 32 x \, dx$$ yields $$\begin{align*}
\int \frac{\sqrt{16x^2 - 9}}{x} \, dx
&= \int \frac{u}{u^2 + 9} u \, du \\
&= \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 3
} |
Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of
$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$
First I try like
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}... | Using $\omega^7=1$ the second sum computes to
$$
\frac{3(\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1)}{\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1}=3
$$
For the first sum I do not obtain $5/3$. I obtain
$$
3\cdot\frac{5\omega^6+ 2\omega^5 + 11\omega^4 - 4\omega^3 + 11\omega^2 + 2\omega + 5}{9\om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfr... | Note that
$$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=
\frac{\sin^3\frac x2 -\cos^3\frac x2}{\sin\frac x2 - \cos\frac x2}
=1+\frac12\sin x
$$
which, along with $\frac{3\cos2x+6}{10}= \frac{9-6\sin^2 x}{10} $, leads to
$$6\sin^2 x +5\sin x +1=0$$
Then, $\sin\alpha =-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle
Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.
So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it... | Let us have a solution based on an alternative idea. We consider on the triangle $T$ the one-form
$$
\omega=\frac 12\cdot \frac {x\; dy - y\; dx}{1+x^2+y^2}\ .
$$
Then
$$
\begin{aligned}
2d\omega
&=
\frac\partial{\partial x}\left(\frac x{1+x^2+y^2}\right)
dx\wedge dy
+
\frac\partial{\partial x}\left(\frac {-y}{1+x^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y... | Below is a different solution. It's not as easy as other methods to which this equation might be amenable, but it generalizes nicely to degree two polynomial Diophantine equations in two variables and has a nice geometric interpretation. The general idea is to parameterize all rational solutions and then restrict them ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Does this double integral exists? I've calculated $\iint_D\frac{y}{x^2+y^2}\,dA$ where D is bounded by $y=x$, $y=2x$, $x=2$, in this way:
$$\iint_D\frac{y}{x^2+y^2}\,dA=\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx=\cdots = \ln \frac{5}{2}.$$
However, I wonder if the fact that the $\frac{y}{x^2+y^2}$ is not bounded on D... | $$\int \frac{ydy}{x^2+y^2} = \frac{1}{2} \int \frac{d(y^2+x^2)}{x^2+y^2} =\frac{1}{2} \ln(x^2+y^2) + C$$
So
$$\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx= \frac{1}{2} \int_{0}^{2}\ln(x^2+y^2)|_{x}^{2x} dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$M = \int_0^{\pi/2} \frac{\cos x}{(x+2)} dx$ and $N = \int_0^{\pi/4} \frac{\sin x \cos x}{(x+1)^2} dx$, then value of $M - N$ is? Options are :
(A) $\pi\quad$ (B) $\frac{\pi}{4}\quad$ (C) $\frac{2}{\pi + 4}\quad$ (D) $\frac{2}{\pi - 4}$
I have solved it using Taylor's expansion of the numerators.
E.g, $\cos x =... | Using Integration by Parts, we can write N as
$$N = -\frac{\sin x\cos x}{x+1}\vert_0^{\frac{\pi}{4}}+\int_0^\frac{\pi}{4}\frac{\cos^2x-\sin^2x}{x+1}dx=-\frac{2}{\pi+4}+\int_0^{\frac{\pi}{4}}\frac{\cos2x}{x+1}dx$$
In the second integral, let $x \to 2x \implies dx \to 2dx$
$$\therefore N=-\frac{2}{\pi+4}+\int_0^\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$ $$\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$$
I will give my solution, which, to my surprise, turned out to be erroneous:
$S_n=\frac{a_1+a_n}{2}n=\frac{\frac{n}{1+n^2}+\frac{n}{... | Divide the general term $\dfrac{n}{k^2+n^2}$ by $n^2$ we have : $\dfrac{1}{n}\cdot \dfrac{1}{1+ \frac{k^2}{n^2}} \to \displaystyle \int_{0}^{1} \dfrac{dx}{1+x^2}= \dfrac{\pi}{4}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the unit digit of: $\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$
Find the unit digit of:
$$\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$$
I am completely clueless about how to deal with such huge powers.
