Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.
And here is my method:
$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$
Then, I define a new function here
$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$
We have
$$\begin{align*}
\lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\
&=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\
&=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\
&=0.
\end{align*}$$
This implies that
$$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$
Therefore,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\
&=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\
&=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\
&
\end{align*}$$
Here, we define two functions:
$$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad
g(x)=\frac{y(x)+1}{2}.
$$
We deduce that,
$$
\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.
$$
Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.
Finally we get,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}
&=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\
&=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\
&=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&&
\end{align*}$$
Is the method I use correct?
I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?
Thank you!
| By squaring, we can verify that
$$
x\le\sqrt{x^2+3}\le x\left(1+\frac3{2x^2}\right)\tag1
$$
and
$$
x+1\le\sqrt{x^2+2x+3}\le(x+1)\left(1+\frac1{x(x+1)}\right)\tag2
$$
Adding $(1)$ and $(2)$ gives
$$
2x+1\le\sqrt{x^2+2x+3}+\sqrt{x^2+3}\le(2x+1)\left(1+\frac3{2x^2}\right)\tag3
$$
Multiplying numerator and denominator by $\sqrt{x^2+2x+3}+\sqrt{x^2+3}$ gives
$$
\sqrt{x^2+2x+3}-\sqrt{x^2+3}=\frac{2x}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\tag4
$$
Bernoulli and cross multiplying yield
$$
1-\frac3{2x}\le\left(1-\frac3{2x^2}\right)^x\le\left(1+\frac3{2x^2}\right)^{-x}\tag5
$$
Therefore $(3)$, $(4)$, and $(5)$ yield
$$
\left(\frac{2x}{2x+1}\right)^x\left(1-\frac3{2x}\right)\le\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x\le\left(\frac{2x}{2x+1}\right)^x\tag6
$$
The Squeeze Theorem then says
$$
\lim_{x\to\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x=e^{-1/2}\tag7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
} |
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x)))=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{4}(x), r_{4}(x)=o(x)$$
That is why fifth expression of Taylor's is $\frac{2}{15}x^5$ for me. However Mathematica say that: $$\sin \tan x=x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^5)$$ Where have I a mistake?
| Taylor's expansions up to the same order can be composed. So you have to consider the Taylor polynomial of degree $5$ for $\sin x$, and substitute $x$ with the degree $5$ Taylor polynomial for $\tan x$, truncating the successive powers of the latter polynomial at degree $5$.
As a start-up, here is the computation for degree $3$:
$$\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)^2=x^2+\tfrac19x^4+o(x^5),$$
\begin{align}
&\text{so that}\hspace{6em} &\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)^3&=\bigl(x^2+\tfrac19x^4+o(x^5)\bigr)\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)&\hspace{6em} \\
&&&=x^3+\tfrac13x^5+\frac19x^5+o(x^5)=\color{red}{x^3+\tfrac49x^5}+o(x^5).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the following manner:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}
This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.
How can I extend the proof for (1) where $r$ is any real number?
When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)
Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}
My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.
| \begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &= \lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n = \lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^n = \lim_{n\to\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} \\ &= \lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n-1}\right)^n} = \lim_{n\to\infty} \left(\frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} \cdot \frac{1}{1 + \frac{1}{n-1}} \right) \\
&= \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} = \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n}\right)^{n}} = \frac{1}{e} \\
&= e^{-1}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
a tricky trig questions from Step If $\theta + \phi + \psi = \pi/2$, show that $\sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi = 1$.
By taking $\theta = \phi =\pi/5$ in this equation, or otherwise, show that $\sin(\pi/10)$ satisfies the equation
$$8x^3 + 8x^2 − 1 = 0$$
I got stuck in the first part. I want to prove this by making connection
with $\sin(\theta + \phi + \psi)=1$,but I failed.
| $$\begin{align*}& \sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi
\\& = \sin^2 \theta + \sin^2 \phi + \sin^2 \left ( \frac {\pi} {2} - (\theta + \phi) \right ) + 2 \sin \theta \sin \phi \sin \left (\frac {\pi} {2} - (\theta+ \phi) \right ) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 (\theta + \phi) + 2 \sin \theta \sin \phi \cos (\theta + \phi) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi - {2 \cos \theta \cos \phi \sin \theta \sin \phi} + {2 \sin \theta \sin \phi \cos \theta \cos \phi} - 2 \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + (1-\sin^2 \theta) \cdot (1 - \sin^2 \phi) - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi +1 - \sin^2 \theta - \sin^2 \phi + \sin^2 \theta \sin^2 \phi - \sin^2 \theta \sin^2 \phi \\ & = 1. \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve for x: $x^3-\lfloor x\rfloor=5$
Solve for x:$$x^3-\lfloor x\rfloor=5$$
My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here
| Write $\lfloor{x}\rfloor$ as $x-\epsilon$ where $\epsilon$ is the fractional part, so $0\le\epsilon<1$. Now consider the function $f(x)=x^3-x-5+\epsilon$, note that $f$ has at least one root since $f$ is cubic polynomial. Now differentiate and get $f’(x)=3x^2-1$ which has two roots $\pm\frac{1}{\sqrt{3}}$.
Now:
$f(\frac{-1}{\sqrt{3}})=-\frac13+\frac{1}{\sqrt{3}}-5+a<0$ thus no solution for $x\le \frac{-1}{\sqrt3}$
$f(\frac{1}{\sqrt{3}})=\frac13-\frac{1}{\sqrt{3}}-5+a<0$ thus no solution for $-\frac{1}{\sqrt3}\le x \frac{1}{\sqrt3}$
Finally $f$ has one single root. Since $f(1)<0$ and $f(2)>0$ the root is located in $(1,2)$ thus $\lfloor{x}\rfloor=1$ wich gives $x=\sqrt[3]{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Decomposing bivariate quadratic form into sum of two squares I would like to decompose
$$ax_1^2 + bx_2^2 + 2cx_1x_2$$
into two expressions, each involving only one variable. I'm trying to use a transform like $x_1 = x_+ + x_-$ and $x_2 = x_+ - x_-$ to hopefully get something like
$$\left( bx_+ + \frac ac \right)^2 + \left(ax_ - + \frac bc \right)^2$$
or something of that form. Is this possible?
| You can write quadratic expressions as $x^T Q x$; in this case $x^T = (x_1, x_2), Q = \begin{pmatrix}a & c \\ c & b\end{pmatrix}$.
Then, if you write $Q$ as $Q = A^TDA$, with $D$ diagonal, you have $x^TQx = x^TA^TDAx = (Ax)^TD(Ax)$: $Ax = (y_1, y_2)^T$ are the new variables you are searching, $D = \begin{pmatrix}d_1 & 0 \\ 0 & d_2\end{pmatrix}$ are the new coefficients; so you can rewrite $ax_1^2 + bx_2^2 + 2cx_1x_2 = d_1 y_1^2 + d_2y_2^2$.
EDIT: another (somewhat equivalent) trick is to "complete the squares":
$$\begin{aligned} ax_1^2 + bx_2^2 + 2cx_1x_2 &= ax_1^2 + 2cx_1x_2 + \frac{c^2}{a}x_2^2 + \left( b - \frac{c^2}{a} \right)x_2^2 \\ &= \left( \sqrt{a}x_1 + \frac{c}{\sqrt{a}}x_2 \right)^2 + \left(b - \frac{c^2}{a} \right)x_2^2 \end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Trigonometric and Exponential Integration
If $\displaystyle I_{n} =\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx.$ Then $\displaystyle \frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx$$
$$I_{n}=-\cos^{n}(x)\cdot e^{-x}\bigg|^{\infty}_{\frac{\pi}{2}}-n\int^{\infty}_{\frac{\pi}{2}}\cos^{n-1}(x)\sin(x)\cdot e^{-x}dx$$
$$I_{n}=n\int^{\infty}_{\frac{\pi}{2}}\cos^{n-1}(x)\sin(x)\cdot e^{-x}dx$$
Could some help me to solve it, Thanks
| As you say,
$$
\frac{I_n}{n} = - \int_{\pi/2}^{\infty} \mathrm{e}^{-x} \cos^{n-1}(x)\sin(x)\ \mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
\begin{align*}
\frac{\mathrm{d} }{\mathrm{d} x} \, \cos^{n-1}(x)\sin(x) &= \cos^n(x) - (n-1)\cos^{n-2}(x)\color{blue}{\sin^2(x)} \\
& = \cos^n(x) - (n-1)\cos^{n-2}(x)\left[\color{blue}{1-\cos^2(x)}\right] \\
& = \color{red}{\cos^n(x)} - (n-1) \cos^{n-2}(x) + (n\color{red}{-1})\cos^n(x) \\
& = n \cos^n(x) - (n-1) \cos^{n-2}(x).
\end{align*}
Then the whole integral becomes
$$
\frac{I_n}{n} = - \displaystyle \color{red}{\left[ -\mathrm{e}^{-x} \cos^{n-1}(x)\sin(x)\right]^{\infty}_{\pi/2}} - \int_{\pi/2}^{\infty} \mathrm{e}^{-x} \left[ n \cos^n(x) - (n-1) \cos^{n-2}(x)\right]\ \mathrm{d}x
$$
with the red term being zero, we obtain
\begin{align*}
\frac{I_n}{n} &= -nI_n + (n-1)I_{n-2}, \\
I_n & = - n^2 I_n + n(n-1)I_{n-2}, \\
(n^2 + 1)I_n &= n(n-1)I_{n-2}, \\
\frac{I_n}{I_{n-2}} &= \frac{n(n-1)}{n^2+1}, \\
\end{align*}
In particular,
$$
\frac{I_{2018}}{I_{2016}} = \frac{2018 \times 2017}{2018^2+1} = \frac{4070306}{4072325}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solve this equation using Bernoulli's equation How do you solve this equation:
$$\frac{dy}{dx} + \frac{1}{xy} = y^3$$
using Bernoulli's equation.
I've tried everything but I think there may be an error with the actual equation.
| Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{1}{2y}\dfrac{du}{dx}+\dfrac{1}{xy}=y^3$
$\dfrac{du}{dx}=2y^4-\dfrac{2}{x}$
$\dfrac{du}{dx}=2u^2-\dfrac{2}{x}$
Which reduces to a Riccati ODE.
Let $u=-\dfrac{1}{2v}\dfrac{dv}{dx}$ ,
Then $\dfrac{du}{dx}=\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}$
$\therefore\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=2\left(-\dfrac{1}{2v}\dfrac{dv}{dx}\right)^2-\dfrac{2}{x}$
$\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{2}{x}$
$\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=\dfrac{2}{x}$
$x\dfrac{d^2v}{dx^2}-4v=0$
$v=C_1\sqrt{x}I_1(4\sqrt{x})+C_2\sqrt{x}K_1(4\sqrt{x})$ (e.g. according to https://www.wolframalpha.com/input/?i=x+v%22-4v%3D0)
$\therefore u=-\dfrac{\dfrac{d}{dx}(C_1\sqrt{x}I_1(4\sqrt{x})+C_2\sqrt{x}K_1(4\sqrt{x}))}{2C_1\sqrt{x}I_1(4\sqrt{x})+2C_2\sqrt{x}K_1(4\sqrt{x})}$
$y^2=-\dfrac{2C_1I_0(4\sqrt{x})-2C_2K_0(4\sqrt{x})}{2C_1\sqrt{x}I_1(4\sqrt{x})+2C_2\sqrt{x}K_1(4\sqrt{x})}$ (according to https://www.wolframalpha.com/input/?i=d%2Fdx(x%5E(1%2F2)besseli(1,4x%5E(1%2F2)) and https://www.wolframalpha.com/input/?i=d%2Fdx(x%5E(1%2F2)besselk(1,4x%5E(1%2F2)))
$y^2=-\dfrac{I_0(4\sqrt{x})-CK_0(4\sqrt{x})}{\sqrt{x}I_1(4\sqrt{x})+C\sqrt{x}K_1(4\sqrt{x})}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a$ and $b$ are rational numbers If $a+b$, $a^2+b$ and $b^2+a$ are rational numbers and $a+b\neq 1$ then $a$ and $b$ are rational.
I try, sum the expresions but I only got that $a^2+b^2$ and $ab$ are rational.
Any suggestion?
| Write : $a^2 + b - (b^2 + a) =(a-b)(a+b-1)$ , this is rational. Since $a+b$ is rational, so is $a+b-1$. It is non-zero, therefore $a-b$ is rational being the quotient of two rationals, the latter non-zero.
Of course, if $a+b$ and $a-b$ are rationals, so are $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a linear system of reciprocals. Solve for $\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$
I turn the equations into $\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$
Not sure if I am doing fine
| how about this, let:
$$X=\frac1x,\,Y=\frac1y,\,Z=\frac1z$$
and it is much easier to solve the following:
$$\begin{pmatrix}
1&1&1\\
4&3&2\\
3&2&4
\end{pmatrix}
\begin{pmatrix}
X\\
Y\\
Z
\end{pmatrix}=
\begin{pmatrix}
0\\5\\-4
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$a$, $b$ and $c$ are three positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Here's what I did.
We have that
$$\left(\frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2}\right)\left(\frac{1}{b} + \frac{1}{c} + \frac{1}{a}\right) \ge \left(\sqrt{\frac{a}{b^3}} + \sqrt{\frac{b}{c^3}} + \sqrt{\frac{c}{a^3}}\right)^2$$
But because of $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$.
$$\implies \frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{b}\sqrt{\frac{a}{c^3}} + \frac{1}{c}\sqrt{\frac{b}{a^2}} + \frac{1}{a}\sqrt{\frac{c}{b^2}}\right)$$
And I am stuck, I can't think anymore.
| Let $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$, so $\sum_{cyc} x=1$. Then we have
$(LHS-RHS)\cdot xyz=(\sum_{cyc}x)\cdot(\sum_{cyc}y^3z)-3\sum_{cyc}x^3yz=\sum_{cyc}x^4y+\sum_{cyc}x^3y^2-2\sum_{cyc}x^3yz$
$=\frac{1}{7}\sum_{cyc}((7x^4y+4z^3x^2+2x^3y^2+y^3z^2)-14x^3yz)\geq0.$
The last inequality come from the GM-AM by viewing $7x^4y+4z^3x^2+2x^3y^2+y^3z^2$ as 14 terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Found a three line solution. Do not always be fooled by the quadratics lol.
If we let the original integral be $I$,note that $I=\int \frac{dx}{(2x+3)\sqrt{\frac{1}{4}(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(2x+3)^2-1}} $.
Now you know what...Put $2x+3=z \implies 2dx=dz \implies dx=\frac{dz}{2}$. Hence, $I=\int \frac{dz}{z\sqrt{z^2-1}}=sec^{-1}z=sec^{-1}(2x+3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Show that ${{2m} \choose {m}} \leq \frac{2^{2m}}{\sqrt{2m}}$. The question is as it is in the title:
Show that
$${{2m} \choose {m}} \leq \frac{2^{2m}}{\sqrt{2m}}$$
for all $m \in \mathbb{N}$.
I've had various attempts at this question but it never seems to lead anywhere fruitful. The hint we have been given is in the question is
Consider the square of the product
$$ \frac{(2m)!}{2^{2m}(m!)^2} = \frac{3 \times 5 \times 7 \times \dots \times (2m-1)}{2 \times 4 \times 6 \times \dots \times (2m)}. $$
From the hint it's not too hard to spot where one would go next - the LHS of the original question is disguised as some cheeky factorials - however induction keeps on failing for me and I'm struggling to see a more direct argument. Any light shed on this problem would be appreciated. I feel like I'm missing something obvious.
Thanks in advance.
| This is an incomplete solution, maybe somebody else can see where to go from here.
Use induction with induction statement
$$P(m): {{2m} \choose {m}}\leq \dfrac{2^{2m}}{\sqrt{2m}}$$
Such that the base $(m=1)$ case is
$$P(1): {{2(1)} \choose {(1)}}\leq \dfrac{2^{2(1)}}{\sqrt{2(1)}}$$
This is easily shown because ${{2}\choose{1}} =2 =\dfrac{2\sqrt{2}}{\sqrt{2}}\leq\dfrac{4}{\sqrt{2}}$ and we know $\sqrt{2}<{2}$. Then, assuming $P(m)$ is true, we must show $P(m+1)$ which states
$$P(m+1): {{2(m+1)} \choose {(m+1)}}\leq \dfrac{2^{2(m+1)}}{\sqrt{2(m+1)}}$$
Start with the left hand side
$$\begin{align}
{{2(m+1)} \choose {(m+1)}} &= \dfrac{(2(m+1))!}{(m+1)!(2(m+1)-(m+1))!}
\\
&=\dfrac{(2m+2)!}{(m+1)!(m+1)!}
\\
&=\dfrac{(2m+2)(2m+1)(2m)!}{(m+1)(m+1)m!m!}
\\
&=\dfrac{(2m+2)(2m+1)}{(m+1)(m+1)}\Bigg(\dfrac{(2m)!}{m!m!}\Bigg)
\\
&=\dfrac{(2m+2)(2m+1)}{(m+1)(m+1)}{{2m} \choose {m}}
\\
&\leq\dfrac{(2m+2)(2m+1)}{(m+1)(m+1)}\dfrac{2^{2m}}{\sqrt{2m}}
\\
&=\dfrac{2(m+1)(2m+1)}{(m+1)(m+1)}\dfrac{2^{2m}}{\sqrt{2m}}
\\
&=\dfrac{(2m+1)}{(m+1)}\dfrac{2^{2m+1}}{\sqrt{2m}}
\end{align}$$
Now, since $m \in \mathbb{N}$, it is obvious that
$$m^2(m +2)>1 \implies 2m^3 + 4m^2 >2 \implies 2m^3 + 4m^2 +2m >2m + 2$$
Then,
$$2m^3 + 4m^2 +2m >2m + 2 \implies (m^2+2m+1)(2m)>2(m+1) \implies (m+1)\sqrt{2m}>\sqrt{2(m+1)}$$
Now, since the denominator is larger, we can say
$$\begin{align}
{{2(m+1)} \choose {(m+1)}} &\leq\dfrac{(2m+1)}{(m+1)}\dfrac{2^{2m+1}}{\sqrt{2m}}
\\
&\leq(2m+1)\dfrac{2^{2m+1}}{\sqrt{2(m+1)}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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$1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.9998 \\
\alpha^{11} & \approx 16238.0000 \\
\vdots &
\end{align*}
(values truncated, not rounded) then the sequence $\{ \alpha^n \}$ is "almost" an unbounded integer sequence. What is going on here? Is this property secondary to $\alpha$ being a unit in $\mathbb{Z}[\sqrt{2}]$? Can $\alpha$ be interpreted as a type of approximate eigenvalue? Does this behavior have to do with the continued fraction representation?
Other numbers such as $1+\sqrt{3}$ have a similar property, but they don't appear to "become integer" as quickly.
| The key fact here turns out to be that the conjugate (in this case in $\Bbb Q[\sqrt{2}]$) of $1 + \sqrt{2}$, namely, $1 - \sqrt{2}$, has absolute value less than $1$.