I noticed that the numerator is $({100^{100}})^{100}$, which brings some sort of similar... | $$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }$$
"Now the first two terms combined are divisible by $x+3$". In fact, you have done the critical step!
$$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }= \frac{x^{100}-3^{100}}{ x+3 }+\frac{3^{100}}{ x+3 }$$
Since $\dfrac{3^{100}}{ x+3 }=\dfrac{3^{100}}{ 100^{100}+3 }<1$, we see that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding Inverse of a 4*4 matrix Find the inverse of the following matrix
$$
\begin{array}{c}
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
\text { where } c_{0}=\frac{1+\sqrt{3}}{4 \s... | Posting as an answer as it the MathJax for the matrix will not fit in a comment. Mathematica gives a rather nasty expression for the inverse of the matrix:
$$
\left(
\begin{array}{cccc}
\frac{1}{560} \left(8 \sqrt{6}-29\right) & \frac{1}{560} \left(99-8 \sqrt{6}\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the permutation matrix P of PLU decomposition unique? Let $A$ be a square matrix. Then there exists a permutation matrix $P$ such that $A=PLU$, where $L$ is a lower triangular matrix and $U$ is an upper triangular matrix. To further ensure the uniqueness, we assume that the main diagonal of $L$ (or $U$) to be 1. So,... | No, the permutaion is not unqiue. Here is an example:
$$
\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} =\underbrace{\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{U}\\
\underbrace{\begin{pmatrix}
0 & 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of the following approximation for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant1+x+x^2$ I was going through the text Introduction to Algorithms by Cormen et. Al. where I came across the following approximation...
for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant 1+x+x^2$
Now I could prove the... | For $0\le x\le1$, $e^x=1+x+\sum_{n\ge2}\frac{x^n}{n!}\le1+x+\sum_{n\ge2}\frac{x^2}{2^{n-1}}=1+x+x^2$. For $-1\le x<0$, write $y:=-x$ so we want to prove $0<y\le1\implies e^{-y}\le1-y+y^2$. Indeed, $e^{-y}=1-y+\sum_{n\ge2}\frac{(-1)^ny^n}{n!}\le1-y+\sum_{n\ge2}\frac{y^2}{2^{n-1}}=1-y+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$. Prove: $\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{3}{2}$ with $a, b, c$ are positive real numbers.
Let $a \le b \le c$:
\begin{align*}
\sqrt{\dfrac{bc}{a(3b+a... | After replacing $a$ at $\frac{1}{a},$ $b$ at $\frac{1}{b}$ and $c$ at $\frac{1}{c}$ we need to prove that:
$$\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\geq\frac{3}{2}.$$
Now, by Holder $$\left(\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\right)^2\sum_{cyc}ac(3a+b)\geq(a+b+c)^3.$$
Id est, it's enough to prove that
$$4(a+b+c)^3\geq9\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric inequalities in a triangle What is the proof of
$~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$
in an acute triangle ?
This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when
$$T \geq \sin^k A+ \sin^k B+ \sin^k C~ ,~~~~ k \geq 3$$
| We have
\begin{align}
\cos^2A+\cos^2B+\cos^2C&=\frac12(1+\cos2A)+\frac12(1+\cos2B)+\cos^2C\\
&=\frac12(2+\cos2A+\cos2B)+\cos^2C\\
&=\frac12(2+2\cos(A-B)\cos(A+B))+\cos^2C\\
&=1+\cos(A-B)\cos(\pi-C)+\cos^2C\\
&=1-\cos(A-B)\cos C+\cos^2C\\
&=1-\cos C(\cos(A-B)-\cos(\pi-(A+B)))\\
&=1-\cos C(\cos(A-B)+\cos(A+B))\\
\cos^2A+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i... | Notice: \begin{align}a \cdot (a + \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) +a\sqrt{a^2-b^2} \\&=
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} +a)\end{align}
and \begin{align}a \cdot (a - \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) -a\sqrt{a^2-b^2} \\ &=
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} -a)\end{align}
So \begin{align} \lim_{b \to a} \dfrac{a \cdot (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series wh... | This series can be evaluated with some generating function techniques. It will be related to the well-known generating function
$$ H(x) = \sum_{n=1}^\infty \frac{1}{n} x^n $$
which has closed form found by:
$$ H'(x) = \sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$
$$ H(x) = \int\! \frac{dx}{1-x} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a... |
\begin{align} x+y+z&=k \tag{1}\label{1},\\ x^2+y^2+z^2&=2k\tag{2}\label{2}. \end{align}
Expressing $y,z$ in terms of $x,k$ gives
\begin{align}
y,z&=
\tfrac12\,(k-x\pm\sqrt{-k^2+2 x k-3x^2+4k})
,
\end{align}
so we must have
\begin{align}
-k^2+2 x k-3x^2+4k\ge0
,
\end{align}
which leads to expression of $k$ in terms of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| God knows why do you want to use L'Hopital rule only. Either you can use brute force Differentiation or you can simply put aside $\cos x$ as others have pointed out, because it is not causing any problem in the limit.
However, a creative way to do this limit-
Let $L= \displaystyle \lim_{x\to 0 } \frac{\tan x - \sin x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Factoring $ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$
Factor $$ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$$
I used the factor theorem to get factors
$$f(a, b, c)=(a-b)(b-c)(a-c)\;g (a, b, c)$$
for some polynomial $g (a, b, c)$.
How can I continue using this method?
(sorry for the previously messed up question I'm ne... | @JohnBentin's method is the slickest, but if you don't want to check one example to get the scaling factor, let $x:=b^2-a^2,\,y:=c^2-b^2$ so your sum is$$\begin{align}abx+bcy-ca(x+y)&=a(b-c)x+c(b-a)y\\&=a(b-c)(b-a)(b+a)+c(b-a)(c-b)(c+b)\\&=(b-a)(c-b)(c(c+b)-a(b+a))\\&=(b-a)(c-b)(c^2+bc-ab-a^2)\\&=(b-a)(c-b)(c-a)(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| In the comments, you recognized that $12^1+12^2+12^3+\cdots+12^{100}$
$\equiv \underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{... | With the change of variable $x=\sin 2t$, we have
$$\sqrt{1+x}-\sqrt{1-x}=\sqrt{\cos^2t+2\cos t\sin t+\sin^2t}-\sqrt{\cos^2t-2\cos t\sin t+\sin^2t}=2\sin t.$$
Then
$$\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int\frac{2\cos2t}{2\sin t}dt=\int\left(\frac1{\sin t}-2\sin t\right)dt
\\=\text{arcoth}(\cos t)+2\cos t+C.$$
From th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_... | There is also the following way.
Since for $\prod\limits_{cyc}(a+b-c)\leq0$ the inequality is true,
it's enough to prove our inequality for $\prod\limits_{cyc}(a+b-c)>0.$
Now, if $a+b-c<0$ and $a+c-b<0$ so $a+b-c+a+c-b<0,$
which is a contradiction.
Thus, it's enough to assume that $z=a+b-c>0$, $y=a+c-b>0$ and $x=b+c-a>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{b}{b-1}\right)}}$ for all $b>2$. While solving a bigger problem, I've reduced it to an inequality $$\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{\color{red}b}{b-1}\right)}}$$for $b>2$, which lo... | This is a partial answer, at least. I'm planning to keep thinking about this tomorrow, but maybe someone else can step in and finish what I have. The argument is almost complete, but I have other things to work on right now.
First we log both sides to get the equivalent inequality
$$
b \log \left ( 1+2^{b^{\frac{1}{b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all... | hint
Multiply the first term by $ b-c$ to get in the numerator
$$b^3+c^2b-a^2b-b^2c-c^3+a^2c$$
and the others, by permutation
$$c^3+a^2c-b^2c-c^2a-a^3+b^2a$$
$$a^3+b^2a-c^2a-a^2b-b^3+c^2b$$
The result is $ -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to prove that $f(x)-x f'(x)\neq 0$? I have this function:
$f(x)=\cosh ^{-1}\left(\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2 x^2+1\right)^2}\right),$
where $0<x<\frac{1}{a}$ and $a$ is a positive real number. I want to prove that $\;f(x)-x f'(x)>0$, or at least to prove that $f(x)-x f'(x)\n... | Partial answer
Remarks: It appears that $h''<0$ on $(0, \frac{1}{a})$, if $a \ge 1$ (I think that proof is not very difficult, according to the expression of $h''$ below).
The difficult case is $a < 1$ in which $h''$ can be positive.
Let us prove that if $a \ge 2$, then $f(x) - xf'(x) > 0$ on $(0, \frac{1}{a})$.
We hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\the... | You're almost there.
$$\int \tan \theta \sec^3 \theta d \theta = \int \frac {\sin \theta}{\cos^4 \theta} d \theta = -\int \frac{du}{u^4}=\frac 13 u^{-3} = \frac {\sec^3 \theta}{3}+ C$$
Since $x= \tan \theta, x^2+1=\sec^2 \theta$, so
$$\frac{\sec^3 \theta}{3} + C = \frac{\sqrt{(x^2+1)^3}}{3}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi... | Yes, any polynomial identity involving $\cos(mx + c)$ and $\sin(mx+c)$ for various constants $c$ and integers $m$ can be written in the form
$R(z) = 0$ where $z = e^{ix}$ and $R$ is a rational function involving the $e^{ic}$. For this to be true, the numerator of $R(z)$ must simplify to the polynomial $0$.