If we expand
$$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$$
and apply the Binomial Theorem to both of the two powers, we find all of the nonintegral terms cancel, so $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer for all $n$. But since $|1 - \sqrt{2}| < 1$, for large $n$ the term $(1 - \sqrt{2})^n$ tends to $0$ and so $$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n \approx (1 + \sqrt{2})^n ,$$ and in particular $(1 + \sqrt{2})^n$ is close to an integer.
We can identify the sequence of integers that occurs, too: $(1 \pm \sqrt{2})$ are the roots of the polynomial $r^2 - 2 r - 1$, so $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ satisfies the recurrence relation $$a_{n + 2} = 2 a_{n + 1} + a_n .$$ Evaluating at, say, $n = 0, 1$ gives $a_0 = a_1 = 2$, and these values and the recursion relation are enough to identify the sequence,
$$2, 2, 6, 14, 34, \ldots .$$
This sequence is OEIS A002203, where its members are called the companion Pell numbers. (Its entries are exactly twice those of OEIS A001333, $1, 1, 3, 7, 17, \ldots$, the numerators of the continued fraction convergents of $\sqrt{2}$.) Since the initial values are integers, as are the coefficients in the recursion relation, this gives an alternative way of seeing that $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer for all $n$.
Finally, working backward, we can indeed interpret $1 + \sqrt{2}$ (and its conjugate) as eigenvalues of a suitable matrix: The companion matrix associated to the polynomial $r^2 - 2 r - 1$ above is $\pmatrix{0&1\\1&2}$, and its eigenvalues are $1 \pm \sqrt{2}$. Then, a straightforward induction shows that
$$\pmatrix{0&1\\1&2}^n \pmatrix{2&2\\2&6} = \pmatrix{a_n&a_{n + 1}\\a_{n + 1}&a_{n+ 2}} .$$
All of these phenomena, by the way are in close analogy to the more familiar case of the Golden Ratio, $\phi := \frac{1}{2}(1 + \sqrt{5})$. (You might work these details out to check that you understand the above arguments.) Indeed, $1 + \sqrt{2}$ has a geometric interpretation analogous to that of the Golden Ratio, and by analogy it is sometimes called the Silver Ratio; in turn these are the first two (nontrivial) members of the infinite family $\frac{n + \sqrt{n^2 + 4}}{2}$, $n = 0, 1, 2, \ldots$, of metallic means, all of which (except for the trivial case $n = 0$) have the key property re the conjugate.
Like you've observed, $1 + \sqrt{3}$ also has the key property, since $|1 - \sqrt{3}| < 1$. But $|1 - \sqrt{3}| > |1 - \sqrt{2}|$, so $|1 - \sqrt{3}|^n \to 0$ more slowly than $|1 - \sqrt{2}|^n \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to do it regular way and I looked for the maximum/maximas finding partial derivatives and setting it to zero and I got $$
\begin{cases}
0=-4x^3+-2x+3x^2y-y^3 \\ 0=-4y^3-2y-3y^2x-x^3\\
\end{cases}
$$
easy to see (0,0) is a solution however I do not know how to prove that it is the only one
| EDIT
As my first answer was mistaken, a correction is coming now.
$$\begin{aligned}x^4+y^4+x^2+y^2+x^3y+xy^3&=\underbrace{x^4+y^4{\color{red}{-2x^2y^2}}}+x^2+y^2\underbrace{{\color{red}{+2x^2y^2}}+xy(x^2+y^2)}\\
&= (x+y)^2(x-y)^2 +x^2+y^2 \quad +xy(x+y)^2 \\
&=(x+y)^2\left((x-y)^2+xy\right)+x^2+y^2\\&=(x+y)^2(x^2-xy+y^2)+x^2+y^2\\
&=(x+y)^2\left((x-{y\over 2})^2+\frac{3y^2}{4}\right)+x^2+y^2\end{aligned}$$
which is non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
a inequality with condition For $a,b,c \ge 0$ and $a+b+c \ge m$, I have to prove that $ 6(a^3+b^3+c^3)+9abc \ge m^3$. By power mean inequality
$ 6(a^3+b^3+c^3) \ge \frac {2(a+b+c)^3}{3} \ge \frac{2m^3}{3}$
Then I stuck
| By Schur's inequality (look at https://en.wikipedia.org/wiki/Schur%27s_inequality for its statement and a standard proof), the basic inequality $3(a^2+b^2+c^2)\geq (a+b+c)^2$ (which can be proved by Cauchy-Schwarz if we write $3=1^2+1^2+1^2$) and the non-negativity of $a,b$ and $c$ we have that
$$a^3+b^3+c^3+3abc\geq a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\Rightarrow\\ 2(a^3+b^3+c^3)+3abc\geq (a+b+c)(a^2+b^2+c^2)\geq \frac{(a+b+c)^3}{3}\Rightarrow\\ 6(a^3+b^3+c^3)+9abc\geq m^3.$$
The first line is Schur's inequality. At the second line we just added $a^3+b^3+c^3$ at both sides and did the algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maclaurin expansion of $\exp(x)/(1-\exp(x))^2$ I am trying to expand following function $f(x)=\frac{e^x}{(1-e^x)^2}$ using the Maclaurin series.
According to wolframalpha it should be:
$\frac{1}{x^2}-\frac{1}{12}+\frac{x^2}{240}-\frac{x^4}{6048}+O(x^6)$
However, I am struggling to arrive at this result following a standard formula for Maclaurin expansion:
$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...$
In our case $\lim(x \to 0)\frac{e^x}{ (1 - e^x)^2 } = \infty$.
0 is a singular point for the $f(x)$ and zero appears in the denominator of every n'th order derivative as well.
So, how can I expand this function?
| Bernoulli numbers $B_n$ are defined in a way that:
$$\frac{x}{e^x-1}= \sum_{n=0}^{\infty} \frac{B_n x^n}{n!}\tag{1}\label{eq1}$$
Divide \eqref{eq1} by $x$:
$$\frac{1}{e^x-1}= \sum_{n=0}^{\infty} \frac{B_n x^{n-1}}{n!}\tag{2}\label{eq2}$$
Calculate the first derivative of both sides of the \eqref{eq2}:
$$-\frac{e^x}{(1-e^x)^2}= \sum_{n=0}^{\infty} B_n \frac{(n-1)x^{n-2}}{n!}\tag{3}\label{eq3}$$
Here are first few Bernoulli numbers: $B_0 = 1, B_1 = -\frac{1}{2}, B_2 = \frac{1}{6}, B_3 = 0, B_4 = -\frac{1}{30}, B_5 = 0, B_6 = \frac{1}{42}, B_7=0$
Substitute these in \eqref{eq3}:
$$\frac{e^x}{(1-e^x)^2}= -B_0x^{-2}-B_2\frac{1}{2!}x^0-B_4\frac{3}{4!}x^2-B_6\frac{5}{6!}x^4+O(x^6)=\frac{1}{x^2}-\frac{1}{12}+\frac{x^2}{240}-\frac{x^4}{6048}+O(x^6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Double sum factorial manipulation $$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$
This is driving me nuts! Is there anyway to reduce
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{A!B!(n-A-B-1)!} \frac{A}{A+B+1}$$
beyond
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{(A-1)!B!(n-A-B-1)!} \frac{1}{A+B+1}$$
I can't figure out how to deal with that $\frac{1}{A+B+1}$ term in any way that brings it inside the $\frac{1}{n-A-B-1}$ term. Is this not possible?
| Wolfram Alpha found this sequence which seems to match the sum for all positive integers...
$$ a_n = \frac{3^{n-1}}{2} - \frac{3^n - 1}{4n} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given a scalar field, how to show that the eigenvalues of its Hessian are bounded?
Show that the eigenvalues of the Hessian of $$f(x_1,x_2) := x_1^2+x_1x_2+x_2^2+\ln (1+2e^{x_2})$$ are bounded, i.e., $$1 \leq \lambda_{\min} \left(\nabla^2 f(x)\right) \leq \lambda_{\max}\left(\nabla^2 f(x)\right) \leq 4$$
What I tried is finding gradient and then Hessian
$$
\nabla f(x) =
\begin{bmatrix}
2x_1+x_2 \\
x_1+2x_2+\frac{2e^{x_2}}{1+2e^{x_2}}
\end{bmatrix}
$$
and Hessian is:
$$
\nabla^2 f(x) =
\begin{bmatrix}
2 & 1 \\
1 & 2
+\frac{2e^{x_2}}{(1+2e^{x_2})^2}
\end{bmatrix}
$$
Let $y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}$. Then
$$
\nabla^2 f(x) =
\begin{bmatrix}
2 & 1 \\
1 & 2
+y
\end{bmatrix}
$$
Then,
$$
\lambda_{1,2} = \frac{4+y\pm \sqrt{y^2+4}}{2}
$$
My questions are:
*
*Am I mistaken?
*Is this function a well-known function?
| Hint: Show that $y$ is bounded for all $x_2\in\mathbb{R}$:
Part 1:
\begin{align}y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}< \frac{1+2e^{x_2}}{(1+2e^{x_2})^2}=\frac{1}{1+2e^{x_2}}<1\end{align}
Part 2:
Observe that
\begin{align} \frac{2e^{x_2}}{(1+2e^{x_2})^2}=\frac{-1+1+2e^{x_2}}{(1+2e^{x_2})^2}=-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}\end{align}
but since $(1+2e^{x_2})^2>1+2e^{x_2}$ we have $$\frac{1}{1+2e^{x_2}}>\frac{1}{(1+2e^{x_2})^2}$$
and thus
$$-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}>0$$
Consequently
\begin{align}y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}=-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}>0 \end{align}
Now you can do the final part with $0<y<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$ Let $a,b\in R^+$ such that $ab+bc+ca=1$. Find the maximize of $$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$$
By Wolframalpha i can see that if $a=b=2-\sqrt 3;c=\sqrt 3$ we will have $P=\dfrac 1 4$
WLOG $a\le b$. I proved that $$P=f(a,b,c)\le f(a,a,c)\le 0$$
$$\Leftrightarrow -(a^2-4a+1)^2\le 0$$
My proof is based on the value of $P$ so it's inconvenient if i dont know value of $P$. This inequality is not homogeneous and symmetric so i dont any idead to solve it
Can you help me solve it without using the equality and value of $P$?
| Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\left\{\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right\}\subset(0^{\circ},90^{\circ}).$
Thus, $\alpha+\beta+\gamma=180^{\circ}$ and
$$P=\frac{1}{2}\sin\alpha+\frac{1}{2}\sin\beta-\cos^2\frac{\gamma}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}-\cos^2\frac{\gamma}{2}=$$
$$=\cos\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}-\cos^2\frac{\gamma}{2}\leq\cos\frac{\gamma}{2}-\cos^2\frac{\gamma}{2}$$
Can you end it now?
I got that the maximal value it's $\frac{1}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum of Fourth Powers of cosine series has closed form solution. A problem posed in the 1988 Irish Mathematical Olympiad asks for a proof of the following
$$\sum\limits_{k=1}^{n} \cos^{4}\Big(\frac{k\pi}{2n+1}\big) = \frac{6n-5}{16}$$
Can anyone give me a heads-up on how to proceed to prove this very interesting result ?
| Using the hints given in answers, we can make the result more general
$$\color{blue}{\sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) = 4^{-p} \binom{2 p}{p}\,n+4^{-p} \binom{2 p-1}{p}-\frac{1}{2}}$$ and get the beautiful
$$\left(
\begin{array}{cc}
p & \sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) \\
1 & \frac{2 n-1}{4} \\
2 & \frac{6 n-5}{16} \\
3 & \frac{10 n-11}{32} \\
4 & \frac{70 n-93}{256} \\
5 & \frac{126 n-193}{512} \\
6 & \frac{462 n-793}{2048} \\
7 & \frac{858 n-1619}{4096} \\
8 & \frac{12870 n-26333}{65536}
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the Minimum Value of $x^2+y^2$ Given:
$x+y=2\sin{a}-\cos{b}\\
xy=2\cos{a}+\sin{b}$
Find the minimum value of $x^2+y^2$.
Attempt:
$\begin{aligned}
x^2+y^2&=(x+y)^2-2xy\\
&=(2\sin{a}-\cos{b})^2-2(2\cos{a}+\sin{b})\\
&=4\sin^2{a}-4\sin{a}\cos{b}+\cos^2{b}-4\cos{a}-2\sin{b}
\end{aligned}$
$\begin{aligned}
f_{a}(a,b)&=0\\
8\sin{a}\cos{a}-4\cos{a}\cos{b}+4\sin{a}&=0\\
\cos{a}\cos{b}&=2\sin{a}\sin{b}+\sin{a}\\
\end{aligned}$
$\begin{aligned}
f_{b}(a,b)&=0\\
4\sin{a}\sin{b}-2\sin{b}\cos{b}-2\cos{b}&=0\\
2\sin{a}\sin{b}&=\sin{b}\cos{b}+\cos{b}\\
\end{aligned}$
Tried to plot it on a graph, the answer should be -6.
| Write $b:={\pi\over2}+c$. Then the equations are
$$x+y=2\sin a+\sin c,\qquad xy=2\cos a+\cos c\ ,$$
and instead of $f$ you now have the function
$$g(a,c)=(2\sin a+\sin c)^2-4\cos a-2\cos c$$
representing $x^2+y^2$. One sees that for $a=c=0$ the function $g$ assumes the value $-6$, but it is impossible for $g$ to assume a smaller value. It follows that $f$ assumes the minimum $-6$ at $(a,b)=\bigl(0,{\pi\over2}\bigr)$.
Note that both $x$ and $y$ are complex at the extremal point. If it is an additional condition that $x$ and $y$ should be real we have a more difficult problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . Find number of solutions and the solutions for this equation in $[0,\pi]$ I tried to solve the equation by doing this-
$$2\sin(\frac{\theta+5\theta}{2})\cos(\frac{\theta-5\theta}{2})=\sin(3\theta)\\
2\sin(3\theta)\cos(2\theta) = \sin(3θ)\\
\cos(2θ) = \frac{1}{2}\\
1-2\sin^2(θ) = \frac{1}{2}\\
\sin^2(θ) = (\frac{1}{2})^2\\
∴ θ = nπ ± α\\
Answer = \frac{π}{6},\frac{5π}{6} $$
But in the solution there are 6 solutions and in step 3 instead of dividing $\sin(3θ)$ by $\sin(3θ)$ they have taken it as common and made "$\sin(3θ)(2\cos(2θ)-1)$"
| $$\sin3\theta=2\sin\dfrac{5\theta+\theta}2\cos\dfrac{5\theta-\theta}2$$
$$\implies\sin3\theta(2\cos2\theta-1)=0$$
If $\sin3\theta=0,3\theta=m\pi$ where $m$ is any integer
We need $0\le\dfrac{m\pi}3\le\pi\iff 0\le m\le3$
If $2\cos2\theta-1=0,\sin^2\theta=\sin^2\dfrac{\pi}6$
$\theta=n\pi\pm\dfrac{\pi}6$
We need $0\le n\pi\pm\dfrac{\pi}6\le\pi\iff0\le 6n\pm1\le6$
Taking '+' sign, $0\le 6n+1\le6\implies-1<-\dfrac16\le n\le\dfrac56<1\implies n=0$
Take '-' sign
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show power series converges for every $x$.
Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$
Now I need to explain that the power series that define $f(x)$ converges for every $x$.
This problem is introduced by two smaller proofs which I will give (or my attempts for them) so the track of thought is clear.
First I found a simple expression for $\frac{a_{2n}}{a{2n-2}}$ for every $n \in \mathbb{N}$. Since this fraction is $\frac{1}{2}$ when $n=1$, it is $\frac{1}{4}$ when $n=2$, it is $\frac{1}{6}$ when $n=3$ and so on, I let
$$\frac{a_{2n}}{a_{2n-2}}=\frac{1}{2n}.$$
Next I had to show that for every fixed $x \in \mathbb{R}$ there exists a $N \in \mathbb{B}$ such that $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ whenever $n \geq N$. This is my attempt:
Let $x \in \mathbb{R}$ be given and let $N > x^{2}$, then for all $n \geq N : \frac{1}{n} < \frac{1}{x^{2}}.$ Then
$$\frac{|a_{2n}|}{|a_{2n-2}|}=\frac{1}{2} \frac{1}{|n|} < \frac{1}{2} \frac{1}{|x^{2}|}=\frac{1}{2} \left| \frac{x^{2n-2}}{x^{2n}} \right|\\
\Longrightarrow\\
|a_{2n}||x^{2n}| < \frac{1}{2} |x^{2n-2}||a_{2n-2}|.
$$
I expect to have to use $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ to show the convergence, but I have no idea how. Any suggestions?
| We can write the differential equation as
$$
\sum_{n\ge1}na_{n}x^{n-1}=\sum_{n\ge0}a_nx^{n+1}
$$
Changing indices, we obtain
$$
a_1+\sum_{n\ge1}(n+1)a_{n+1}x^{n}=\sum_{n\ge1}a_{n-1}x^n
$$
so that $a_1=0$ and
$$
(n+1)a_{n+1}=a_{n-1}
$$
Therefore all odd terms are zero, and $(n+2)a_{n+2}=a_n$ for even $n$, starting with $a_0=1$. The first terms are
$$
a_2=\frac{1}{2},\qquad
a_4=\frac{1}{4}\frac{1}{2},\qquad
a_6=\frac{1}{6}\frac{1}{4}\frac{1}{2}
$$
and we can conjecture that
$$
a_{2n}=\frac{1}{2^nn!}
$$
Indeed, $(2n+2)a_{2n+2}=a_{2n}$ and so
$$
a_{2n+2}=\frac{1}{2n+2}\frac{1}{2^nn!}=\frac{1}{2^{n+1}(n+1)!}
$$
Using the ratio test, we have
$$
\left|\frac{a_{2n+2}x^{2n+2}}{a_{2n}x^{2n}}\right|=\frac{2^nn!}{2^{n+1}(n+1)!}|x^2|=
\frac{x^2}{2(n+1)}
$$
whose limit is $0$. Thus the series converges everywhere.
By the way, $f(x)=e^{x^2/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explained in the textbook. The long division I need to do is this one:
$$\frac{3x^3-x^2-13x-13}{x^2-x-6}$$
The book states that the result is:
$$3x+2+\frac{7x-1}{x^2-x-6}$$
But the mechanics of the long division are never explained. Can I please get some help with this one?
| For long division, you essentially keep reducing the degree of the numerator until it's lower than the denominator.
In the example you gave, $\displaystyle \frac{3x^3 - x^2 - 13x - 13}{x^2 - x- 6}$, we first look at the leading term, $3x^3$. We now want to remove this from the numerator, so we multiply $x^2-x-6$ by $3x$ to obtain $3x^3 - 3x^2 - 18x$. We now write the numerator with this term in it: $$\displaystyle \frac{3x^3 - x^2 - 13x - 13}{x^2 - x- 6} = \frac{(3x^3 - 3x^2 - 18x) + 2x^2 + 5x - 13}{x^2 - x- 6} = 3x + \frac{2x^2 + 5x - 13}{x^2 - x- 6}.$$
We now repeat the process again; we remove the leading term $x^2$on the numerator by multiplying $x^2-x-6$ by $2$ to obtain $2x^2 - 2x - 12$, and then write the numerator with this term:
$$3x + \frac{2x^2 + 5x - 13}{x^2 - x- 6} = 3x + \frac{(2x^2 -2x - 12) + 7x - 1}{x^2 - x- 6} = 3x + 2 + \frac{7x-1}{x^2-x-6}.$$
The numerator now has lower degree than the denominator, and so we stop.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$ Some time ago a question was asked here regarding the value of the sum $$\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$$.