For example... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does n... | Let the terms of the sum be $a_n$. You need to find:
$$a_{n+1}+a_{n+2}+\cdots+a_{2n}=\\
(a_1+\cdots+a_{n}+a_{n+1}+\cdots+a_{2n})-(a_1+\cdots+a_n)=\\
S_{2n}-S_n$$
In your first method, you are subtracting the term $a_{n+1}$ and losing it.
Addendum: Note the correct formula to use is:
$$S_n=\sum_{k=1}^n k(k+1)=\frac{2n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {x... | You can proceed like this: $$(a-x)(a-y)(a-z) \leqslant \left( \dfrac{(a-x)+(a-y)+(a-z)}{3} \right) ^3 =\dfrac{8}{27} a^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
To prove that (as I believe) a particular pair of diophantine equations has no solutions in positive integers. The diophantine equation $$a^2+b^2+ab=d^2$$ has positive integer solutions. If, in addition,$$ a^2+b^2=c^2$$ I suspect that it does not. Is there an easy way of seeing this?
| \begin{cases}
\begin{split}
a^2+b^2+ab&=d^2\\
a^2+b^2&=c^2\\
\end{split}
\end{cases}
Parametric of solution of second equation is given below.
$a = m^2-n^2$
$b = 2mn$
$c = m^2+n^2$
Substitute above ${a,b}$ to first equation, then we get
$$d^2 = m^4+2nm^3+2m^2n^2-2n^3m+n^4$$
Let $x=\frac{m}{n}, y=\frac{d}{n^2}$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
find function $f$ such that $f(x)=xf(x-1)$ and $f(1) = 1$
Find function $f$ such that $f(x)=xf(x-1)$ and $f(1) = 1$.
I can prove that there is just one function as $f$ (see Proof1).
I know that there exists a pi function $\Pi(z) = \int_0^\infty e^{-t} t^z\, dt$ that fits as $f$ so it is the only solution.
My problem... | No, it is not true. Given any solution $f$ and any function $g$ that is periodic with period $1$ (i.e. $g(x+1) = g(x)$) and $g(1)=1$, then $h(x) = f(x) g(x)$ also satisfies your equations. For example, you could take $g(x) = 1 + \sin(2\pi x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real ro... | $$\frac{x^9+6x^6+31x^3-1}{x^3-3x^2+4x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$ The sequence $a_n$ is defined by $a_1 = \frac{1}{2}$ and $a_n = a_{n - 1}^2 + a_{n - 1}$ for $n \ge 2.$
Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$
| You have
$$\frac{a_k}{a_{k-1}}=a_{k-1}+1$$ and
$$a_k - a_{k-1}=a_{k-1}^2$$
Hence,
$$\sum_{k=1}^n\frac 1{a_k+1}= \sum_{k=1}^n\frac{a_{k}}{a_{k+1}}= \sum_{k=1}^n\frac{a_{k}^2}{a_{k}a_{k+1}}$$
$$= \sum_{k=1}^n\frac{a_{k+1}-a_k}{a_{k}a_{k+1}}=\sum_{k=1}^n\left(\frac{1}{a_{k}} - \frac 1{a_{k+1}}\right)$$
I leave the last s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}... | Let $f=\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}$. Then
$$\begin{align}f&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-(2n+1)\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Roze... | The third problem.
By C-S and AM-GM twice we obtain:
$$\sum_{cyc}\frac{d}{1+a^2b}=\sum_{cyc}\frac{d^2}{d+a^2bd}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(a+a^2bd)}=$$
$$=\frac{16}{4+ab(ad+bc)+cd(bc+ad)}=\frac{16}{4+(ab+cd)(ad+bc)}\geq$$
$$\geq\frac{16}{4+\left(\frac{ab+cd+ad+bc}{2}\right)^2}=\frac{16}{4+\left(\frac{(a+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Sum of a diagonal matrix and a symmetric positive semidefinite matrix Let the following symmetric matrix
$A = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right]$,... | Note that $A$ is positive semidefinite is not positive definite. In other words, $A$ is positive semidefinite and singular.
First, note that if all three $d_{ii}$ are positive, then $A + D$ will necessarily fail to be positive semidefinite.
So, we must suppose that at least on of these is zero. Without loss of generali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$6$-digit permutation problem How many $6$-digit different numbers greater than $400000$ can be formed from the digits $2$, $2$, $5$, $6$, $7$, $7$, $8$?
Note: $22$ and $77$ are allowed in the number. For instance, $622775$ is allowed.
My approach: $5$ ways to choose the first digit, $6$ ways to choose the second, $5$ ... | Since the number must exceed $40000000$, the leading digit may be $5$, $6$, $7$, or $8$. Of these digits, only $7$ may be repeated. Therefore, it makes sense to handle the case in which the leading digit is $7$ separately from the others.