But it was deleted later by the OP. I went around it but didn't find a solution.
Some common combinatoric identities I know of are $4^i=2^{2i}=\sum_{j=0}^{2i} {2i\choose j}$ also that ${2i \choose i}=\sum_{j=0}^{i} {i\choose j}^2$. But they were hardly of any use. I also thought of individual term as probability but turns out I can't think of it as anything sensible.a quick WA check gave the answer as $${k+\frac{1}{2} \choose k}=\frac{(2k+1)!}{k!^24^k}$$
Help would be appreciated!
| We prove that $$\sum_{k=0}^{n} \frac{{2k \choose k}}{4^k} = \frac{(2n+1)!}{n!^24^n}.$$
We proceed by induction. The base case is easily verifiable. Denote the RHS of the above equation as $a_n$.
Now consider $$a_{n+1} = \frac{(2n+3)!}{(n+1)!^24^{n+1}} = \frac{(2n+2)(2n+3)}{4(n+1)^2} \cdot \frac{(2n+1)!}{n!^24^n} = \frac{(2n+3)}{2n+2} a_n.$$
So we now have $$a_{n+1} - a_n =\frac{a_n}{2n+2} = \frac{(2n+1)!}{2(n+1)\cdot n!^24^n} = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot 2\\ = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot \frac{2n+2}{n+1} = \frac{(2n+2)!}{(n+1)!^24^{n+1}} = \frac{{2(n+1) \choose n+1}}{4^{n+1}},$$
which is the added term in the summation. This completes the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.
| It is good that you managed to factor!
Here is one way to do it:
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)=\\
x^3+(a+b+c)x^2+(ab+bc+ca)x+abc+\\
2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
ab(a+b+c)+ac(a+b+c)+bc(a+b+c)=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
(a+b+c)(ab+bc+ca)=\\
x^2(x+a+b+c)+(ab+bc+ca)(x+a+b+c)=\\
(x+a+b+c)(x^2+ab+bc+ca).$$
Now, write the given equation in this form:
$$(x+2)(x-3)(x-1)+4=0 \iff \\
(x+2)(x-3)(x-1)+(2-3)(-3-1)(-1+2)=0 \iff \\
(x+2-3-1)(x^2+2(-3)+(-3)(-1)+(-1)2)=0 \iff \\
(x-2)(x^2-5)=0 \Rightarrow x_1=2, x_{2,3}=\pm \sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \{2,4\}, & \{2,5\}, & \{2,6\},\\
& & \{3,3\}, & \{3,4\}, & \{3,5\}, & \{3,6\},\\
& & & \{4,4\}, & \{4,5\}, & \{4,6\},\\
& & & & \{5,5\}, & \{5,6\},\\
& & & & & \{6,6\}
\end{array}
but when I roll two different dice, then the probability of getting sum of 7 is $1/6$
\begin{array}{c c c c c c}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array}
How this is possible?
| If you roll two dice then there are $6*6=36$ possible outcomes.
6 of those outcomes add up to 7.
So yes , the probability of rolling the sum of 7 is $\frac{6}{36}=\frac{1}{6}$
But in your first statement you considered $(2,1)$ and $(1,2)$ as one outcome. These are Different possible outcomes. If you consider those as one outcome then there are 21 possible outcomes. In other words, there are 21 different possible number combinations that the dice could show. But in reality you must treat these separately to calculate probabilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \sqrt{b^2 - 1})$$
$$e^a = b + \sqrt{b^2 - 1}$$
$${1 \over e^a} = {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = b + \sqrt{b^2 - 1} + {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b(b + \sqrt{b^2 - 1}) + \sqrt{b^2 - 1}(b + \sqrt{b^2 - 1}) + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b^2 + b\sqrt{b^2 - 1} + b\sqrt{b^2 - 1} + b^2 - 1 + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {2b^2 + 2b\sqrt{b^2 - 1} \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = 2b$$
This is all well and good, but for the life of me I can't see: how would someone ever know in advance that adding $b + \sqrt{b^2 - 1}$ to its own reciprocal would clean things up so nicely? Is this just a "special case" which needs to be memorized, or is there some general principle behind it which can be used when solving similar equations?
| Lets call $c=\sqrt{(b^2-1)}$. Note that $b^2-c^2=1$ so $(b+c)(b-c)=1$ and $b-c={1\over b+c}$ , so that $b+c +{1\over b+c} = b+c + b-c =2b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Definite integral of $1/(2\sin^4x + 3\cos^2x)$ I have $f = \frac 1 {(2\sin^4x + 3\cos^2x)}$ which area should be calculated from $0$ to $\frac{3\pi}2 $.
I noticed that $$\int_0^{\frac{3\pi}2} f \,dx= 3\int_0^{\frac{\pi}2} f \,dx$$
I tried to calculate this integral with the Weierstrass substitution:
$t = \tan{\frac x2}$
I got this integral:
$$6\int_0^1 \frac{(1+t^2)^3}{32t^4+3(1-t^4)^2} \,dt$$
My second try: I divided and multiplied on $\cos^4x$.
$$\int_0^{\frac{\pi}2} f \,dx = \int_0^{\frac{\pi}2} \frac1{(2\tan^4x + \frac3{\cos^2x})\cos^4x} \,dx$$
$t = \tan x, dt = \frac{dx}{\cos^2x}, \cos^2x = \frac1{\tan^2 x +1}$
I got:
$$ \int_0^\infty \frac{t^2+1}{2t^4 + 3t^2 + 3} \,dt$$
Nothing of this helps me to calculate the area by getting primitive function.
| Thank you, Dr Zafar Ahmed DSc, for a good idea, but your answer that I calculated is not correct. I tried to solve it myself with getting these coefficients (A, B, etc.). This is my solution that based on yours:
$$ \int_0^\infty \frac {1+\frac1{t^2}}{2t^2+\frac3{t^2}+3} \,dt$$
I. We can write $2t^2+\frac3{t^2}+3$ as
$(\sqrt2t+\frac{\sqrt3}t)^2 - C^2$, where $C = 2\sqrt6 - 3$
or $(\sqrt2t+\frac{\sqrt3}t)^2 + D^2$, where $D = 2\sqrt6 + 3$
II. $1+\frac1{t^2}$ can be represented as $A*(\sqrt2+\frac{\sqrt3}{t^2}) + B(\sqrt2-\frac{\sqrt3}{t^2})$
We can find A and B by opening brackets and solving the system of two linear equations.
$A = \frac{\sqrt3+\sqrt2}{2\sqrt6}, B = \frac{\sqrt3-\sqrt2}{2\sqrt6}$
III. Ok, now we can write our integral as:
$$ A\int_0^{\infty} \frac{\sqrt2 + \frac{\sqrt3}{t^2}}{(\sqrt2t-\frac{\sqrt3}{t})^2+D^2} \,dt + B\int_0^{\infty} \frac{\sqrt2 - \frac{\sqrt3}{t^2}}{(\sqrt2t+\frac{\sqrt3}{t})^2-C^2} \,dt$$
Look at the first integral in this sum:
$u = \sqrt2t-\frac{\sqrt3}{t}; du = \sqrt2+\frac{\sqrt3}{t^2}$
We have:
$$A\int_{-\infty}^{\infty} \frac{du}{u^2+D^2} = \frac{A}D\arctan{\frac{u}D}\Big|_{-\infty}^{\infty} = \frac{A}D\pi$$
Look at the second integral:
$u = \sqrt2t+\frac{\sqrt3}{t}; du = \sqrt2-\frac{\sqrt3}{t^2}$
We have
$$B\int_{\infty}^{\infty} \frac{dv}{v^2-C^2}=0$$
Then, the result is $$\frac{A}D\pi = \frac{\sqrt3+\sqrt2}{2\sqrt6\sqrt{2\sqrt6+3}}\pi$$
The answer to my based task is:
$$3\frac{A}D\pi = 3\frac{\sqrt3+\sqrt2}{2\sqrt6\sqrt{2\sqrt6+3}}\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$ Prove
$$0.9999^{\!101}<0.99<0.9999^{\!100}$$
I think its original idea is
$$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$
For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.
Furthermore $0.99^{\!-99}\times0.99^{\!100}=0.99<1.01^{\!100}\times0.99^{\!100}=0.9999^{\!100}$.
For $x=-100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,1.01^{\!101}<0.99^{\!-100}$.
Furthermore $1.01^{\!101}\times0.99^{\!101}=0.9999^{\!101}<0.99^{\!-100}\times0.99^{\!101}=0.99$.
| $$0.99<0.9999^{100}$$ it's
$$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli:
$$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$
$$0.9999^{101}<0.99$$ it's
$$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or
$$\left(1-\frac{1}{100^2}\right)^{100}\left(1+\frac{1}{100}\right)<1$$ or
$$\left(1-\frac{1}{100^2}\right)^{100}<1-\frac{1}{101},$$ which is true because
$$\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}<1-\frac{1}{101}.$$
I used the following lemma.
Let $0<x<1$ and $n>2.$ Prove that:
$$(1-x)^n<1-nx+\frac{n(n-1)}{2}x^2.$$
Proof.
We need to prove that $f(x)>0,$ where
$$f(x)=1-nx+\frac{n(n-1)}{2}x^2-(1-x)^n.$$
Indeed, $$f'(x)=-n+n(n-1)x+n(1-x)^{n-1}$$ and
$$f''(x)=n(n-1)-n(n-1)(1-x)^{n-2}=n(n-1)\left(1-(1-x)^{n-2}\right)>0.$$
Thus, $$f'(x)>f'(0)=0,$$
$$f(x)>f(0)=0$$ and the lemma is proven.
Now, take $x=\frac{1}{100^2}$ and $n=100.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove a summation equals to one How to prove that
$$
S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1,
$$
For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$.
Also, WolframAlpha gave the result
$$
S = 2 - \binom{2K-1}{K-1} \frac{(xy)^K}{(x+y)^{2K}} \left\{ {}_2F_1\left(1,2K;K+1;\frac{a}{a+b} \right) + {}_2F_1\left(1,2K;K+1;\frac{b}{a+b} \right) \right\},
$$
but $S = 1$ only when $0 < \frac{a}{a+b} < 1$ and $0 < \frac{b}{a+b} < 1$.
| Write $x={a\over a+b},\ J=K-1$, and
$$f(x,J) = (1-x)^{J+1}\sum_{k=0}^J{J+k\choose k}x^k$$ Then we want to show $$f(x,J)+f(1-x,J)=1,\ J=0,1,2,\dots\tag{1}$$
We proceed by induction on $J$. When $J=0$, $(1)$ says $x+(1-x)=1.$
Suppose that $J>0$ and that $(1)$ is true for $J-1$. Then
$$\begin{align}
f(x,J)&=(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k}x^k+(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k-1}x^k\tag{2}
\end{align}$$
The first term on the right of $(2)$ is $$
(1-x)^{J+1}\sum_{k=0}^{J-1}{J-1+k\choose k}x^k+(1-x)^{J+1}{2J-1\choose J}x^J$$ or
$$(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}\tag{3}$$
The second term on the right of $(2)$ is
$$\begin{align}
(1-x)^{J+1}\sum_{k=1}^J {J-1+k\choose k-1}x^k&=(1-x)^{J+1}\sum_{k=0}^{J-1}{J+k\choose k}x^{k+1}\\
&=xf(x,J)-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{4}
\end{align}$$
Now $(2),(3),\text{ and }(4)$ give
$
(1-x)f(x,J)=(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{6}
$
By symmetry,
$
xf(1-x,J)=xf(1-x,J-1)+x^{J+1}(1-x)^J{2J-1\choose J}-x^{J+1}{2J\choose J}(1-x)^{J+1}\tag{7}
$
Multiply $(6)$ by $x$, multiply $(7)$ by $1-x$, add the results and apply the induction hypothesis to get
$$
x(1-x)(f(x,J)+f(1-x,J))= x(1-x)\tag{8}$$ after verifying that $${2J\choose J}=2{2J-1\choose J}$$
This proves $(2)$ for $x\neq0,1$, but since it is a polynomial identity, it is true for these values also.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime.
Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$
Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equation, but it doesn't work. I have been taught Congruence
| Suppose $x,y\geq 0$ and let $g:= \gcd(x^2-2x+2,x^2+2x+2)$
Let prime $q\mid g$ then $q\mid 4x$. If $q\ne 2$ then $q\mid x$ but then $$q\mid x^2+2x \implies q\mid 2 \implies q=2$$
So the only prime which can divide $ \gcd(x^2-2x+2,x^2+2x+2)$ is 2.
*
*$\color{red}{Case\; 1}$: $2\mid g$ then $x$ is even, so $x=2z$ then we see $4\nmid g$ and thus $g=2$. But then $2\mid y$ so $16\mid 4(4z^4+1)$ which is impossible.
*$\color{red}{Case\; 2}$: $2\nmid g$, so $g=1$, then $$x^2-2x+2 = pa^4 \;\;\;\wedge \;\;\;x^2+2x+2=b^4$$ where $a,b$ are relatively prime or $$x^2-2x+2 = a^4 \;\;\;\wedge \;\;\;x^2+2x+2=pb^4$$
*$\color{red}{Case\; 2.1}$: $$(x+1)^2<x^2+2x+2 < (x+2)^2$$
so no solution.
*
*$\color{red}{Case\; 2.2}$: If $x>1$ then $$(x-1)^2<x^2-2x+2 < x^2$$
so no solution. So $x=1$ ...
| {
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Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1}\right)-\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\sum_{k=1}^{n}\left(\sqrt{k+2}-\sqrt{k+1}\right)-\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\left(\sqrt{n+2}-\sqrt{2}\right)-\left(\sqrt{n+1}-1\right)\right]\\
&=\lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n+2}}+1-\sqrt{2}\right)\\
&=1-\sqrt{2}.
\end{align*}
But how to determine the convergence directly by comparison and without evaluating?
| Just an alternative answer and using a different test from
$$\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}=\\
(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})=\\
\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
and using alternating series test the series converges, where $|a_n|=\frac{1}{\sqrt{n}+\sqrt{n-1}}$.
| {
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Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\frac{3}{2}}}{8a^3}+\frac{3b^2\sqrt{b^2-4ac}}{8a^3}-\frac{3b\left(b^2-4ac\right)}{8a^3}$$ multiplying by $ 8a^3$ and simplifying we get $$4a^3=-b^3+3abc+\left(b^2-4ac\right)^{\frac{1}{2}}\left(b^2-ac\right)$$ Now $a$ , $b$ and $c$ are integers and the left hand side is an integer and so must the right hand side be and this happens only if the term $b^2-4ac$ is a perfect square. Let $b^2-4ac = m^2$. This implies that $$x=\frac{m-b}{2a}$$ Substituting $ x =2^{\frac{1}{3}}$ and multiplying by $2$ yields $$2^{\frac{4}{3}}=\frac{m-b}{a}$$ so $2=\left(\frac{m-b}{a}\right)^{\frac{3}{4}}$ which implies that all powers of two can be written in this form. choosing $ 2^4=16$ shows that $16^{\frac{1}{3}}=\frac{h}{a}\ $ where $h=m-b$. Now all I have to do is prove that $16^{\frac{1}{3}} = 2\ \left(2^{\frac{1}{3}}\right)$ is an irrational number. Now $h=2a\left(2^{\frac{1}{3}}\right)$. the number $a$ is the product of some primes and this product must contain a $\left(2^{\frac{r}{3}}\right)$ where $ r \equiv 2 mod(3)$ but that would mean that $r$ is not divisible by $3$ which means that the term $\left(2^{\frac{r}{3}}\right)$ can't exist in the prime factorization of any integer which shows that the initial assumption ( that $a$ , $b$ and $c$ are integers) is wrong.
The same argument works for $x=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ but the signs are different.
edit: Guys this is proof verification. please don't suggest alternative methods without mentioning why the proof is wrong.
| hint: Write $$a\cdot2^{2/3}+b\cdot2^{1/3}=-c$$ and raise to the power three both sides.
| {
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The area enclosed by the locus of point C Triangle ABC is such that AB = 4, BC = 2, and AC = 3. If vertex A is confined to the x-axis and vertex B is confined to the y-axis, what is the area of the region enclosed by the locus of all points point C could possibly be?
| You have some sort of a "do-nothing machine"
The midpoint of $AB$ traces a circle.
Lets start with $B = (0,0), A = (4,0)$ Where is $C$
Area $= \sqrt {(\frac 92)(\frac 52)(\frac 32)(\frac 12)} = \frac {\sqrt {135}}{4}$
$h = \frac {\sqrt {135}}{8}$
And from Pythagoras theorem we get that the altitude splits the base into segments $\frac {11}{8},\frac {21}{8}$
Note there are 2 ways we can orient the triangle at this time.
$(\frac {11}{8},\frac {\sqrt {135}}{4})\\
(\frac {11}{8},-\frac {\sqrt {135}}{4})\\
$
Lets pick $C = (\frac {11}{8},\frac {\sqrt {135}}{4})$ for now.
$A = (4\cos\theta, 0)\\
B = (0, 4\sin\theta)\\
C = (\frac {11}{8}\cos\theta + h\sin\theta, \frac {21}{8}\sin\theta + h\cos\theta)$
$C = (2\cos(\theta - \arctan\frac {\sqrt{135}}{11}), 3\sin (\theta + \arctan \frac{\sqrt {135}}{21}))$
That will trace out something that is roughly an ellipse. I don't think it will be exactly an ellipse.
And swapping the different starting position of $C$ will change the sign on the angle of orientation.
| {
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Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c^3\equiv3abc$(mod 6). Now I am not sure how to show $3abc\equiv 0$(mod6), and even doing that, we have only half of the proof because then we need to prove that the converse is also true.
| Among $a$, $b$, $c$ there are at least two even or two odd numbers. In any case $(a+b)(b+c)(c+a)$ is even.
| {
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Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
And I got stuck at this point.
Second (desperately), I expanded
$$(a_1 x + b_1y)(a_2 x + b_2 y)(a_3 x + b_3 y)(a_4 x + b_4 y)$$
and compared it to $x^4 + y^4 - x^2y^2$, but let's just say it didn't go smoothly.
I think there should be a nice way to do it. Anybody?
| Note that $x^2 - y^2 - ixy = y^2 \left( (\frac{x}{y})^2 - i(\frac{x}{y}) -1 \right)$, which is simply a quadratic in $(\frac{x}{y})^2$, multiplied by $y^2$. Factorise the quadratic and then multiply $y^2$ back in. Do the same with the other factor to finish.
| {
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Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $
Solve the inequality
$$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$
Attempt:
I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes
$$ y - 4 \sqrt{y + 4} < 1 $$
$$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$
so from here I get that $0 < y < 9 - 2\sqrt{14} \: \:\cup \:\: y > 9 + 2\sqrt{14}$. This is for $y>0$ or $x(2x+1)>0 \implies x > 0 \:\: \cup \:\: x < -1/2$
I may continue to solve for $x$.. but the answer key says that the solution is $ -7/2 < x < 3$. I doubt that the solution set will be that tidy. Are there better approaches to solve the inequality?
| Let $y= \sqrt{2x^{2} + x + 4} \geq 0$, (notice that $2x^2+x+4>0$ for all real $x$) then we get $$
y^2-4-4y<1\implies y^2-4y-5 = (y-5)(y+1)<0$$
So $$y\in (-1,5)$$
and so
$$ \sqrt{2x^{2} + x + 4} < 5 \implies 2x^2+x+4<25$$
so $$2x^2+x-21= (2x+7)(x-3)<0 \implies x\in (-{7\over 2},3)$$
| {
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Least value of $ab-cd$ is
If $a,b,c,d\in\mathbb{R}$ and $a^2+b^2=c^2+d^2=4$
$ac+bd=0$. Then least value of $ab-cd$ is
Plan
$(ac+bd)^2+(ab-cd)^2=a^2c^2+b^2d^2+a^2b^2+c^2d^2\geq 0$
$a^2(b^2+c^2)+d^2(b^2+c^2)\geq 0$
How do i solve it Help me please
| Geometric interpretation of conditions above with $U=\binom{a}{b}$, $V=\binom{c}{d}$ give :
$\|U\|=\|V\|=2$ and dot product $U . V=0$, i.e., $U$ and $V$ are orthogonal.
Thus $a = 2\cos \alpha, b=2\sin \alpha, c = 2\cos \beta, d=2\sin \beta \tag{1}$
for a certain $\alpha$ ($-\pi < \alpha \leq \pi$) with $\beta=\alpha\pm \pi/2,$ expressing orthogonality.
As $$ab-cd=4 \sin \alpha \cos \alpha-4 \sin \beta \cos \beta$$
$$ab-cd=2 \sin 2 \alpha-2 \sin 2 \beta $$
with $2 \beta=2 \alpha\pm \pi$.
$$ab-cd=2 \sin 2 \alpha-2 \sin (2 \alpha\pm \pi)$$
$$ab-cd=2 \sin 2 \alpha+2 \sin 2 \alpha=4 \sin 2 \alpha \tag{2}$$
with minimal value $-4$ for $\alpha=-\pi/4$, i.e., using (1) with $\beta=\alpha+\pi/2=\pi/4$ with :
$$a=\sqrt{2}, b=- \sqrt{2}, c= \sqrt{2}, d=\sqrt{2}.$$
More generaly with :
$$a=s_1 \sqrt{2}, b=- s_1 \sqrt{2}, c= s_2 \sqrt{2}, d=s_2 \sqrt{2} \ \text{with} \ s_1, s_2 = \pm 1$$
| {
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Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :-
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{2 \sqrt{2}} \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} dx - \frac{1}{2 \sqrt{2}} \int \frac{x - \sqrt{2}}{x^2 - \sqrt{2}x + 1} dx $$
Now, to the first integral, I tried making a u-substitution:-
$$ \text{Let }x^2 + \sqrt{2}x + 1 = u \\
\frac{du}{dx} = 2x + \sqrt{2} \\
\implies du = (2x + \sqrt{2}) dx \\ $$
As you can see, it is not the same as the numerator, which is $$ (x + \sqrt{2}) dx $$
Any hints on how to proceed ?
| $x+\sqrt2 = \frac{1}{2}(2x+2\sqrt2) = \frac{1}{2}(2x+\sqrt2)+\frac{1}{2}\sqrt2$
Let $x^2+\sqrt2 \ x+1 = t \implies (2x+\sqrt2)dx = dt $
So, the integral is
$$I = \frac{1}{2}\int\frac{dt}{t}+\frac{1}{\sqrt2}\int\frac{dx}{(x^2+\sqrt 2 \ x+1)}$$
The first part gets evaluated into $\frac{1}{2}\ln t$. Convert the second part into $u^2+ a^2$ form which gets evaluated into $\frac{1}{a}\arctan(\frac{u}{a})$
| {
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Inequality for optimization. I found this problem:
Find the minimum value of:
$f(a,b,c)=a+b+c+\frac{9a}{(a+2b+c)^2}+\frac{9b}{(b+2c+a)^2}+\frac{9c}{(c+2a+b)^2}$
Where $a,b,c >0$.
After looking for some minimum values, I think the lowest one is for the equality case $a=b=c=3/4$.Obviously,
$f(a,a,a)=3a+\frac{27}{16a} \geq 2\sqrt{\frac{81}{16}}=9/2$
I got inspired from the equality case to "substitute" $a=(a+b+c)/3$ and then having this inequality:
$f(a,b,c) \geq 3(\frac{a+b+c}{3})+\frac{27}{16((a+b+c)/3)}=3(\frac{a+b+c}{3})+\frac{81}{16(a+b+c)}$.
Subtracting on the both sides by $a+b+c$ and simplifying by $9$, we get
$\frac{a}{(a+2b+c)^2}+\frac{b}{(b+2c+a)^2}+\frac{c}{(c+2a+b)^2} \geq \frac{9}{16(a+b+c)}$
This last inequality is really the problem
I`m pretty sure it is true after giving some arbitrary values and also looking at cases when some variables tend to infinity.
Still, I cannot prove it. I tried Titu's lemma but end up with an invalid inequality.
I'd like to see a nice proof for this inequality.
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\left(a+\frac{9}{(a+2b+c)^2}\right)\geq6\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{(a+2b+c)^2}}\geq6\sum_{cyc}\frac{a}{a+2b+c}=$$
$$=6\sum_{cyc}\frac{a^2}{a(a+2b+c)}\geq\frac{6(a+b+c)^2}{\sum\limits_{cyc}a(a+2b+c)}=\frac{6(a+b+c)^2}{\sum\limits_{cyc}(a^2+3ab)}\geq\frac{9}{2},$$ where the last ineqyality it's just $\sum\limits_{cyc}(a-b)^2\geq0.$
The equality occurs for $a=b=c=\frac{3}{4},$ which says that we got a minimal value.
| {
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Show that $\left| z + \sqrt{z^2 + c^2} \right| > c$ for $\operatorname{Re}(z) > 0$ I have encountered this seemingly obvious yet bizarre-to-show inequality: $$\left| z + \sqrt{z^2 + c^2} \right| > c$$ for $z \in \Bbb C$ such that $\operatorname{Re}(z) > 0$ and $c > 0$. I've tried the reverse triangular inequality but it hasn't helped. How do you show this?
| I'll assume that the branch of $\sqrt{z^2 + c^2}$ is chosen such that $\operatorname{Re} \sqrt{z^2 + c^2} > 0$ (otherwise the claim is wrong).
Then
$$
\left| z + \sqrt{z^2 + c^2} \right|^2 = |z|^2 + |z^2+c^2| + 2 \operatorname{Re} \left( \bar z \cdot\sqrt{z^2 + c^2}\right) \\
\ge |z|^2 + (c^2 - |z|^2) + 2 \operatorname{Re} \left( \bar z\cdot\sqrt{z^2 + c^2}\right) \\
= c^2 + 2 \operatorname{Re} \left( \bar z\cdot\sqrt{z^2 + c^2}\right)
$$
so that is is sufficient to show that
$$ \tag{*}
\operatorname{Re} \left( \bar z \cdot\sqrt{z^2 + c^2}\right) > 0 \, .
$$
Now observe that if $z$ lies in the first quadrant then $\sqrt{z^2 + c^2}$ is in the first quadrant as well, and $\bar z$ is in the forth quadrant, so that the product $\bar z \cdot\sqrt{z^2 + c^2}$ is in the right half-plane. A similar argument works if $z$ lies in the forth quadrant, or on the positive real axis. In any case, $(*)$ holds.
| {
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Finding $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ given $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$ I am supposed to find the value of $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ and I have been provided with the information that $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$.
I tried to approach this using vectors. We can consider three unit vectors that add up to $0$. Unit vectors because the coefficients of the $\sin$ and $\cos$ terms are $1$. For the sake of simplicity, let one of the vectors $\overline{a}$ be along the $x$-axis. Let the angles between $\overline{b}$ and $\overline{c}$ be $\alpha$, between $\overline{a}$ and $\overline{b}$ be $\gamma$ and between $\overline{a}$ and $\overline{c}$ be $\beta$. Then we have:
$$\begin{aligned}\overline{a}&=\left<1,0\right>\\ \overline{b}&=\left<-\cos\gamma, -\sin\gamma\right>\\ \overline{c}&=\left<-\cos\beta, \sin\beta\right>\end{aligned}$$
$$\begin{aligned}\cos\gamma+\cos\beta &=1\\ \sin\beta&=\sin\gamma\end{aligned}$$
Now, $\cos \gamma$ and $\cos\beta$ must have the same sign. So we get $\sin\alpha=-\sqrt{3}/2$, $\sin\beta=\sqrt{3}/2$ and $\sin\gamma=\sqrt{3}/2$. This contradicts with the answer key provided according to which $\sum_{cyc}\sin^2\alpha=3/2$. What am I doing wrong?
This was the picture I had in mind with $\overline{a}$ aligned with the horizontal.
| From Clarification regarding a question,
$\alpha-\beta=\dfrac{2\pi}3+2\pi a,\beta-\gamma=\dfrac{2\pi}3+2\pi b$
$\implies\alpha-\gamma=\dfrac{4\pi}3+2\pi c=-\dfrac{2\pi}3+2\pi d$
If $S=\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
$$2S=3-(\cos2\alpha+\cos2\beta+\cos2\gamma)$$
Method$\#1:$
$$\cos2\alpha+\cos2\beta=\cos2\left(\gamma-\dfrac{2\pi}3\right)+\cos2\left(\gamma+\dfrac{2\pi}3\right)=2\cos2\gamma\cos\dfrac{2\pi}3=-\cos2\gamma$$
Method$\#2:$
If $\cos3x=\cos3A$
$3x=2n\pi\pm3A\implies x=\dfrac{2n\pi}3\pm A$
Now $\cos3A=\cos3x=4\cos^3x-3\cos x$
$\implies4\cos^3x-3\cos x-\cos3A=0$
$\implies\sum_{r=-1}^1\cos\left(x+r\dfrac{2r\pi}3\right)=\dfrac04$
| {
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Solve$(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ We need to solve the following equation $y=(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ and it is equal to 2 while I am getting the value in excel I am not able to solve it manually eventhough the values are conjugate
I tried $y=a-b$
$a=(10+6\sqrt3)^{\frac{1}{3}}$ & $b=(-10+6\sqrt3)^{\frac{1}{3}}$
$y^3=(a-b)^3$
$y^3=a^3-b^3-3ab(a-b)$ after this step I am struck
| Since $(1\pm\sqrt3)^3=10\pm6\sqrt3$,$$\sqrt[3]{10+6\sqrt3}+\sqrt[3]{10-6\sqrt3}=1+\sqrt3+1-\sqrt3=2.$$
| {
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Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$
*
*Find the quotient of $P(x)$ divided by $x^2 + x + 1$
*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers.
I've done the first task and came up with $P(x)= (x+1)^5 + x = (x^3+4x^2+5x+1)(x^2+x+1)$ which can be written as $A =BC +D$ where $D=0$. So, for the second task I think I have to prove that B or C or both are prime, but I don't know how to continue.
| *
*If $a,b$ are a roots of $x^2+x+1=0$ then $a^3=b^3 =1$ and by Vieta we have $a+b=-1$, so $a+1 =-b$. Now since $$p(a) = (a+1)^5+a = -b^5+a = -b^2+a = (b+1)+a = 0$$
and the same for $p(b)=0$ we see that $x^2+x+1$ divide $p$, so the remainder is $0$.
*
*Since for $n\geq 2$ we have $p(n)=\underbrace{(n^2+n+1)}_{\geq 7}\underbrace{(n^3+4n^2+5n+1)}_{\geq 35}$
both parts are divisible by some prime.
Let $d$ be a gcd of both factors. So $d\mid n^2+n+1$ and $d\mid n^3+4n^2+5n+1$ then $d\mid n^3-1$ and so $d\mid n-2$ so $d\mid 7$. So at least one factor must be also divisible by some prime $p\ne7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find every $a>0$ such that $\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges We know that $\forall x \in [2,+\infty)$,
$$\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$
And it's easy to prove that $\int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx$ converges if and only if $a>3$.
But is it legitimate to conclude that our first integral converges if and only if $a>3$? The Comparison Criterion says that if $0 \leq f\leq g$, then, if $g$ converges, $f$ converges. But the converse isn't valid. Are we losing any possible values for $a$?
For example, the following equivalence holds:
$$\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx = \int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx - \int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$$
Shouldn't we prove that $\int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges to $0$?
I'm confused.
| It is true that $\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$. What is also true is that $\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a}$. Altogether, this inequality is $$\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$
The left-hand side diverges for $a \leq 3$. The right-hand side converges for $a > 3$. Therefore, the integral converges if and only if $a > 3$.
| {
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How to find GCD of those two complex polynomials? I have a polynomial
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$
I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $f(x)$ and its derivative $f'(x)$.
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^4+3x^2+1)-8x^3$$
$$f'(x) = i(6x^5-12x^3+6x)+(6x^5+12x^3+6x)-24x^2$$
I know I should find the GCD as $(x-i)(x-1)$ but this is where I'm stuck. What would the next step be?
| By inspection, $f(1)=0$ so $x-1$ is a factor of $f(x)$.
Since $f'(x)= 3i(x^2-1)^2(2x)+3(x^2+1)^2(2x)-24x^2$, we also have that $f'(1)=0$, so $x-1$ is a factor of $f'(x)$.
This means $(x-1)^2$ is a factor of $f(x)$.
One also has that by inspection $f(i)=0$ so $x-i$ is a factor of $f(x)$.
We also have that $f'(i)=0$, so $x-i$ is a factor of $f'(x)$.
So we also have that $(x-i)^2$ is a factor of $f(x)$.
| {
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Exponential equations, solve for x I'm preparing for uni entrance exam. I've been struggling with this problem for about 90 minutes, tried everything I could think of. Can anybody explain how to solve for x step by step?
$$3^{\frac{x-1}{2}}-2^{\frac{x+1}{3}}=2^{\frac{x-2}{3}}+3^{\frac{x-3}{2}}$$
| Bring same exponents to one side:
$$3^{\frac{x-1}{2}}-3^{\frac{x-3}{2}}=2^{\frac{x+1}{3}}+2^{\frac{x-2}{3}}$$
Factor out $3^{\frac{x-3}2}$ on the LHS and $2^{\frac{x-2}3}$ on the RHS:
$$3^{\frac{x-3}2}(3-1)=2^{\frac{x-2}3}(2+1)$$
Bring the $2$ and $3$ over again:
$$3^{\frac{x-3}2}\cdot2=2^{\frac{x-2}3}\cdot3$$
$$3^{\frac{x-5}2}=2^{\frac{x-5}3}$$
Take sixth powers, for neatness:
$$3^{3(x-5)}=2^{2(x-5)}$$
$$27^{x-5}=4^{x-5}$$
But the only place where graphs of two different exponential functions $a^x$ and $b^x$ intersect is at $x=0$. Hence, $x-5=0$ and $x=5$.
| {
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Completing the square (diagonal matrix/bilinear form) Let $A=\begin{pmatrix} 2 & 2 & 5 \\ 2 & 1 & 3 \\ 5 & 3 & \frac{17}{2} \end{pmatrix} \in M_3(\mathbb{R})$.
I want to find an invertible matrix $C$ such that $C^TAC=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, since there are three eigenvalues which are postive, negative and zero.
I tried to use completing the square with the bilinear form $s(v,v)=\langle v,Av \rangle$:
$s(v,v)= \langle v,Av \rangle=(v_1,v_2,v_3)\begin{pmatrix} 2 & 2 & 5 \\ 2 & 1 & 3 \\ 5 & 3 & \frac{17}{2}\end{pmatrix}\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}=2v_1^2+4v_1v_2+10v_1v_3+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$
Now I used completing the square with respect to $v_1$:
$2(v_1^2+2v_1v_2+5v_1v_3)+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$
$=2(v_1^2+v_1(2v_2+5v_3))+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$
$=2(v_1^2+v_1(2v_2+5v_3)+\frac{1}{2}(2v_2+5v_3)^2-\frac{1}{2}(2v_2+5v_3)^2)+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$
$=2(v_1+(2v_2+5v_3))^2-\frac{1}{2}(2v_2+5v_3)^2+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$
$=2(v_1+2v_2+5v_3)^2-\frac{1}{2}(2v_2+5v_3)^2+(v_2+3v_3)^2-\frac{1}{2}v_3^2$
Now there is one summand too much so I can't set a matrix $C$ which is the inverse of the matrix with the entries in the braces.
I think I made a mistake with the completing the square but I don't see where.
| $$2x^2+4xy+10xz+y^2+6yz+\frac{17}2z^2=2\left(x+y+\frac52z\right)^2-y^2-4yz-4z^2=$$
$$=2\left(x+y+\frac52z\right)^2-\left(y+2z\right)^2$$
and there you go: one positive, one negative, one zero...
| {
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For $ f_{n+1} = \sin f_n(x) $, $f_n(x) \gt f_2(x)/(n-1) $?
Let $f_0(x) = x, f_{n+1} = \sin f_n(x) (0 < x < \pi)$.
Then for $n \ge 3$,
$$f_n(x) \gt \frac{f_2(x)}{n-1} ? $$
Here is what I have shown:
For $x > 0$, $\sin x > x - x^3 / 6$.
So to show the proposition, it suffices to show that, by induction, for $n$, $n f_n(x)^2 < 6$.
And so it suffices to show that $n f_n(1)^2 < 6$.
And using a calculator, this inequality seems to be true.
| Analogously to $\sin(x)<x-x^3/6$, we have $\sin(x)>x-x^3/6+x^5/120$ (one additional term in the Taylor Series). Additionally, $\sin([0, 1])\subseteq [0,1]$ so that $0\leq f_n(1)\leq 1$ for all $n$.
We prove by induction that $nf_n(1)^2<6$. Base case: $f_1(1)^2=\sin(1)<6$. Now let $n\geq 2$ and assume the statement holds for $n-1$. Since $\sin(x)$ is increasing on $[0,1]$, we have
$$nf_n(1)^2=n\sin^2(f_{n-1}(1))<n\sin^2\left(\sqrt{\frac{6}{n-1}}\right)<n\left(\sqrt{\frac{6}{n-1}}-\frac{1}{n-1}\sqrt{\frac{6}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{6}{n-1}}\right)^2=6\left(\sqrt{\frac{n}{n-1}}-\frac{1}{n-1}\sqrt{\frac{n}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{n}{n-1}}\right)=6g(n),$$
where
$$g(x)=\sqrt{\frac{x}{x-1}}-\frac{1}{x-1}\sqrt{\frac{x}{x-1}}+\frac{3}{10(x-1)^2}\sqrt{\frac{x}{x-1}}.$$
We have to prove $g(n)\leq 1$ for all $n\geq 2$.