Since there are a total of seven available digits, six of them must be used. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\binom{n}{0}\binom{n+1}{n} +\binom{n}{1}\binom{n}{n-1} +\binom{n}{2}\binom{n-1}{n-2} +\cdots +\binom{n}{n}\binom{1}{0} = 2^{n-1}(n+2)$
Prove the below:
$$\binom{n}{0}\cdot\binom{n+1}{n} +\binom{n}{1}\cdot\binom{n}{n-1} +\binom{n}{2}\cdot\binom{n-1}{n-2} +\cdots +\binom{n}{n}\cdot\binom{1}{0} = 2^{n-1}\cdot... | Hint:
$$\binom nr\cdot\binom{n+1-r}{n-r}=\binom nr(n+1-r)=(n+1)\binom nr-r\binom nr$$
For $r\ge1,$ $$r\binom nr=n\binom{n-1}{r-1}$$
Now put $m=n,n-1$ one after another in $$(1+1)^m=\sum_{r=0}^m\binom mr$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not s... | After square completion, there is a simple solution just using the triangle inequality for the Euklidean distance:
Using $(x-1)^2 + (y+1)^2 =4$ we get
$$x^2+y^2= \left(\left|\binom xy\right|\right)^2 \stackrel{\triangle-ineq.}{\leq} \left(\left|\binom{x-1}{y+1}\right| + \left|\binom{-1}{1}\right|\right)^2= (2+\sqrt 2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Interpreting $a^2b \div \frac13a^2b^3$. The maths book I'm using shows:
$$ a^2b \div \frac13a^2b^3 $$
Which would be something like:
$$ a^2 \cdot b ÷ \frac13 \cdot a^2 \cdot b^3 $$
My understanding of order of operations it would equal:
$$ 3a^4b^4 $$
However in the book the second term is evaluated before the division ... | In this form
$$a^2b \div \frac13a^2b^3$$
my interpretation is to consider the two terms as a whole, that is
$$\left(a^2b\right) \div \left(\frac13a^2b^3\right)=\frac{3a^2b}{a^2b^3}=\frac 3 {b^2}$$
Otherwise, in this form
$$a^2 \times b ÷ \frac13 \times a^2 \times b^3$$
whithout parenthesis my interpretation would be
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Inequality with a High Degree Constraint This question-
Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$
The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes i... | A solution due to AoPS user @realquarterb:
By AM-GM, we have $1 + 19 x^{100/19} \ge 20 x^5$.
By AM-GM, we have $10\frac{x^4}{y^3} + 3x^{10} + 6x^5y^5 \ge 19x^{100/19} \ge 20x^5 - 1$.
Thus, we have $10 \sum_{\mathrm{cyc}} \frac{x^4}{y^3} + 3(x^5+y^5+z^5)^2 \ge 20(x^5+y^5+z^5) - 3$. (Q. E. D.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(a, b,c, d): 2a - b = 0, 2a - 3c - d = 0\}$. Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(x, y,z, w): 2x - y = 0, 2x - 3z - w = 0\}$ where $x_1 = (1,2,1, -1), x_2 = (2,4,1,1), x_3 = (-1,-2,-2,-4) \text{ and } x_4 = (3,6,2,0)$.
My approach:
We have ... | I suspect it should be $x_3 = (-1,-2,-2,4)$. Then the correct reduced matrix is
$$\left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
2 & 4 & -2 & 6 & y\\
1 & 1 & -2 & 2 & z\\
-1 & 1 & 4 & 0 & w\\
\end{array}\right] \sim \left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
0 & 0 & 0 & 0 & 2x-y\\
0 & 1 & 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $S$ is the unit square with opposite corners $(0, 0)$ and $(1, 1)$, show $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2})$ Let $S$ be the unit square with opposite corners $(0, 0)$ and $(1, 1)$. Show $$\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2}).$$
What I tried:
Since $\iint_S \frac{y^2}{\sqrt{(... | Under the variable interchange $x\leftrightarrow y$ we have that
$$I = \iint\limits_{[0,1]^2} \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}dA = \iint\limits_{[0,1]^2} \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}dA$$
and adding the two integrals gets us
$$2I = \iint\limits_{[0,1]^2} \frac{x^2+y^2}{(x^2+y^2)^{\frac{3}{2}}}dA \implies I = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $A_n A_{n+1}=A_{n}+2 A_{n+1}$ to disagree with a question Solving $$A_n A_{n+1}=A_{n}+2 A_{n+1} \tag1$$ to disagree with a question attached below:
Let us re-write *1) as
$$A_{n+1}(A_n-2)=A_n \tag 2$$
Let $A_n-2=B_n/B_{n-1}$ in (2) to simplify and to get
$$B_{n+1}-2B_{n-1}+B_n=0 \tag3$$
Let $B_n=x^n$, we get ... | This is a Ricatti recurrence:
$\begin{align*}
w_{n + 1}
= \frac{a w_n + b}{c w_n + d}
\end{align*}$
with $c \ne 0$ and $a d - b c \ne 0$. There are several ways to solve them. Brand "A Sequence Defined by a Recurrence Relation", AMM 62:7 (1955), pp 489-492 goes as follows.