But the only local extremum of $g(x)$ in $\mathbb{R}^+$ is between $x=1$ and $x=2$ and there $g(x)<1$. Furthermore, we easily see
$\lim\limits_{x\rightarrow\infty} g(x)=1,$
so that $g(x)\leq 1$ for all $x\geq 2$. Thus, the claim is proved.
| {
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Roots of $z^4 + 2z^3 + 20z + 12 = 0$
Find the roots of $$z^4 + 2z^3 + 20z + 12 = 0.$$
I still haven't been able to solve it after hours of trying. There are no rational roots and factoring seems impossible. Who of you, mathematical geniuses, can help me? :-)
| The rational roots test fail so there is no linear real factor but there may be some $(z^2 + az + b)(z^2 + cz + d)$ where
$a+c = 2$ and $b+ ac + d =0$ and $ad+bc = 20$ and $bd=12$
which .... isn't obvious but.... oh heck. If we assume $\{b,d\} = \{\pm 1,12\}, \{\pm 2,6\}, \{\pm 3,4\}$ then $ac =-b-d= \mp 13, \mp 8, \mp 7$. But $a+c = 2$ so (assuming integers) $a$ and $c$ are opposite signs. $13$ and $7$ are primes larger than $3$ so a difference of $2$ between their factors is not possible. SO the nicest is $\{b,d\} = \{2,6\}$ and $ac = -8$. And as $a+c = 2$ so $a=4; c=-2$. (That's the nicest; it's not a guarentee.)
So $ad+bc= 4*(2,6)-2*(6,2) = 20$ is solved with $d=6; b= 2$.
Hurray! $(z^2 +4z +2)(z^2 -2z + 6)=z^4 + 2z^3 + 20z +12$
And we can use quadratic formula to solve:
$z = -2 \pm \sqrt 2$ and $z= 1 \pm \sqrt{5}i$.
| {
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial.....but it didnt help either.
| We factor $x^4-12x^2+8x+3$.
We see, that $x=1$ is a root of $f(x)=x^4-12x^2+8x+3$, since $f(1)=1-12+8+3=0$
After long division $(x^4-12x^2+8x+3)\div (x-1)=x^3+x^2-11x-3$
We see, that x=3 is a root of the polynomial $g(x)=x^3+x^2-11x-3$
Again after long division we achieve:
$(x^3+x^2-11x-3)\div (x-3)=x^2+4x+1$
Which gives us the solution $k=4$.
| {
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If I know that $b$ is a multiple of $32$ then how can i show that $b+5$ and $b^2 + 7$ are coprime? I'm using the polynomial division method and i got a remainder of 32 but how do I explain in words or mathematically how this justifies my solution.
The previous question has answers based on Properties of Divisibility but I want to know how I can use long polynomial division in my method.
Thanks
| Let $d = gcd(b+5, b^{2}+7) = gcd(32k + 5, 32^{2}k^{2} + 320k + 32)$, for some integer $k$. Hence, $d | (b^{2} + 7) - b$, which implies $d | b^{2} + 7 - (b + 5)r$, when $r \in Z.$
Now, we want to choose $r$ such that $(b + 5)r = b^{2} + c$, for some integer $c$; specifically, let $r = (b - 5)$ as then $(b + 5)r$ will result in the difference of two perfect squares $b^{2} - 25$ and permit us to express the difference $b^{2} + 7 - (b - 5)r$ as an integer; that is, $d | b^{2} + 7 - (b^{2} - 25) = 32$.
Certainly $d | b^{2} + 7$ as $b^{2} + 7 = 32^{2}k^{2} + 320k + 32$; however, since $b + 5 = 32k + 5$ and $b = 32k$, it follows that $d | b + 5 - b = 5$; and this allows us to conclude that $gcd(5, 32) = 1.$
Therefore, $gcd(b + 5, b^{2} + 7) = 1$; and so, the two integers contained therein are coprime.
| {
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The number of triples that sum to a constant Problem:
How many triples are there of the form $(x_0,x_1, x_2)$ where $x_0 \in I$, $x_1 \in I$, $x_2\in I$ $x_0 \geq 0$,
$x_1 >= 0$, $x_2 >= 0$ and $n = x_0 + x_1 + x_2$ where $n \in I$?
Answer:
Let $c(n)$ be the number of tuples we can have for a given $n$. For $n = 0$, the only valid triple is $(0,0,0)$, hence $c(0) = 1$.
For $c(1) = 3$, the set of valid triples is: $(0,0,1 ), (0,1,0), (0,0,1)$ Hence $c(1) = 3$.
For $c(2) = 6$, the set of valid triples is:
$$ (1,0,1 ), (0,1,1 ), (0,0,2 ), (1,1,0), (0,2,0), (0,0,2)$$
Hence $c(2) = 6$.
Using the information on this URL:
How many $k-$dimensional non-negative integer arrays $(x_1,\cdots,x_k)$ satisfies $x_1+x_2+\cdots+x_k\le n$
I find the answer to be:
$$ c(n) = {{n+3}\choose{3}} - {{n+2}\choose{3}} $$
$$ c(n) = \frac{(n+3)!}{3!n!} - \frac{(n+2)!}{3!(n-1)!} $$
$$ c(n) = \frac{(n+3)(n+2)(n+1) - (n+2)(n+1)(n)}{6} $$
$$ c(n) = \frac{(n+2)(n+1)(n+3 - n)}{6} $$
$$ c(n) = \frac{3(n+2)(n+1)}{6} $$
$$ c(n) = \frac{(n+2)(n+1)}{2} $$
Do I have that right?
Thanks,
Bob
| This is a prototypical “stars and bars” problem. Here we have $n$ stars and $2$ bars, with $\binom{n+2}2 = {(n+2)(n+1)\over2}$ ways to choose among the $n+2$ possible positions for the two bars.
Another way: The number of partitions of $n$ into three non-negative integers is equal to the coefficient of $z^n$ in the formal power series $(1+z+z^2+z^3+\cdots)^3$. This coefficient can be found using the generalized binomial theorem: $$(1+z+z^2+z^3+\cdots)^3 = \frac1{(1-z)^3} = \sum_{k=0}^\infty \binom{-3}{k} z^k = \sum_{k=0}^\infty \binom{k+2}2 z^k.$$ The number of partitions is therefore $\binom{n+2}2$ as before.
| {
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Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$. This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 6(d)
Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$.
I tried to prove it in the following way but I'm not sure if the proof makes sense, and the author uses another method.
Proof: Let $n=k+1$, $k$ is odd such that $x^{k+1}=y^{k+1}$. This would only be possible if $x^k=y^k$ or $x^k=-y^k$. This is because, if $x^k=y^k$ and $k$ is odd, then $x=y$ (I proved this on a previous exercise). Therefore, $x^k\cdot x = y^k \cdot x \Rightarrow x^{k+1}=y^{k+1}$.
Similarly, $x^k=-y^k \Rightarrow x=-y \Rightarrow x^{k+1}=y^{k+1}$. Thus, I have proved that $x^{k+1}=y^{k+1}$ is possible only if $x^k=y^k$ or $x^k=-y^k$. As I have already proved, $x^k=-y^k \Rightarrow x=-y$ and $x^k=y^k \Rightarrow x=y $
$\therefore x^n=y^n$ and $n$ is even $\Rightarrow x=y$ or $x=-y$
| Bear with me.
If $0 \le x < y$ then $x^n < y^n$.
Pf: By induction. Base case is $x < y$. Induction step is If we assume $x^{n} < y^{n}$ then $x^{n+1} = x*x^{n} < x*y^{n} < y*y^{n-1} = y^{n+1}$.
So if we know $x$ and $y$ are non negative and we are told $x^n = y^n$ then we know $x = y$. (By contradiction. If $x \ne y$ then one is larger than the other and $x^n \ne y^n$)
Finally $(-x)^n = (-1*x)^n = (-1)^n x^n$ and $(-1)^n = 1$ if $n$ is even. (Because $(-1)^{2k} = [(-1)^2]^k=1^k = 1$.) So $(-x)^n = x^n$.
So.... we are ready:
If $x^n= y^n$ there are four cases:
1: $x \ge 0$ and $y \ge 0$. We've shown that means $x = y$.
2: $x \ge 0$ and $y < 0$ then $-y > 0$ and $(-y)^n = y^n$ so $x^n = (-y)^n$ so $x = -y$.
3: $x < 0$ and $y\ge 0$. Then $-x > 0$ and $(-x)^n=x^n = y^n$ so $-x = y$ and $x = -y$.
4) $x < 0$ and $y < 0$ then $-x>0$ and $-y > 0$ so $(-x)^n = x^n = y^n = (-y)^n$ so $-x =-y$ and $x = y$.
C'est tout.
| {
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How to manipulate $\frac{1}{\sin x}-\frac{1}{x}$ into $\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi}\right)(-1)^k$? 1) How do you get to $\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi} \right)(-1)^k$ from $\frac{1}{\sin x}-\frac{1}{x}$? (as demostrated here)
If $\pi\cot\pi x=\frac{1}{x}+\sum_{k\geq 1}\left(\frac{1}{x-k}+\frac{1} {x+k}\right)$ multiplying by cosine or $(-1)^k$ gets the $\frac{1}{\sin x}$ but I cannot see how it does not make $\pi\csc\pi x=(-1)^k[\frac{1}{x}+\sum_{k\geq 1}\left(\frac{1}{x-k}+\frac{1}{x+k}\right)]$
| Not a full solution, but an attempt to use special functions here instead of complex analysis.
We are interested in the series:
$$f(x)=\frac{2x}{\pi^2} \sum_{k=1}^\infty \frac{(-1)^{k+1}}{(k-x/\pi)(k+x/\pi)} $$
Let's consider an auxiliary function:
$$g(x,y)= -\frac{x^2}{\pi^2} \sum_{k=0}^\infty \frac{y^k}{(k-x/\pi)(k+x/\pi)}$$
$$g(x,0)=1$$
If we denote the terms $a_k$, we have:
$$\frac{a_{k+1}}{a_k}=\frac{(k+1)(k-x/\pi)(k+x/\pi)}{(k+1-x/\pi)(k+1+x/\pi)} \frac{y}{k+1}$$
This means that:
$$g(x,y)={_3 F_2} \left(1, \frac{x}{\pi},-\frac{x}{\pi};1+\frac{x}{\pi},1-\frac{x}{\pi}; y \right)$$
So we have:
$$f(x)=\frac{2}{x} (g(x,-1)-1)$$
Which means we need to prove that:
$$\frac{2}{x} \left({_3 F_2} \left(1, \frac{x}{\pi},-\frac{x}{\pi};1+\frac{x}{\pi},1-\frac{x}{\pi}; y \right)-1 \right)=\frac{1}{\sin x}-\frac{1}{x}$$
$${_3 F_2} \left(1, \frac{x}{\pi},-\frac{x}{\pi};1+\frac{x}{\pi},1-\frac{x}{\pi}; -1 \right)=\frac{1}{2} \left(1+\frac{x}{\sin x} \right)$$
In the next part for brevity, let's introduce a new variable $x=\pi z$:
$${_3 F_2} \left(1, z,-z;1+z,1-z; -1 \right)=\frac{1}{2} \left(1+\frac{\pi z}{\sin \pi z} \right)$$
Note that from the Gamma function reflection theorem:
$$\frac{\pi z}{\sin \pi z}=z \Gamma(z) \Gamma(1-z)=z B(z,1-z)$$
Using the Euler's integral transform for the hypergeometric function, we have:
$${_3 F_2} \left(1, z,-z;1+z,1-z; -1 \right)= \frac{\Gamma(1+z)}{\Gamma(z)} \int_0^1 t^{z-1} {_2 F_1} (1,-z ; 1-z; -t ) dt$$
$${_3 F_2} \left(1, z,-z;1+z,1-z; -1 \right)=z \int_0^1 t^{z-1} {_2 F_1} (1,-z ; 1-z; -t ) dt$$
On the other hand:
$$1= z \int_0^1 t^{z-1} dt$$
$$z B(z,1-z)= z \int_0^1 t^{z-1} (1-t)^{-z} dt$$
So we need to prove that:
$$\int_0^1 t^{z-1} \left(2 ~{_2 F_1} (1,-z ; 1-z; -t )-1-(1-t)^{-z} \right) dt=0$$
While I can't prove it right now, I can prove a particular case:
$$z=\frac{1}{2} \\ {_2 F_1} \left(1,-\frac{1}{2} ; \frac{1}{2}; -t \right)=1+ \sqrt{t} \arctan \sqrt{t}$$
So we have:
$$\int_0^1 \frac{1}{\sqrt{t}} \left(1+2\sqrt{t} \arctan \sqrt{t}-\frac{1}{\sqrt{1-t}} \right) dt$$
Doing each part separately (all are elementary integrals) we have:
$$2 + (\pi-2) -\pi=0$$
So at least for $z=1/2$ we have our proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve the equation $2\sin(3x)\cos(4x)=1$? How do I solve the equation $2\sin(3x)\cos(4x)=1$? Normally, I'd use the $\sin(2x)=2\sin(x)\cos(x)$ rule, but here you have two different values for $x$, so I'm not sure how to add them together.
| Let $f(x)=2\sin(3x)\cos(4x)-1$ and solve $f(x)=0$.
The strategy will be to solve this using a Chebychev polynomial of the first type.
First, get a similar function entirely in terms of cosine by shifting the graph of $f$ horizontally to the right by $\frac{\pi}{2}$ units. This will be an even function with symmetric solutions.
\begin{eqnarray}
g(x)&=&f\left(x-\frac{\pi}{2}\right)-1\\
&=&2\sin\left(3x-\frac{3\pi}{2}\right)\cos\left(4x-2\pi\right)-1\\
&=&2\cos(3x)\cos(4x)-1
\end{eqnarray}
We will find solutions for $g(x)=0$ then subtract $\frac{\pi}{2}$ units to obtain the solutions for $f(x)=0$.
Using the identities $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ and $\cos(-\theta)=\cos(\theta)$, this becomes
$$ g(x)=\cos(7x)+\cos(x)-1 $$
If we let $u=\cos(x)$ then we can solve $g(x)=0$ by solving
$$ T_7(u)+u-1=0 $$
where $T_7(u)$ is the seventh Chebychev polynomial of type 1. From Wolfram,
$$ T_7(u)=64u^7-112u^5+56u^3-7u $$
so we need the solutions for
$$ 64u^7-112u^5+56u^3-6u-1=0 $$
Wolfram gives three real and four complex solutions. The real solutions are
$u=\frac{1}{2},\,0.759756,\,0.975695$.
So the solutions for $g(x)=0$ are
$$ \left\{2\pi n\pm\frac{\pi}{3},\,2\pi n\pm\arccos(0.759756),\,2\pi n\pm\arccos(0.975695) \right\} $$
To obtain the corresponding solutions for $f(x)=0$ we subtract $\frac{\pi}{2}$ from each of these six solutions.
$$ \left\{\left(2n-\frac{1}{2}\right)\pi\pm\frac{\pi}{3},\,\left(2n-\frac{1}{2}\right)\pi\pm\arccos(0.759756),\,\left(2n-\frac{1}{2}\right)\pi\pm\arccos(0.975695) \right\} $$
Letting $n=1$ gives the six solutions in the interval $[0,2\pi]$
$$\left\{\frac{11\pi}{6},\,\frac{7\pi}{6},\,5.4202475,\,4.0045304,\,4.9333148,\,4.4914632 \right\} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$
$ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
$ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator?
is there also alternative proof using jensen inequality?
| The inequality you are interested in is equivalent $$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$which is just AM-HM inequality.
You can also prove it using CS inequality.
| {
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"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Possible Jordan Canonical Forms: Intuition As I was reviewing linear algebra before I head off to grad school in the fall, I came across a question about Jordan Canonical Forms. It reads:
"Suppose that A is a square complex matrix with characteristic polynomial $c_A(x) =
(x−1)^4(x+ 3)^5$.
Assume also that $A−I$ has nullity 4 and $A+3I$ has nullity 1, where $I$
is the identity matrix of the same size as $A$. Find, with justification, all possible Jordan
canonical forms of $A$, and give the minimal polynomial for each."
I believe that there will be 2 Jordan Blocks, for each eigenvalue of $A$. Since the rank of the null space of the linear operator $A-I$ is 4, then the dimension of the eigenspace $E_1$ will be four. So there will be four linearly independent eigenvectors with eigenvalue 1. Thus there will one 4 by 4 Jordan Block with no ones in the super-diagonal. Since the dimension of the null space of $A+3I$ has rank 1, then there is only one linearly independent eigenvector with eigenvalue -3. If $K_{-3}$ is the generalized eigenspace, then $dim(K_{-3})=5$. Does this implies that there are 5 linearly independent generalized eigenvectors with corresponding to -3? Is so, then there will be one 5 by 5 Jordan block for the eigenvalue -3 with ones in the superdiagonal. So we get
\begin{bmatrix}
1& 0 &0 &0 &0 &0 &0 &0 &0 \\
0& 1 &0 &0 &0 &0 &0 &0 &0 \\
0& 0 &1 &0 &0 &0 &0 &0 &0 \\
0& 0 &0 &1 &0 &0 &0 &0 &0 \\
0& 0 &0 &0 &-3 &1 &0 &0 &0 \\
0& 0 &0 &0 &0 &-3 &1 &0 &0 \\
0& 0 &0 &0 &0 &0 &-3 &1 &0 \\
0& 0 &0 &0 &0 &0 &0 &-3 &1 \\
0& 0 &0 &0 &0 &0 &0 &0 &-3
\end{bmatrix}
How do I determine the minimal polynomial? Are there any other options for the Jordan blocks corresponding to the eigenvalue -3? Is it possible for a different linear transformation with the same characteristic polynomial to have the dimension of the null space of $A-3I$ to be 2,3 or 4? Is so, how does this affect the Jordan form and why?
| We have a matrix $A$ whose characteristic polynomial is
$$
\chi_A(t)=(t-1)^4(t+3)^5
$$
This immediately tells us that $A$ is a $9\times 9$ matrix whose eigenvalues are $1$ and $-3$. Moreover, the algebraic multiplicities of these eigenvalues are
\begin{align*}
\operatorname{am}_A(1) &= 4 & \operatorname{am}_A(-3)=5
\end{align*}
The eigenspaces of $A$ are
\begin{align*}
E_1 &= \operatorname{Null}(1\cdot I_9-A) & E_{-3} &= \operatorname{Null}(-3\cdot I_9-A)
\end{align*}
The geometric multiplicities of the eigenvalues are the dimensions of these eigenspaces, which are
\begin{align*}
\operatorname{gm}_A(1) &= 4 & \operatorname{gm}_A(-3) &= 1
\end{align*}
The algebraic and geometric multiplicities of the eigenvalues of $A$ give us partial information about the Jordan canonical form $J$ of $A$. For each eigenvalue $\lambda$, we have
\begin{align*}
\operatorname{am}_A(\lambda) &= \text{number of $\lambda$'s on diagonal of $J$} &
\operatorname{gm}_A(\lambda) &= \text{number of Jordan blocks in $J$ corresponding to $\lambda$}
\end{align*}
So, in our example, $J$ will have
*
*four Jordan blocks corresponding to $\lambda=1$ whose sizes sum to four
*one Jordan block corresponding to $\lambda=-3$ whose size is five
So, it turns out that there is only one possible Jordan form
$$
J =
\left[\begin{array}{r|r|r|r|rrrrr}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 0 & 0 & 0 & -3 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -3 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -3 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -3 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -3
\end{array}\right]
$$
Finally, the minimal polynomial is $\mu_A(t)=(t-1)(t+3)^5$ since the largest Jordan block corresponding to $\lambda=1$ has size one and the largest Jordan block corresponding to $\lambda=-3$ has size five.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $x \in \mathbb R$ and $x>2$ then $y+ \frac{1}{y} = x$ will have real solution.
Prove that for real number $x$, if $x > 2$ then there is real number $y$ such that $y + \frac{1}{y} = x$
My attempt:
Rewriting equation, we have:
$$\tag1y + \frac{1}{y} = x$$
$$\tag2 \frac{y^2+ 1}{y} = x$$
$$\tag3 y^2+ 1 = yx$$
$$\tag4 y^2 - yx + 1 = 0$$
We know that for quadratic equations of the format $aq^2 + bq + c$ ($a,b,c$ are constants and q is variable), if discriminant is positive, then equation will have real solutions.
Let discriminant be denoted by $D$, then
$D = \sqrt{b^2- 4ac}$
Substituting our values into the equation, we have:
$D = \sqrt{x^2 - 4}$
It can be seen that if $x > 2$ then $D$ is positive, hence $y + \frac{1}{y} = x$ will have real solution.
Is it correct?
| Multiplying both sides by $y$ and putting the result in standard form, we have
$$y^{2} - xy + 1 = 0.$$
Hence, the discriminate of this quadratic equation is $x^{2} - 4$; and so, if $x > 2$, we certainly get two distinct real solutions. But also, if $x < -2$ we may draw the same conclusion as well. Finally, if $x = \pm 2$, then we get exactly one real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Can I split $\frac{1}{a-b}$ into the form $f(a)+f(b)$? I was wondering if I can split apart the following fraction
\begin{align}
\frac{1}{a-b}
\end{align}
into the form:
\begin{align}
f(a)+f(b)
\end{align}
where $f(a)$ and $f(b)$ is some function in terms of $a$ and $b$
| In fact, we can't even split it as
$$\frac{1}{a-b}=f(a)+g(b)$$
for any functions $f,g$.
Suppose instead that functions $f,g$ from $\mathbb{R}$ to $\mathbb{R}$ are such that
$$f(a)+g(b)=\frac{1}{a-b}$$
for all $a,b\in\mathbb{R}$ with $a\ne b$.
Then we would have
\begin{align*}
&
\begin{cases}
f(x+1)+g(x)={\Large{\frac{1}{(x+1)-x}}}={\Large{\frac{1}{1}}}=1\\[4pt]
f(x-1)+g(x)={\Large{\frac{1}{(x-1)-x}}}={\Large{\frac{1}{-1}}}=-1\\
\end{cases}
\\[6pt]
\implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=2
\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\end{align*}
but we would also have
\begin{align*}
&
\begin{cases}
f(x+1)+g(x-2)={\Large{\frac{1}{(x+1)-(x-2)}}}={\Large{\frac{1}{3}}}\\[4pt]
f(x-1)+g(x-2)={\Large{\frac{1}{(x-1)-(x-2)}}}={\Large{\frac{1}{1}}}=1\\
\end{cases}
\\[6pt]
\implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=\frac{1}{3}-1=-\frac{2}{3}\\[4pt]
\end{align*}
contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
How is the 'Answer' achieved in the 'Problem' after x,y-axis reflection? Problem: $f(x) = x^3 - x^2 + x – 1$ is reflected in both the $x$-axis and $y$-axis. What is the final equation to represent this transformation?
Answer: $x^3 + x^2 + x +1$
My solution: let $g(x)=-f(x)$ << reflection across $x$-axis and $h(x)$ be $g(-x)$ << reflection across $y$-axis.
I ended up having:
$g(x) = -x^3 +x^2 -x+1$ << multiplied every term in $f(x)$ by $-1$
$h(x) = g(-x) = -(-x^3 +x^2 -x)+1 $
MY FINAL ANSWER: $h(x)=x^3 +x^2 -x+1$ which is different from the equation at 'Answer', how do I achieve the equation in 'Answer'?
| $$-f(-x) = - [(-x)^{3} - (-x)^{2} + (-x) - 1 ] = - [-x^{3} - x^{2} - x - 1] = x^{3} + x^{2} + x + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$ .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.
| Note that adding up the three equations gives you:
$$x+y+z =x+y+z - \frac{1}{x} - \frac{1}{y} - \frac{1}{z}$$
, so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$
Try it from there
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Value of $\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$ The value of the $$\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$$ is?
The answer given is $1/4$.
I rationalized and got $$\lim\limits_{x\to-\infty} \frac {-x}{|x|[(4- \frac {1}{x})^{1/2}-2]}$$ how to proceed further?
| \begin{align*}
(4x^2-x)^{1/2}+2x
&=\sqrt{4x^2-x}+2x
\\&=\frac{(\sqrt{4x^2-x}+2x)(\sqrt{4x^2-x}-2x)}{\sqrt{4x^2-x}-2x}
\\&=\frac{(4x^2-x)-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{(2x)^2(1-\frac{1}{4x})}-2x}
\\&=\frac{-x}{2|x|\sqrt{1-\frac{1}{4x}}-2x}
\\&\qquad [x<0]
\\&=\frac{-x}{-2x\sqrt{1-\frac{1}{4x}}-2x}
\\&=\frac{1}{2\sqrt{1-\frac{1}{4x}}+2}
\\&\to\frac{1}{2\sqrt{1+0}+2}
\\&=\frac{1}{4}.
\end{align*}
(as $x\to-\infty$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
| Hint: It is
$$4\sin(x)\cos(2x)\sin(3x)-1=$$$$-2 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(2
x+\frac{\pi }{4}\right) (2 \cos (2 x)-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| I am now going to add one more for fun and wish you can enjoy it. $$
\begin{aligned}
&(\cos x+\sin x)^{2}=1+2 \sin x \cos x=2 \\
\Rightarrow & \cos x+\sin x=\pm \sqrt{2} \ldots(1) \\
&(\cos x-\sin x)^{2}=1-2 \sin x \cos x=0 \\
\Rightarrow & \cos x-\sin x=0 \quad \cdots(2)
\end{aligned}
$$
(1) $\pm$ (2) yields $$\quad \cos x=\sin x=\pm \frac{1}{\sqrt{2}}$$
$$
x=n \pi+\frac{\pi}{4}
$$
where $n \in \mathbb{Z}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
} |
Showing $\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $
Show that
$$\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $$
I tried to find a polynomial that had its roots, but the degree grows too fast and I'm getting lost.
| Sketch of the proof using @saulspatz ninth degree equation above, though all computations that follow from that should be easily done by trigonometry too:
Let $a,b,c, \omega a, \omega b, \omega c, \omega^2 a, \omega^2 b, \omega^2 c$ the roots of the 9th degree equation $8x^9-6x^3+1=0$, where $a,b,c$ are the real ones, so $\sqrt[3]{\cos(\frac{2\pi}{9})},\sqrt[3]{\cos(\frac{4\pi}{9})},\sqrt[3]{\cos(\frac{8\pi}{9})}$ and $w^3=1$ is a primitive cubic root of unity.
Let $d=a+b+c, q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Clearly $a^3+b^3+c^3=0$ as it is a third of the cubic sum of the roots of the degree $9$ equation above and that has no coefficients in degrees $6,7,8$ (so by Newton formulas) and similarly $\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=6$ using the reciprocal equation and Newton's formulas - obviously direct trigonometric proofs should be easy and available too
Now $p=abc=-\frac{1}{2}$ (the product of the 9 roots is $-\frac{1}{8}$ and $abc$ is the real cube root of that by inspection) or again by trigonometry.
Using $d^3=a^3+b^3+c^3+3abcdq-3abc$ and the same equation for $q^3=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\frac{dq}{abc}-3\frac{1}{abc}$, we get the following system in $d,q$:
$2d^3+3dq-3=0$,
$q^3+6dq-12=0$
Now this leads to a cubic in $d^3$ or we can "see" that the only real roots are $d^3=\frac{3\sqrt[3]9-6}{2}, q^3=6(1+\sqrt[3]9), dq=3-\sqrt[3]9$ using the OP's identity
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determining the range of $6^n+ 6^{-n} +3^n +3^{-n}+2$
I have to solve for range of the function $$6^n+ 6^{-n} +3^n +3^{-n}+2$$
The textbook solves it as
$$\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \tag{1}$$
i.e., $$(a-b)^2+(a-b)^2$$
which will always be greater than $6$, so the range is $(6,\infty)$ (since other two terms are squared). But, if we take
$$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{2}$$
or $$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{3}$$
instead of $(1)$, we get that the range is $(-2,\infty)$ or $( 2,\infty)$, respectively.
So, how do we know what range is correct?
| Note that:
$$6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \ \ \text{(textbook)}\\
6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 \overbrace{\require{cancel}\cancel{\color{red}+}}^{-} 2\ \ \text{(yours)}\\
$$
Note that by AM-GM:
$$x+\frac1x\ge 2, x>0,$$
the equality occurs for $x=1$.
Hence:
$$6^n+6^{-n}\ge 2, 3^n+3^{-n}\ge 2,\\
\sqrt{6^n}+\sqrt{6^{-n}}\ge 2,\sqrt{3^n}+\sqrt{3^{-n}}\ge 2,$$
the equality in each of the four inequalities occurs for $n=0$.
Thus, your method must be:
$$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 -2 \ge 6,$$
the equality occurs for $n=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:
$ n^2+n =p^2$
$ n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$ (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$ (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$
I am not able to proceed further from here. Any suggestion on what to do next?
| There are only $2$ integer solutions. Neither are greater than $0$.
Assume $n\ge0$.
Without loss of generality, $p\ge0$. Then since $n\ge0$, $p^2=n^2+n\implies p\ge n$. Note that
$$
n(n+1)=p^2\implies n=p^2-n^2\tag1
$$
If $p=n$, then $n=p^2-n^2=0$.
If $p\gt n$, then
$$
n(n+1)=p^2\implies n=\overbrace{\ (p-n)\ }^{\ge1}\overbrace{\ (p+n)\ }^{\gt2n}\gt2n\implies n\lt0\tag2
$$
Thus, $n\lt0$, which contradicts $n\ge0$.
Therefore, the only non-negative solution is $n=0$. Symmetry then says the only other integer solution is $n=-1$.
| {
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On completing the square and symmetry The quadratic form $f(x,y)=x^2+xy+y^2$ appears quite often on the site here. It is provably positive definite by completing the square: $$f(x,y)=\left(x+\frac y2\right)^2+\frac {3y^2}{4}$$
and this is the only analysis ever really given. I have never seen the symmetric version $$f(x,y)=\frac 34(x+y)^2+\frac 14(x-y)^2$$quoted in an answer (either to the expression being positive, or to any other question on the site). This refers the original expression to principal axes and retains the symmetry between the two variables, so in principle ought to be preferred.
Clearly the first version is adequate for many purposes - but is there a reason why the symmetric version does not feature?
Note: This is related to other questions on quadratic forms - the general theory identifies canonical decomposition, but pragmatic methods seem to dodge this, and find other expressions adequate.
| It is revealing to recast both expressions in terms of linear algebra. We may write $$f(x,y)=x^2+xy+y^2 \equiv v^\top A v = \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$
If we look to the expression $f(x,y)=(x+y/2)^2+\dfrac34 y^2$, this may be expressed in matrix form as
$$f(x,y)= \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1/2 & \sqrt{3}/2 \end{pmatrix} \begin{pmatrix} 1 & 1/2 \\ 0 & \sqrt{3}/2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\equiv v^\top L L^\top v.$$ This corresponds to the Cholesky decomposition $A=LL^\top$. By contrast, the spectral decomposition $$A=VDV^\top = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2}\end{pmatrix}
\begin{pmatrix} 3/2 & 0 \\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2}\end{pmatrix}$$ yields the second formula. So the two expressions simply reflect different decompositions of the matrix $A$, and both establish that $A$ is positive-definite.
| {
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Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$ Let $ a, b \in (1, \infty)$. Prove that:
$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$
I tried switching the bases on each of the logaritm but I got stuck:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$
| By C-S and AM-GM
$$\log_a\frac{a+b}{2}+\log_b\frac{a+b}{2}=\frac{1}{\ln_{\frac{a+b}{2}}a}+\frac{1}{\ln_{\frac{a+b}{2}}b}\geq$$
$$\geq\frac{(1+1)^2}{\log_{\frac{a+b}{2}}a+\log_{\frac{a+b}{2}}b}=\frac{4}{\log_{\frac{a+b}{2}}ab}\geq\frac{4}{\log_{\frac{a+b}{2}}\left(\frac{a+b}{2}\right)^2}=2.$$
| {
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Simplify $\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $ Simplify
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $$
Attempt:
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} = \sqrt{10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})} $$
let $X =\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}}$, then
$$ X^{2} = 10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15}) $$
How to continue?
| You can continue so:
$$x^2=10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=2+5+3+ 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=(\sqrt2+\sqrt5+\sqrt3)^2.$$
Can you end it now?
| {
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Is this true for all polynomials I took $3$ random polynomials with non zero roots one having even degree and two having odd degrees
*
*$f(x)=\color{red}{4}x^2-(4\sqrt3+12)x+12\sqrt3$ having roots $\color{blue}{3,\sqrt3}$ and leading coefficient $\color{red}{4}$ and calculated values of $xf'(x)$$(f'(x)$ is the derivative of $f(x))$ at both roots which are $3f'(3)$ and $\sqrt3f'(\sqrt3)$ and then sum of their reciprocals $\frac1{3f'(3)}+\frac1{\sqrt3f'(\sqrt3)}=\frac{-1}{12\sqrt3}=\frac{-1}{\color{red}{4}}\left(\frac{1}{\color{blue}{3\cdot\sqrt3}}\right)$ then repeated same thing for
*$g(x)=\color{red}{1}x^3-\frac{20}{3}x^2-12x+\frac{32}{3}$ having roots $\color{blue}{8,-2,\frac{2}{3}}$
$\frac1{8g'(8)}+\frac1{-2g'(-2)}+\frac1{\frac{2}{3}g'(\frac2 3)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{8\cdot-2\cdot\frac2 3}}\right)$
*$h(x)=\color{red}{1}x^5+41x^4+137x^3-1601x^2-1818x+3240 $ having roots $\color{blue}{1,-2,5,-9,-36}$
$\frac1{1h'(1)}+\frac1{-2h'(-2)}+\frac1{5h'(5)}+\frac1{-9h'(-9)}+\frac1{-36h'(-36)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{1\cdot-2 \cdot5\cdot-9\cdot-36}}\right)$
Is this true for all polynomials? Is there any known result?
| This results from partial fraction decomposition.
Suppose $g(x)$ has no repeated roots, and without loss of generality is monic (leading coefficient $1$). Then it factors as $\displaystyle g(x)=\prod_{r}(x-r)$ over all roots $r$ and
$$ \frac{1}{g(x)}=\sum_{r}\frac{c(r)}{x-r} $$
for some constants $c(r)$, one for each root $r$. To find the constant $c(s)$ for a specific root $s$, first multiply the equation by the factor $(x-s)$,
$$ \frac{x-s}{g(x)}=c(s)+\sum_{r\ne s} c(r)\frac{x-s}{x-r} $$
then "evaluate" i.e. take the limit $x\to s$ to obtain
$$ \frac{1}{g'(s)}=c(s). $$
Therefore we may plug $x=0$ into
$$ \frac{1}{g(x)} =\sum_{r} \frac{1}{g'(r)(x-r)} $$
and manage negative signs to get
$$ \frac{(-1)^{\deg g}}{\displaystyle \prod r} = -\sum_{r} \frac{1}{rg'(r)}. $$
When $\deg g$ is odd, all signs can go away.
| {
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Let $W = \textit{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$... Let $W = \text{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$, where \begin{equation*}
\mathbf{w}_1 = \begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix}, \quad \mathbf{w}_2 = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}, \quad \mathbf{w}_3 = \begin{pmatrix}
2 \\
2 \\
2
\end{pmatrix}.
\end{equation*}
1. Find a basis $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$ for $W$.
Let
\begin{equation*}
A = \begin{pmatrix}
1 & 1 & 2 \\
2 & 0 & 2 \\
1 & 1 & 2
\end{pmatrix}
\end{equation*}
be a matrix that contains the vectors in $W$. Row reducing $A$ leads to $\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{pmatrix}$. We see that the first two columns are pivots (meaning they are linearly independent) so the basis for $W$ is
\begin{equation*}
\left\{\begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix},\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}\right\}.
\end{equation*}
2. Find a basis $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ for $W^{\perp}$.
So do we transpose $A$ then reduce and then find a basis for the null space?
3. Write each $x\in \mathbb{R}^3$ uniquely in the form
\begin{equation*}
\mathbf{x} = a_1\mathbf{v}_1+\ldots+a_n\mathbf{v}_n+b_1\mathbf{u}_1+\ldots +b_m\mathbf{u}_m.
\end{equation*}
Here the coeffieients $a_j$ and $b_k$ will depend on $\mathbf{x}$.
Not sure how to do that one. Please help!
| For $2)$, you could take the cross product of $(1,2,1)$ and $(1,0,1)$.
Get $\begin {vmatrix} i&j&k\\1&2&1\\1&0&1\end{vmatrix}=(2,0,-2)$.
For $3)$, given $\vec x=(x_1,x_2,x_3)$, we need to solve $\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pmatrix}\vec y=\vec x$.
Thus, $\vec y=\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pmatrix}^{-1}\vec x$, where $\vec y=(a_1,a_2,b_1)$.
| {
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How to solve this? (Argument of the complex number in complex plane) Let the $z \in C$ s.t. $|z-10i|= 6 $
$\newcommand{\Arg}{\operatorname{Arg}}$
Say the $\theta = \Arg(z)$
Find the maximum and minimum value of the $8\sin \theta + 6\cos \theta$
My trial)
Trying to solve by picture and graph, I considered $8i\sin \theta + 6\cos \theta$ instead of the $8\sin \theta + 6\cos \theta$
Then, locus of the $8i \sin \theta + 6\cos \theta$ is the tangent line of the circle $|z-10i|= 6 $
(Surely $ {-4\over 3} \leq \tan \theta \leq {4 \over 3} $ )
So the question is I can't figure out the next step.
P.s.)