Define:
$\begin{align*}
y_{n + 1}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x+\frac{1}{y+\frac{1}{z+\frac13}}=\frac{118}{51}$ find $xyz$. I'm having a difficult time figuring out how to solve this algebra problem.
If $$x+\cfrac 1 {y+\cfrac 1 {z+\cfrac 1 3}}=\frac{118}{51}$$
find $xyz$.
| I will presume that $x,y,z$ are intended to be positive integers.
\begin{align}
\frac{118}{51} & = 2 + \frac{16}{51} = 2 + \frac 1 {\left( \frac{51}{16} \right)} = 2 + \cfrac 1 {3 + \cfrac 3 {16}} \\[15pt]
& = 2 + \cfrac 1 {3 + \cfrac 1 {\left( \frac{16} 3 \right)}} = 2 + \cfrac 1 {3 + \cfrac 1 {5 + \cfrac 1 3}}.
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Challenging limit: $\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$ Here is a challenging limit proposed by a friend:
$$\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$$
and he claims that the close... | Let $f \colon (0,\infty) \to \mathbb{R},$
\begin{align}
f(\alpha) &= \frac{1}{2\alpha} - \int \limits_1^\infty \frac{\mathrm{d} x}{\sinh(\pi \alpha x) \sqrt{x^2 -1}} \stackrel{x = \cosh(t)}{=} \frac{1}{2 \alpha} - \int \limits_0^\infty \frac{\mathrm{d} t}{\sinh(\pi \alpha \cosh(t))} \\
&= \frac{1}{2 \alpha} - \int \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
An uncertainty for $a^3+b^3+c^3-3abc$ I have a doubt regarding $a^3+b^3+c^3-3abc$.
For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see-
$$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$
AM-GM inequ... | When $a+b+c=0$, then $a,b,c$ all of $a,b,c$ are not positive, So AM_GM will not be used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Use infinite series to evaluate $\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$ Use infinite series to evaluate $\displaystyle\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$
I know the limit is $-\dfrac{5}{3}$ after rationalizing the expression, but I don't know how to prove it using Ta... | Taylor series ... Take
$$
(x^3-5x^2+1)^{1/3} = x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3}
$$
Taylor series for $1-5z+z^3$ near $z=0$ is $1 - \frac{5}{3}z + o(z)$ as $z \to 0$. So as $x \to \infty$,
\begin{align}
(x^3-5x^2+1)^{1/3} - x &= x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3} - x
\\ &= x\left(1 - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Obtain sum of a sequence from sum of its odd terms. I would like to compute the sum
$$
\sum_{k=1}^\infty \frac{1}{k^4}
$$
by using the Fourier series of $f(x)=|x|$ over $(-\pi,\pi)$. Coefficients $b_k$ are all $0$ because $f$ is even. Doing the integration stuff, I obtained:
$$
a_0 = \pi
$$
and
$$
a_k = \frac{2}{k^2}\... | Observe that what you have is that $2\sum_{k=0}^{\infty} \frac 1{(2k+1)^4}=\frac {\pi^4}{48}$. Calling $\sum_{k=0}^{\infty} \frac 1{k^4}=S$ you have that $\sum_{k=0}^{\infty} \frac 1{(2k)^4}=\frac 1{16} S$ and finally you have $S-\frac 1{16}S=\frac 12 \frac {\pi^4}{48}$ from which $S=\frac {\pi^4}{90}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_... | Observe that $x_1 = 0$, $x_2 = 0$, $x_3 = 1$, $x_4 = \frac{4}{3}$. We can prove by induction that $x_n <2$ for all $n$. Suppose that the inequality is true for $x_1, x_2,\ldots, x_{n+1}$. Then
$$
x_{n + 2} = \frac{1}{3}x_{n + 1} + \frac{1}{6}x_n + 1 < \frac{2}{3} + \frac{2}{6} + 1 = 2.