The answer sheet said "it is clear that when the case $\cos \theta = \pm{3 \over 5} $ is maximum and minimum"
But this word totally unclear for me. Why does that case have a max or min value?
| Let $z=r(\cos\theta+i\sin\theta).$
Thus, $$r^2\cos^2\theta+(r\sin\theta-10)^2=36$$ or
$$\sin\theta=\frac{r^2+64}{20r}\geq\frac{2\sqrt{64r^2}}{20r}=\frac{4}{5},$$ which gives
$$\arcsin\frac{4}{5}\leq\theta\leq\pi-\arcsin\frac{4}{5}.$$
Also, write $$8\sin\theta+6\cos\theta=10\sin\left(\theta+\arccos\frac{4}{5}\right).$$
Can you end it now?
Actually, I assumed that $0\leq\arg{z}<2\pi.$
I got $$2.8\leq8\sin\theta+6\cos\theta\leq10.$$
I used the following reasoning.
By C-S $$8\sin\theta+6\cos\theta\leq\sqrt{8^2+6^2)(\sin^2\theta+\cos^2\theta)}=10.$$
The equality occurs for $(\sin\theta,\cos\theta)||(4,3)$, which gives $\theta=\arctan\frac{4}{3}\in\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right]$, which says that we got a maximal value.
Also, $$f(x)=8\sin{x}+6\cos{x}=10\sin\left(x+\arccos\frac{4}{5}\right)$$ is a concave function on $\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right],$ which says
$$\min\limits_{\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right]}f=\min\left\{f\left(\arcsin\frac{4}{5}\right),f\left(\pi-\arcsin\frac{4}{5}\right)\right\}=f\left(\pi-\arcsin\frac{4}{5}\right)=2.8.$$
| {
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If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be able to cancel anything out. I am also struggling with other methods like AM-GM since $a,b$ are not necessarily positive. Any help would be appreciated.
| Equivalently, you want that
$$
f = \frac18((a^2+b^2)^3 - 32(a^3+b^3)(ab-a-b)) \ge 0
$$
Let $a = x+y$ and $b = x-y$. Then we have $x \ge 0$ and arbitrary $y$.
Putting in these variables you have that
$$
f = (x - 4)^2 x^4 + y^2 (3 ((x- 8/3)^2 + 80/9) x^2 + y^2 (x (3 x + 24) + y^2))
$$
which is clearly $\ge 0$.
| {
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Solving $\frac53\left(3-\frac{x}{5}\right) = x$ Although I had already solved the following equation, I can't figure out why I fail when passing thru a $15$ denominator I get a bad result.
$$\frac53\cdot\left(3-\frac{x}{5}\right) = x$$
Transformation to get $15$ denominator:
$$\begin{align}
\frac{5\cdot 5(15\cdot 3-3x)}{15} &= \frac{15\cdot x}{15} \tag{1}\\[4pt]
25\cdot(45-3x) &= 15x \tag{2}\\
1125 - 75x &= 15x \tag{3}\\
1125 &= 90x \tag{4}\\[4pt]
x &= \frac{1125}{90} \tag{5}
\end{align}$$
Using alternative approaches, $x = 15/4 = 3.75$, which I believe is the right answer.
What am I doing wrong on the above try?
| Your first line of working is wrong because
$$\frac53\left(3-\frac{x}5\right)\ne\frac{5\cdot5(15\cdot3-3x)}{15}$$
you should actually get
$$\frac53\left(3-\frac{x}5\right)=\frac{5\cdot5}{15}\cdot\frac{15\cdot3-3x}{15}$$
| {
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The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$ Given the circle $C: 2x^2 + 2y^2 + 2x + 2y - 13 = 0$. Line $L$, with slope $m$ and passing through the point $P(0,2)$, cuts the circle at points $A$ and $B$ such that $AB=\sqrt{2}$.
*
*Find the equation of $L$.
*Find the equation of the locus of the centers of the circles passing through $A$ and $B$.
The equation of the line is obviously $L:y=mx+2$
The length of the chord $AB$ is $\sqrt{2}$ so the following must be satisfied:
$(x_B-x_A)^2+(y_B-y_A)^2 = 2$
I tried a few things to calculate $m$, but unsuccessfully.
I guess that once I find $m$ the equation of the locus of the centers of the circles passing through $A$ and $B$ is calculated based on the condition that the locus is perpendicular to $AB$ (please correct me if I am wrong).
Any hint would be useful.
| Circle: $(x+0.5)^2 + (y+0.5)^2 = 7 = r^2$
Distance P (0,2) to center C (-0.5,-0.5) = $\sqrt{0.5^2 + 2.5^2} = \sqrt {6.5}$
Distance AB to center = $\sqrt{r^2 - (\frac{AB}{2})^2} = \sqrt{7-{2 \over 4}}= \sqrt {6.5}$
Thus, AB perpendicular to line PC, with slope, $m = -\frac{0.5}{2.5} = -0.2$
Line L: $y = -x/5 + 2$
| {
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How can you show by hand that $ e^{-1/e} < \ln(2) $? By chance I noticed that $e^{-1/e}$ and $\ln(2)$ both have decimal representation $0.69\!\ldots$, and it got me wondering how one could possibly determine which was larger without using a calculator.
For example, how might Euler have determined that these numbers are different?
| Here's a method that only uses rational numbers as bounds, keeping to small denominators where possible. It is elementary, but finding sufficiently good bounds requires some experimentation or at least luck.
We instead establish the inequality $$-\log \log 2 < \frac{1}{e}$$ of positive numbers, which we can see to be equivalent by taking the logarithm of and then negating both sides. Our strategy for establishing this latter inequality (which is equivalent to yet another inequality suggested in the comments) is to use power series to produce an upper bound for the less hand side and a larger lower bound for the right-hand side. To estimate logarithms, we use the identity $$\log x = 2 \operatorname{artanh} \left(\frac{x - 1}{x + 1}\right) ,$$ which yields faster-converging series and so lets us use fewer terms for our estimate.
Firstly,
$$\log 2 = 2 \operatorname{artanh} \frac{1}{3} .$$
Then, since the power series $$\operatorname{artanh} u = u + \frac{1}{3} u^3 + \frac{1}{5} u^5 + \cdots $$ has nonnegative coefficients, for $0 < u < 1$ any truncation thereof is a lower bound for the series, and in particular
$$\log 2 = 2 \operatorname{artanh} \frac{1}{3} > 2\left[\left(\frac{1}{3}\right) + \frac{1}{3} \left(\frac{1}{3}\right)^3 + \frac{1}{5} \left(\frac{1}{3}\right)^5\right] = \frac{842}{1215} .$$
We'll use the same power series to produce an upper bound for $-\log \log 2$, but since we're nominally computing by hand and want to avoid computing powers of a rational number with large numerator and denominator, we'll content ourselves with a weaker rational lower bound for which computing powers is faster: Cross-multiplying shows that
$$\log 2 > \frac{842}{1215} > \frac{9}{13}$$ and so
$$-\log \log 2 < -\log \frac{9}{13} = 2 \operatorname{artanh} \frac{2}{11} .$$
This time, we want an upper bound for $2 \operatorname{artanh} \frac{2}{11}$. When $0 < u < 1$ we have
$$\operatorname{artanh} u = \sum_{k = 0}^\infty \frac{1}{2 k + 1} u^{2 k + 1} < u + \frac{1}{3} u^3 \sum_{k = 0}^\infty u^{2k} = u + \frac{u^3}{3 (1 - u^2)},$$ and evaluating at $u = \frac{2}{11}$ gives
$$2 \operatorname{artanh} \frac{2}{11} < \frac{1420}{3861} < \frac{32}{87} .$$
On the other hand, the series for $\exp(-x)$ alternates, giving the estimate
$$\frac{1}{e} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} > \sum_{k=0}^7 \frac{(-1)^k}{k!} = \frac{103}{280} .$$
Combining our bounds gives the desired inequality
$$-\log \log 2 < \frac{32}{87} < \frac{103}{280} < \frac{1}{e} .$$
| {
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Primitive Root Modular Arithmetic Question 3)Given a prime $p$ and an integer $a$, we say that $a$ is a $\textit{primitive root} \pmod p$ if the set $\{a,a^2,a^3,\ldots,a^{p-1}\}$ contains exactly one element congruent to each of $1,2,3,\ldots,p-1\pmod p$.
For example, $2$ is a primitive root $\pmod 5$ because $\{2,2^2,2^3,2^4\}\equiv \{2,4,3,1\}\pmod 5$, and this list contains every residue from $1$ to $4$ exactly once.
However, $4$ is not a primitive root $\pmod 5$ because $\{4,4^2,4^3,4^4\}\equiv\{4,1,4,1\}\pmod 5$, and this list does not contain every residue from $1$ to $4$ exactly once.
What is the sum of all integers in the set $\{1,2,3,4,5,6\}$ that are primitive roots $\pmod 7$?
I have no clue how to answer this question. Any help will be great.
Thank you very much.
| Modulo $7$, the powers of $1$ are $1,1,1,1,1,1,$ so $1$ is not a primitive root; the powers of $2$ are $2,4,1,2,4,1,$ so $2$ is not a primitive root; and the powers of $3$ are $3,2,6,4,5,1$, so $3$ is a primitive root. Can you take it from here? Test whether $4,5,$ and $6$ are primitive roots, and then compute the sum of the primitive roots as requested.
| {
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Integral of $\frac{1}{\sin^2(x)\cos^2(x)}$ $$ I = \int\frac{1}{\sin^2(x)\cos^2(x)}$$
I have tried the following:
$$\sin^2x = \sin x \cdot \sin x = \frac{1}{2}(1 - \cos2x)$$
$$\cos^2x = \cos x \cdot \cos x = \frac{1}{2}(1 + \cos2x)$$
The integral becomes:
$$I = 4\int\frac{1}{1 - \cos^2(2x)} = 4\int\frac{1}{\sin^2(2x)}$$
Substitute $u = 2x \implies du = 2dx$
$$I = 2\int\frac{1}{\sin^2u} = -2\cot(u)$$
Plugging back x I get:
$$I = -2\cot(2x)$$
I tried plugging in the integral into an integral-calculator and the answer was: $\tan(x) - \cot(x)$. Can you help me identify what I did wrong?
| \begin{align*}
I & = \int\frac{1}{\sin^2(x)\cos^2(x)} \, dx\\
&= \int\frac{\sin^2x+\cos^2x}{\sin^2(x)\cos^2(x)} \, dx\\
&= \int\frac{1}{\cos^2(x)} \, dx + \int\frac{1}{\sin^2(x)} \, dx\\
&= \int\sec^2(x) \, dx + \int\csc^2(x) \, dx\\
&=\tan x -\cot x+c
\end{align*}
But this is same as your answer, so nothing wrong with what you arrived at.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices.
The point E lies within the rectangle ABCD.
If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?
| We want
$$\left\{
\begin{align*}
a^2&+c^2&=p^2& \hspace{4em} (1)\\
a^2&+d^2&=q^2 &\hspace{4em} (2)\\
b^2&+c^2&=r^2 &\hspace{4em} (3)\\
b^2&+d^2&=s^2 &\hspace{4em} (4)
\end{align*}
\right.$$
where $a+b$ and $c+d$ are the side lengths of the rectangle, and $p,q,r,s$ are the distances to the vertices. We want to find a way such that $p,q,r,s$ are distinct integers (i.e. let $\gcd(p,q,r,s)=1$).
\begin{align*}
(1)-(2): && c^2-d^2&=p^2-q^2 &\hspace{4em} (5) \\
(3)-(4): && c^2-d^2&=r^2-s^2 &\hspace{4em} (6) \\
(5)-(6): && 0 & = p^2-q^2-r^2+s^2 \\
&&r^2-s^2 &=p^2-q^2 & \hspace{4em}(7)
\end{align*}
A solution (and the solution containing the smallest numbers) for the diophantine equation $(7)$ is
$$(p,q,r,s)=(8,4,7,1).$$
Therefore, the answer to your question is the sum of $p,q,r,s$, or $20$.
More info: https://oeis.org/A118882
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Bernoulli Numbers are rationals? There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are $B_n$, such that $\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$.
How can I prove that?? or it's equivalence?
| We have
$$\frac{x}{e^x-1} = \frac{x}{x + x^2/2 + x^3/6 + x^4/24 + \cdots} = \frac{1}{1 + x/2 + x^2/6 + x^3/24 + \cdots}.$$
Now, if we let $f(x) := \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$, then
$$\frac{x}{e^x-1} = \frac{1}{1 + f(x)} = 1 - f(x) + (f(x))^2 - (f(x))^3 - \cdots.$$
In this expansion, each power of $f(x)$ has rational coefficients, and since $f(x)$ is divisible by $x$, the final coefficient of $x^n$ is the same as the coefficient of $x^n$ in $1 - f(x) + (f(x))^2 - (f(x))^3 + \cdots + (-1)^n (f(x))^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix} a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix} a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b \\ 0 & 1 \end{pmatrix}$
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time.
| To add on to lhf's hints, do you know induction? If so, try to find patterns in each of the 4 entries individually once you have corrected your examples.
| {
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"url": "https://math.stackexchange.com/questions/3345771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Limit of function $f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$
I tried to calculate limit when $x$ goes to infinity for the following function
$$f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$$
where $a$, $b$, $d$ are some positive constants.
It's easy to see that terms before and after minus sign goes to infinity so that gives me indeterminate symbol. Is there some way to solve this problem?
| Let $t=1/(2 d x)$. Then the limit of interest can be written as
$$\lim_{t\to 0^{\pm}}\left(\sqrt{(2a d/t + d)^2 + b^2 d^2/t^2} - \sqrt{(2a d/t - d)^2 + b^2 d^2/t^2}\right)\\
\hspace{4cm}=2d\,\text{sgn}(t)\times \lim_{t\to 0^{\pm}}\frac{1}{t}\left[\sqrt{(a+t/2)^2+b^2}-\sqrt{(a-t/2)^2+b^2}\right].$$
The definition of the derivative therefore yields the limits for $t\to 0^{\pm}$ as
$$\pm 2d \frac{d}{dt}\left( \sqrt{(a+t)^2+b^2}\right)_{t=0}=\pm \frac{2ad}{\sqrt{a^2+b^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
An identity containing $\exp\left(i\cdot \frac{2k\pi}{n} \right)$ I found the following equation without proof.
$$\frac{1}{(1+x)^n - 1} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{a(k,n)}{x + 1 – a(k,n)}$$
where $$a(k,n) = \exp\left(i \cdot \frac{2k\pi}{n} \right); \quad n = 1,2,\ldots; \quad k=0,1,2,\ldots,n-1.$$
Note that $(a(k,n))^n = 1$.
I tried to proof the above equation but failed. I’d appreciate it if you could help me out.
| We have the following polynomial with its roots:
$$P(x)= (x+1)^n-1=\prod\limits_{j=0}^{n-1}\left(x-x_j\right)$$
From which
$$P'(x)=n(x+1)^{n-1}=\sum\limits_{k=0}^{n-1}\prod\limits_{j=0,j\ne k}^{n-1}\left(x-x_j\right)$$
then
$$\color{blue}{\frac{P'(x)}{P(x)}=\sum\limits_{k=0}^{n-1}\frac{1}{x-x_k}} \Rightarrow \\
\frac{xP'(x)}{P(x)}=\sum\limits_{k=0}^{n-1}\frac{x}{x-x_k} \Rightarrow \\
\frac{xP'(x)}{P(x)}-n=\sum\limits_{k=0}^{n-1}\left(\frac{x}{x-x_k}-1\right) \Rightarrow \\
\frac{xP'(x)}{P(x)}-n=\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}$$
or
$$\frac{xn(x+1)^{n-1}}{(x+1)^n-1}-n=\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k} \Rightarrow \\
\color{green}{\frac{x(x+1)^{n-1}}{(x+1)^n-1}-1=\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}}$$
Altogether
$$\color{red}{\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k+1}{x-x_k}}=
\color{green}{\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}}+
\frac{1}{n}\color{blue}{\sum\limits_{k=0}^{n-1}\frac{1}{x-x_k}}=\\
\color{green}{\frac{x(x+1)^{n-1}}{(x+1)^n-1}-1} + \frac{1}{n}\color{blue}{\frac{n(x+1)^{n-1}}{(x+1)^n-1}}=\\
\frac{(x+1)^{n}}{(x+1)^n-1}-1=
\color{red}{\frac{1}{(x+1)^n-1}}$$
What's left now is to show that the roots of $P(x)$ are $$x_k=a(k,n)-1=e^{\frac{2 k\pi}{n}\cdot i}-1, \space k=\overline{0..n-1}$$
which is an easy exercise. You will see similar techniques being used here, here and here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to solve such an unlinear matrix equation? I have a matrix equation of the form:
$Sx=x^TKxp$, while $
S = \begin{pmatrix}
0 & a & a \\
b & 0 & b \\
c & c & 0 \\
\end{pmatrix}
$, $a$ and $b$ are real, known numbers, $K = \begin{pmatrix}
0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 0
\end{pmatrix}
$, $p = \begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}$ and $x = \begin{pmatrix}
r \\
s \\
t
\end{pmatrix}$ with $r$, $s$ and $t$ as the unknows.
Is it possible to solve this equation system analytically or only numerically? How can a approach be?
| Let $u=x^TKx=rs+st+tr$ (1).
$Sx=up$ is then a straightforward equation which gives $r,s,t$ in terms of $u$.
Substitute into (1) and you have a quadratic equation for $u$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a substitute that will eliminate terms of the degree $3$ and $2$ from the quartic polynomial? The standart form of the quartic equation:
$$ax^4+bx^3+cx^2+dx+e=0$$ where $$a\neq 0.$$
The purpose of the question is not to solve quartic. Resources related to the solution are sufficiently available. Just about special curiosity, I am looking for a special substitute that will transform the following:
$$ax^4+bx^3+cx^2+dx+e=0 \Longrightarrow z^4+pz+q=0$$
where $$x=f(z)$$
Here $f(z)$ can be a linear or nonlinear complex function.
This substitute is known:
$$x\longmapsto x-\frac b{4a}$$
Which gives us
$$x^4+px^2+qx+r=0$$
Is a substitution known to eliminate term of degree $3$ and $2$ together?
| We can assume that $a=1$ by dividing by it, so start with
$$ x^4 + bx^3 + cx^2 + dx + e = 0 $$
(there's no need for $f$, either).
We will use a more general substitution known as a Tschirnhaus substitution: put
$$ y = x^2 + \lambda x + \mu . $$
Then
$$ y^2 = x^4 + 2\lambda x^3 + 2(\mu+\lambda^2) x^2 + 2\lambda\mu x + \mu^2 = (2\lambda-b)x^3 + (\lambda^2+2\mu-c)x^2 + (2\lambda\mu-d)x + (\mu^2-e) \\
y^3 = \dotsb = (-b^3 + 2 b c - d + 3 b^2 \lambda - 3 c \lambda -
3 b \lambda^2 + \lambda^3 - 3 b \mu + 6 \lambda \mu) x^3 + (-b^2 c + c^2 + b d - e + 3 b c \lambda - 3 d \lambda -
3 c \lambda^2 - 3 c \mu + 3 \lambda^2 \mu + 3 \mu^2) + \dotsb \\
y^4 = \dotsb ,
$$
and then since $1,y,y^2,y^3,y^4$ give five equations with four coefficients, we can find a linear combination so that they add up to zero, leaving us with the equation
$$ y^4 + (-b^2 + 2 c + b \lambda - 4 \mu) y^3 + (c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2) y^2 + ()y + () = 0 , $$
where we're not interested in the last two coefficients. (This effectively calculates the resultant of the two polynomials $x^4+bx^3+cx^2+dx+e$ and $x^2+\lambda x-\mu-y$, which is generally best done with a computer.)
To eliminate the first two coefficients, we need
$$
-b^2 + 2 c + b \lambda - 4 \mu = 0 \\
c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2 = 0 .
$$
The first of these is linear, the second quadratic, so it is easy to solve both simultaneously: we find
$$ \lambda = \frac{3 b^3 - 10 b c + 12 d \pm 2 \Delta}{3b^2-8c}, \qquad \mu = \frac{2 b^2 c - 8 c^2 + 6 b d \pm b \Delta}{6b^2-16c} , $$
where $\Delta^2 = 6 b^3 d - 28 b c d - 2 b^2 (c^2 - 6 e) + 4 (2 c^3 + 9 d^2 - 8 c e) $. I've left out a few of the details of the calculation, but they should be easy, if tedious, to fill in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How can I solve $\int e^{2\theta} \sin(3\theta)\, d\theta$ with integration by parts? $\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am not sure how to solve it.
My Steps:
$\int e^{2\theta}\sin(3\theta)d\theta$
Let $u = \sin(3\theta)$ and $dv=e^{2\theta}d\theta$
Then $du = 3\cos(3\theta)d\theta$ and $v = \frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta} \sin(3 \theta)d\theta &= \frac{1}{2} e^{2\theta}\sin(3\theta) - \int\frac{1}{2}e^{2\theta}3\cos(3\theta)d\theta\\
&=e^{2\theta}\sin(3\theta) - \frac{3}{2}\int e^{2\theta}\cos(3\theta)d\theta\\
\end{align*}
$\int e^{2\theta}\cos(3\theta)d\theta$
Let $u = \cos(3\theta)$ and $dv = e^{2\theta}d\theta$
Then $du = -3\sin(3\theta)d\theta$ and $v=\frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta}\cos(3\theta) &= \frac{1}{2}e^{2\theta}\cos(3\theta)-\int (\frac{1}{2}e^{2\theta}\cdot-3\sin(3\theta))d\theta\\
&=\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta
\end{align*}
So you can see I just keep going in circles. How can I break out of this loop?
| Hint
In case integration by parts is not mandatory,
find
$$\dfrac{e^{2x}(a\cos3x+b\sin3x)}{dx}$$ and compare with $e^{2x}\sin3x$ to find the values of $a,b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3351012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$ Then $2A+2B+2C =360$
So $$\sin 2C=-\sin(2A+2B)$$
Putting that in the equation
$$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$
$$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+\cos B-\cos(A+B)+1}$$
I don’t know how to proceed. Please help me continue
| $\begin{align}
\sin(2a) + \sin(2b) + \sin(2c) &= 2 \sin(a+b) \cos(a-b) - \sin(2a+2b) \cr
&= 2\sin(a+b)\;\{ \cos(a-b) - \cos(a+b) \} \cr
&= 2\sin(c)\;\{2 sin(a) sin(b)\} \cr
&= 4\sin(a)\sin(b)\sin(c) \cr
{\sin(2a) + \sin(2b) + \sin(2c) \over 8 \cos{a\over2}\cos{b\over2}\cos{c\over2}}
&= 4 \sin{a\over2} \sin{b\over2} \sin{c\over2} \cr
&= 4 \sin{a\over2} \sin{b\over2} \cos{a+b\over2} \cr
&= 4 \sin{a\over2} \sin{b\over2} \left(\cos{a\over2}\cos{b\over2} - \sin{a\over2}\sin{b\over2} \right)\cr
&= \sin(a)\sin(b) - 4(\sin^2{a\over2})(\sin^2{b\over2}) \cr
&= \sin(a)\sin(b) - (1 - \cos(a))(1 - \cos(b)) \cr
&= \sin(a)\sin(b) - 1 + \cos(a) + \cos(b) - \cos(a)\cos(b) \cr
&= \cos(a) + \cos(b) - \cos(a+b) - 1 \cr
&= \cos(a) + \cos(b) + \cos(c) - 1 \cr\cr
{\sin(2a) + \sin(2b) + \sin(2c) \over \cos(a) + \cos(b) + \cos(c) - 1 } &= 8 \cos{a\over2}\cos{b\over2}\cos{c\over2}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my calculator, entering the right-hand-side and arctan-ing it for $x$.
To my surprise, the angle $x$ comes out at exactly 72 degrees. I did not expect such a neat relationship. Then, I thought I should have solved the equation analytically for the whole-degree angle without the calculator. I spent a good amount of time already and was not able to derive it yet.
Either the equation is not as innocent as it looks, or a straightforward method just eludes me.
| Golden triangles, I use these mnemonic:
$\;\displaystyle \cos 36° = \frac{\sqrt{2+\frac{1}{ϕ}}}{2} = \frac{ϕ}{2}\;,\;
\sin 36° = \frac{\sqrt{2-\frac{1}{ϕ}}}{2}$
$\;\displaystyle \cos 18° = \frac{\sqrt{2+ϕ}}{2} \qquad\;\;,\;
\sin 18° = \frac{\sqrt{2-ϕ}}{2} = \frac{1}{2ϕ}$
$\displaystyle \cos (60°-18°) =
\frac{1}{2} × \frac{\sqrt{2+ϕ}}{2} \;+\;
\frac{\sqrt{3}}{2} × \frac{1}{2ϕ}$
$\displaystyle \sec 42° = \frac{4ϕ}{ϕ\sqrt{2+ϕ} \,+ \sqrt{3}}
= \frac{4ϕ\,(ϕ\sqrt{2+ϕ} \,- \sqrt{3})}{ϕ^2\,(2+ϕ)-3}
= ϕ\sqrt{2+ϕ} \,- \sqrt{3}$
First term: $\;ϕ\sqrt{2+ϕ} \,= \cot 18° = \tan 72°$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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If $\sin\alpha+\sin\beta=a$ and $\cos\alpha-\cos\beta=b$, then what is $\tan(\frac{\alpha-\beta}{2})$? If I square both the equations
$$2+2\sin(\alpha-\beta)=a^2+b^2$$
$$\sin(\alpha-\beta)=\frac{a^2+b^2-2}{2}$$
Since $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$, then $$\sin(\alpha-\beta)=\frac{2\tan\frac{\alpha-\beta}{2}}{1+\left(\tan\frac{\alpha-\beta}{2}\right)^2}$$
It’s obviously the too long to solve, so is there a shorter way to do this?
| Denote $x = \dfrac{\alpha + \beta}{2}$ and $y = \dfrac{\alpha - \beta}{2}$.
Question: Are the any relations between $x, y$ and $\alpha, \beta$?
Answer:
Yes. $$x - y = \dfrac{\alpha + \beta}{2} - \dfrac{\alpha - \beta}{2} = \dfrac{2\beta}{2} = \beta\\x + y = \dfrac{\alpha + \beta}{2} + \dfrac{\alpha - \beta}{2} = \dfrac{2\alpha}{2} = \alpha$$ Therefore, whenever we see $\alpha$ and $\beta$, we can substitute $x + y$ and $x - y$ respectively.
We try to express $a$ and $b$ in terms of trigonometric functions of $x$ and $y$.
We have:
$$\begin{array}{rcl}
a &=& \sin \alpha + \sin \beta\\
&=& \sin \left(x + y\right) + \sin (x - y)\\
&=& (\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)\\
&=& 2 \sin x \cos y\\
&=& 2 \sin \dfrac{\alpha + \beta}{2} \cos \dfrac{\alpha - \beta}{2}
\end{array}$$
Also we have:
$$\begin{array}{rcl}
b &=& \cos \alpha - \cos \beta\\
&=& \cos(x + y) - \cos(x - y)\\
&=& (\cos x \cos y - \sin x \sin y) - (\cos x \cos y + \sin x \sin y)\\
&=& -2\sin x \sin y\\
&=& -2 \sin \dfrac{\alpha + \beta}{2} \sin \dfrac{\alpha - \beta}{2}
\end{array}$$
We consider $\dfrac{b}{a}$. (Why?)
$$\begin{array}{rcl}
\dfrac{b}{a} &=& \dfrac{-\color{red}{2\sin \dfrac{\alpha + \beta}{2}} \sin \dfrac{\alpha - \beta}{2}}{ \color{red}{2\sin \dfrac{\alpha + \beta}{2}} \cos \dfrac{\alpha - \beta}{2}}\\
&=& - \dfrac{\sin \dfrac{\alpha - \beta}{2}}{\cos \dfrac{\alpha - \beta}{2}}
\end{array}$$
Recall that $\dfrac{\sin\left(\star\right)}{\cos \left(\star\right)} = \tan \left(\star\right)$.
$$\begin{array}{rcl}
\dfrac{b}{a} &=& -\color{green}{\dfrac{\sin \dfrac{\alpha - \beta}{2}}{\cos \dfrac{\alpha - \beta}{2}}}\\
&=& - \color{green} {\tan \dfrac{\alpha - \beta}{2}}\\
-\dfrac{b}{a} &=& \tan \dfrac{\alpha - \beta}{2}
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something particularly interesting happens:
$$e^{i z} = 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots \\
=1 + iz - \frac{z^2}{2!} - \frac{i z^3}{3!} + \frac{z^4}{4!} + \frac{i z^5}{5!} + \cdots \\
= (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i (z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots),$$
so $$e^{i z} = \cos z + i \sin z.$$
But I think the author didn't use a rearrangement of the terms at all.
Am I right?
Let
$$c_n := 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots + (-1)^n \frac{z^{2 n}}{(2 n)!},$$
$$s_n := z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots + (-1)^n \frac{z^{2 n + 1}}{(2 n + 1)!},$$
$$e_n := 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots + \frac{(iz)^n}{n!}.$$
Then, $$e_{2 n + 1} = c_n + i s_n,$$
$$\lim_{n \to \infty} e_{2 n + 1} = e^{i z},$$
$$\lim_{n \to \infty} c_n + i s_n = \lim_{n \to \infty} c_n + i \lim_{n \to \infty} s_n = \cos z + i \sin z,$$
so $$e^{i z} = \cos z + i \sin z.$$
| Your statement:
But I think the author didn't use a rearrangement of the terms at all.
He did re-arrange the terms as said...See:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integration using Trig Substitution I was given the following problem: $$\int\sqrt{1-7w^2}\ dw$$ I used the sin substitution - getting $w=\frac1{\sqrt{7}}\sin\theta$. I then needed to change the $dw$ to a $d\theta$, so I got this: $dw=\frac1{\sqrt{7}}\cos\theta\ d\theta$. My new problem looks like this: $$\int\sqrt{1-7(\frac1{\sqrt{7}}\sin\theta)^2}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$
Continuing, I get: $$\int\sqrt{1-\sin^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$ $$=\int\sqrt{\cos^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$$$=\int\frac1{\sqrt7}|\sin^2\theta|\cos\theta\ d\theta$$ If I now set $u=\sin\theta$ I get: $$\frac1{\sqrt7}\int u^2 \ du$$$$=\frac1{3\sqrt7}u^3$$$$=\frac{\sin\theta}{3\sqrt7}+c$$
This is not the correct answer. Why not? Where did I go wrong? What is the proper way to do a problem like this one?
| It may be more desirable to integrate directly without substitution.
$$I= \int \sqrt{1-x^2}dx = x\sqrt{1-x^2} + \int \frac{x^2}{\sqrt{1-x^2}}dx $$
$$=x\sqrt{1-x^2} -I + \int \frac{dx}{\sqrt{1-x^2}}=x\sqrt{1-x^2} -I + \sin^{-1}x+C$$
Thus,
$$I = \frac 12 \left( x\sqrt{1-x^2}+ \sin^{-1}x \right)+C $$
With $x=\sqrt{7}w$, the original integral,
$$\int\sqrt{1-7w^2}\ dw=\frac{1}{\sqrt{7}}I =\frac{1}{2\sqrt{7}} \left( \sqrt{7}w\sqrt{1-7w^2}+ \sin^{-1}(\sqrt{7}w)\right) +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do I find all complex numbers that are solutions of $z^3 +8i=0$? My workbook has answers in the standard form $a+bi$. I would assume that to solve this, I would expand the complex number $z$ into trigonometric form to deal with that exponent. This is what I have so far using De Moivre's Theorem:
$(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$.
However, I do not know where to go from here. Thank you!
| $(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$
So continue:
$(r^3(\cos(3\theta)+i\sin(3\theta)) = -8i= 8(0 + i(-1))$
$(r^3(\cos(3\theta)+i\sin(3\theta)) = 8((\cos \frac {3\pi}2+2k\pi) + i\sin(\frac{3\pi}2+2k\pi))$
So $r^3 = 8$ and $3\theta = \frac {3\pi}2 + 2k\pi$
So $r =2$ and $\theta = \frac \pi 2 + \frac 23k\pi$
So there are three solutions:
1)if $k =0$ then $2(\cos \frac \pi 2 + i\sin \frac \pi 2) = 2i$.
Check. $(2i)^3 = 2^3i^2 = 8*i^2 *i = -8i$. That's good.
2) if $k = 1$ then $2(\cos (\frac \pi 2+\frac {2\pi} 3) + i \sin (\frac \pi 2+\frac {2\pi} 3))= 2(\cos \frac {7\pi}6) + i\sin \frac {7\pi}6) = -\sqrt 3 - i$.
Check: $(-\sqrt 3-i)^3 = -[(\sqrt 3)^3 + 3(\sqrt 3)^2i + 3(\sqrt 3)i^2 + i^3] = $
$-[3\sqrt 3 + 9i - 3\sqrt 3 - i] = 8i$. It works.
and if $k = 2$ then 3) $2(\cos (\frac \pi 2+\frac {4\pi} 3) + i \sin (\frac \pi 2+\frac {4\pi} 3))= 2(\cos \frac {11\pi}6) + i\sin \frac {11\pi}6) = \sqrt 3 - i$.
Check: $(\sqrt 3-i)^3 = (-\sqrt 3)^3 + 3(-\sqrt 3)^2i + 3(-\sqrt 3)i^2 + i^3] = $
$-3\sqrt 3 - 9i +3\sqrt 3 + i = -8i$. It works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 6
} |
Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3\theta=1/2$ are
$$\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}$$
| $$\theta \in \left[0, 2\pi\right]\implies 3\theta \in \left[0, 6\pi\right]$$
$$\sin 3\theta = \dfrac{1}{2}$$
$$3\theta = \frac{\pi}{6}+2n\pi, \dfrac{5\pi}{6}+2n\pi, n \in \Bbb Z$$
But $3\theta \in \left[0, 6\pi\right]$, so $$3\theta = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6},\dfrac{25\pi}{6},\dfrac{29\pi}{6}$$
Therefore
$$\theta = \dfrac{\pi}{18}, \dfrac{5\pi}{18}, \dfrac{13\pi}{18},\dfrac{17\pi}{18},\dfrac{25\pi}{18},\dfrac{29\pi}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Trying to find $c_{1}$ , $c_{2}$ for Big theta notation. Need to find $c_{1},$ $c_{2}$ and $n_{0}$.
\begin{equation}
c_{1}n^3 \leq \frac{n^3}{100} - 100n^2 - 100n + 3\leq c_{2}n^3
\end{equation}
\begin{equation}
c_{1} \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
\begin{equation}
\frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
This is what I have done so far, but I am currently stuck do I need to chose c2 which is positive and greater than $\frac{100}{n} + \frac{3}{n^3}$? And for c1 , I need a c1, which is >0. How do I find these constants?
| The first step is to find a suitable $n_0.$
You know that you need
$$
0 < c_1 \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_2.
$$
Ignore the $c_1$ and $c_2$ for a moment, and you may see that you need
$$
0 < \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3}.
$$
This must be true for every $n > n_0.$ There is no penalty for choosing $n_0$ too large. So guess big and see what happens. If you cannot show the inequality above for every $n$ greater than your guess of $n_0,$ guess bigger.
Once you have a good guess you can start writing your proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the asymptotes to $y^2+2y-4x^2=0$ Calculate the center and asymptotes to the hyperbola $y^2+2y-4x^2=0$, aswell as the intersections of the actual coordinate axis.
The hyperbola takes the form $\left ( \frac{x}{\frac{1}{\sqrt{4}}}\right )^{2}-\left ( {y+1}\right )^{2}=-1$ and I understand how to complete all of the steps except how to find the asymptotes to the hyperbola. My course textbook states that they should be given by the formula $y=\pm \frac{b}{a}x$ if $(\frac {y}{b}^{2})=(\frac{x}{a})^{2}\pm1\approx (\frac{x}{a})^2$. But if I use this formula I get that $y+1=\pm \sqrt4 x $. What did I do wrong?
| I think your hyperbola has the form
\begin{align*}y^2 + 2y {\color{red}+\color{red}1\color{red}-\color{red}1}-4x^2 =0 &\Leftrightarrow (y+1)^2 -1- 4x^2= 0 \\&\Leftrightarrow (y+1)^2 - 4x^2=1
\end{align*}
So $a = \frac{1}{2}$ and $b= 1$.
Then you should get the asymptotes
$$y = \pm 2x-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
find the value $\sum_{i=1}^{n}|a_{i}|-\sum_{1\le ilet $n>3$,if $a_{1}=a_{2}=\cdots=a_{n-1}=1$,and $a_{n}=-2$,find the following value
$$f_{n}=\sum_{i=1}^{n}|a_{i}|-\sum_{1\le i<j\le n}|a_{i}+a_{j}|+\sum_{1\le i<j<k\le n}|a_{i}+a_{j}+a_{k}|-\cdots+(-1)^{n-1}|a_{1}+a_{2}+\cdots+a_{n}|$$
I have find the $n=4$,then
$$f_{4}=5-6-4+3-1=-3$$
$$f_{5}=6-12-4+3\times 4-4-4+2=-4$$
|
First note the following identities
${n}\choose{1}$ - $2$${n}\choose{2}$ + $3$${n}\choose{3}$ - ... $=0$
${n}\choose{1}$ - ${n}\choose{2}$ + ${n}\choose{3}$ - ... $=1$.
The sums which do not include $a_n$ add up to
${n-1}\choose{1}$ - $2$${n-1}\choose{2}$ + $3$${n-1}\choose{3}$ - ... $=0$.
The sums which do include $a_n$ add up to
$2$${n-1}\choose{0}$ - $1$${n-1}\choose{1}$ + $0$${n-1}\choose{2}$ - $1$${n-1}\choose{3}$ + $2$${n-1}\choose{4}$- ...
$=2$ - ${n-1}\choose{1}$ + $2$${n-1}\choose{2}$ - $3$${n-1}\choose{3}$ + $4$${n-1}\choose{4}$- ... - $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...
$=2$ - $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...
$=2$ - $2$${n-1}\choose{1}$ + $2$${n-1}\choose{1}$- $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...
$=2$ - $2$${n-1}\choose{1}$ + $2$
$=6-2n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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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.