$$
Now we show that the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\lim_{x\to\infty} \frac{1}{x} = 1$ is false $$\lim_{x\to\infty} \frac{1}{x} = 1$$
Given $\epsilon > 0$
$$\left|\frac{1}{x} - 1\right| < \epsilon.$$
Rewrite it as
$$-\epsilon < \frac{1}{x} - 1 < \epsilon$$
$$-\epsilon + 1< \frac{1}{x} < \epsilon + 1$$
If epsilon is very small, then on both sides we are getting va... | To show that the statement is false, find an $\epsilon$ which does not have a corresponding $x^\star$, such that whenever $x \geq x\star$, $| \frac{1}{x} - 1 | < \epsilon$ does not hold. Suppose $L = 1$ and let $\epsilon = \frac{1}{2}$.
Consider when $x^\star \geq 1$, then
\begin{align*}
| \frac{1}{x} -1 | \geq \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How can I solve the recurrence relation:$F\left(n\right)=F\left(n-1\right)+2F\left(n-2\right),\:F\left(0\right)=1,F\left(1\right)=1$ I've been trying to solve it using Power Series and Partial Fractions but got stuck with.
| Let $F(n) = x^n$. Then we have $x^n - x^{n-1} - 2x^{n-2} = x^{n-2}(x^2 - x - 2) = 0$. So in order to have solution(s) in the form of $Cx^n$ we should set $$x^2 - x - 2 = (x-2)(x+1)= 0 \implies x = 2 , -1$$ Combining these solutions, the general solution is $F(n) = a_12^n + a_2(-1)^n$. Using initial conditions we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of this geometric serie: $ 3+ \sqrt3 + 1 + ...$ I am trying to find the sum of this geometric series but can't find it:
$ 3+ \sqrt3 + 1 + ...$
The solution I get is:
$S=\frac{3(3+\sqrt{3})}{4}$
but the answer key shows:
$S=\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$
This exercice is from a book called Pre-Calc... | Indeed the common ratio was incorrect, it should be $\frac1{\sqrt3}=\frac{\sqrt3}{3}$
\begin{align}
S &= \frac{3}{1-\frac{\sqrt3}{3}}=\frac{9}{3-\sqrt3}=\frac{9(3+\sqrt3)}{9-3} =\frac{3(3+\sqrt3)}2 = \frac{3\sqrt3(\sqrt3+1)}{2}
\end{align}
Edit: Alternative working:
$$S=\frac{3}{1-\frac1{\sqrt3}}=\frac{3\sqrt{3}}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)... | Here is a straightforward solution based on quadratic inequalities.
For simplicity, denote $A=1-a+a^2$ and $B=1-b+b^2$. We need to show that
$$
3AB(1-c+c^2) \geq 1+abc +a^2b^2c^2.
$$
This is equivalent to showing that
$$
(3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1) \geq 0.
$$
If we regard the left-hand side above as a quadrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$3^{123} \mod 100$
The Question:
Evaluate $3^{123}\mod 100$
My Attempt
So initially I attempted to list the powers of 3 and find a pattern of the last two digits - which, despite much painful inspection did not yield an obvious useful pattern.
So I then attempted to simplify this and use Euler's Generalisation of F... | The OP began by looking for a pattern but stated
that
...despite much painful inspection did not yield an obvious useful pattern.
You can use some light theory to actually predict the form and structure of the pattern.
Observe that if $a \in \{0,2,4,6,8\}$ and $b \in \{1,3,7,9\}$ and
$\quad 3 \times (10 a + b) \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
What function $f(x)$ would this limit, involving 3 integrals be equal to? What function $f(x)$ would the limit
$$\lim_{n\rightarrow{0^+}}\left(-\int_{-\infty}^{-x-n}\frac{xK_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da+\int_{-x+n}^{x-n}\frac{xK_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-\int_{x+n}^\infty\frac{x... | The final answer is $$2\ \text{Chi}(x)\cosh(x) - 2\ \text{Ci}(x)\cos(x) - 2\ \text{Shi}(x)\sinh(x) - 2\ \text{Si}(x)\sin(x)+ \pi\sin(x)$$ for $x > 0$. $\text{Ci}(x)$ is the cosine integral, $\text{Si}(x)$ is the sine integral, $\text{Chi}(x)$ is the hyperbolic cosine integral, and $\text{Shi}(x)$ is the hyperbolic sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that $\frac{D(y^3D^2y)}{y^2}$ is constant If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that
$$\frac{D(y^3D^2y)}{y^2}$$
is a constant, where D denotes the derivative operator.
I have tried expressing the expression above... | First consider
$$Dy = \frac{1}{2y}Dy^2$$
and
$$D^2y = D(\frac{1}{2y}Dy^2) =\frac{-1}{2y^2}(Dy)(Dy^2) + \frac{1}{2y}D^2y^2 = -\frac{1}{4y^3}(Dy^2)^2 + \frac{1}{2y}D^2y^2 $$
Therefore, the numerator in your expression is
\begin{align}
D(y^3D^2y) &= D(-\frac{1}{4}(Dy^2)^2 + \frac{y^2}{2}D^2y^2)
\\
&=-\frac{1}{2}(Dy^2)(D^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+... | Proceeding along your approach:
It suffices to prove that $1 + p^2 - 2q + q^2 - 2pr + r^2 \ge 2q$.
Since $p^2 \ge 3q, q^2 \ge 3pr$, it suffices to prove that
$1 + 3q - 2q + q^2 - \frac{2}{3}q^2 \ge 2q$
or $1 - q + \frac{q^2}{3} \ge 0$
or $\frac{1}{3}(q - \frac{3}{2})^2 + \frac{1}{4} \ge 0$ which is true. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
When its possible to write a number $a+b\sqrt{c}$ as the square of a sum? I was teaching my high school students how to find the roots of 2nd degree equations like the following:
$\frac{x^2}{2}+\sqrt{3}x-\sqrt{2}=0$
In this case, using the formula we get:
$x=-\sqrt{3}\pm\sqrt{3-2\sqrt{2}}$
To simplify I writted $3-2\sq... | This is not as hard as it seems.
[I will slightly alter your notation, and instead of $(m+n)^2$ I will use $(m + n\sqrt{c})^2$.]
$$ (m + n\sqrt{c})^2 = a + b \sqrt{c} $$
Expanding and collecting everything on the left-hand side gives us:
$$ n^2 c + (2mn - b) \sqrt{c} + (m^2 - a) = 0$$
We are given $a, b, c \in \mathbb{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$? The wolframalpha gives the answer $0$:
Wolframalpha culculation
I tried like this:
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$,then:
$\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$ = $\lim \limits_{r→0} \frac{r^4}{r^2 \sin(\theta)\cos(\theta)}$ =... | We have that for $x=y=t\to 0$
$$\frac{(x^2+y^2)^2}{xy}=\frac{4t^4}{t^2}=4t^2 \to 0 $$
but for $x=t\to 0$ and $y=t^3$
$$\frac{(x^2+y^2)^2}{xy}=\frac{(t^2+t^6)^2}{t^{4}}=1+2t^4+t^8 \to 1 $$
therefore the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sq... | $$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Find the sum of series without differentiation Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum
My method does not require differentiation but there is a difficulty.
Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^... | Let
\begin{eqnarray*}
S=\sum_{i=1}^{\infty} i^2 x^i.
\end{eqnarray*}
Now multiply by $(1-3x+3x^2-x^3)$ and note that for $i \geq 2$
\begin{eqnarray*}
(i-3)^2-3(i-2)^2+3(i-1)^2-i^3=0.
\end{eqnarray*}
Examine the lower order terms more carefully, and we have
\begin{eqnarray*}
(1-3x+3x^2-x^3)S=x(1+x)
\end{eqnarray*}
givi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $ \int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $ How can I evaluate
$$ \int_{0}^{1} \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $$
U-substitution has not worked for me. Integration by parts, Differentiation under integral sign, Mathematica is not coming up with a solution... | To calculate this integral I'll use series expansion $$\frac 1 {\sqrt{1+x^2}}=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n}$$ for $|x|\le1$
$$I=\int_{0}^{1}\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n} \ln\left(x+\sqrt{1-x^2}\right)dx$$
By dominated convergence
$$I=\sum_{n=0}^{\infty }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Given the O.D.E. $2yy''=1+(y')²$ by using the transformation $y'=z, y''=z·(dz/dy)$ find the solutions.Why cases for y are not taken? By using this transformation we have:
$$\dfrac {2z\;dz}{z^2+1}=\dfrac {dy}y$$
Then by integrating the book comes to the conclusion that:
$z=±\sqrt {c_1y-1}$ , where $c_1$ is a non zero ar... | It's an autonomous equation.
We need to solve the ODE: $yy''=1+(y')^{2}$.
Now, treating $y$ as the independent variable, let $z=y'$ wich gives $y''=z\frac{dz}{dy}$, so $$2y\frac{dz}{dy}z=z^{2}+1$$Solving for $\frac{dz}{dy}$, we can see that $$\frac{dz}{dy}=\frac{z^{2}+1}{2yz} \overbrace{\implies}^{\div \frac{z^{2}+1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.