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Prove that $1 < \frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}} < 2$ for $a, x > 0$
Let $a>0$, show that for $x>0$, $1<f(x)<2$, where $$f(x)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}.$$
Source (added 20221017): It was the last question in math exam of Jiangxi Province, 2008 China's national college entrance exam (a.k.a. Gaokao).
It was said that no examinee proved $f(x) < 2$. It was seen as one of the hardest math problems in GaoKao history. It attracted much attention in the community of high school math or Olympiad math. Perhaps there are more than 10 proofs. By the way, it was reported that in 2021 more than 10 million students took Gaokao.
I could take the derivative, find the maximum of the function and conclude. But that involves heady algebra and you have to solve a high degree polynomial. Is there an easier way simply using inequalities? Thanks, much appreciated
| Here is my proof for $f(x) < 2$ without calculus:
It suffices to prove that, for all $a, b, c > 0$ with $abc = 8$,
$$\frac{1}{\sqrt{1 + a}} + \frac{1}{\sqrt{1 + b}} + \frac{1}{\sqrt{1 + c}} < 2.$$
WLOG, assume that $a \le b \le c$.
We split into two cases:
*
*$a + b \ge 6$:
We have $b \ge 3$. Thus,
$$\mathrm{LHS} < 1 + \frac{1}{\sqrt{1 + 3}} + \frac{1}{\sqrt{1 + 3}} = 2.$$
*$a + b < 6$:
Using AM-GM, we have
$$\frac{1}{\sqrt{1 + a}} = \frac{\sqrt{1 + a}}{1 + a} \le \frac{\frac{(1 + a) + 1}{2}}{1 + a} = \frac{a + 2}{2 + 2a}$$
and
$$\frac{1}{\sqrt{1 + b}} = \frac{\sqrt{1 + b}}{1 + b} \le \frac{\frac{(1 + b) + 1}{2}}{1 + b} = \frac{b + 2}{2 + 2b}.$$
It suffices to prove that
$$\frac{a + 2}{2 + 2a} + \frac{b + 2}{2 + 2b} + \frac{1}{\sqrt{1 + c}} < 2$$
or
$$\frac{a}{2 + 2a} + \frac{b}{2 + 2b} > \frac{1}{\sqrt{1 + c}}.$$
Using AM-GM, it suffices to prove that
$$ 2\sqrt{\frac{a}{2 + 2a} \cdot \frac{b}{2 + 2b}} > \frac{1}{\sqrt{1 + c}}$$
or
$$4\cdot \frac{a}{2 + 2a} \cdot \frac{b}{2 + 2b} > \frac{1}{1 + c}$$
or
$$\frac{ab(7 - a - b)}{(1 + a)(1 + b)(ab + 8)} > 0$$
which is true.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/1646819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:
*
*The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.
*Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.
For $n=k+1$, there are two ways the number of sixes are even:
a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n]$
b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
I have two questions
*
*How do I get from $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$
*Other than that, is my use of induction correct? Is it rigorous enough to prove the formula?
| You should be trying to get from $\frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^{\bf{n+1}}]$ because that is the formula applied to $n+1$
$$\frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])=\frac 12+\frac 56\cdot \frac 12(\frac 23)^n-\frac 16\cdot \frac 12(\frac 23)^n=\frac 12[1+(\frac 23)^{n+1}]$$
Yes, you were approaching it the right way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying Expression Factorial Expression I'm confused as how I'm meant to simplify this:$$\frac{(n-2)!}{(n-2-r)!}$$
I have other factorial questions where the variable isn't present in the top factorial like the question above and I'm trying to figure out how I simplify.
Thanks
| \begin{align*}
\frac{(n-2)!}{(n-2-r)!} &=\frac{(n-2)!}{(n-(2+r))!} \\
&= \frac{(n-2)(n-(2+1))\dotsm(n-(2+(r-1)))(n-(2+r))!}{(n-(2+r))!}\\
&=(n-2)(n-3)\dotsm(n-1-r)\\
&=\prod_{k = 0}^{r-1}(n-(2+k))\\
&=\prod_{k=0}^{r-1}(n-2-k)
\end{align*}
For example, let $r = 3$. Then
\begin{align*}
\frac{(n-2)!}{(n-2-3)!}&=\frac{(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}\\
&=(n-2)(n-3)(n-4)
\end{align*}
Using the formula
$$\prod_{k = 0}^{3-1} (n-2-k) = \prod_{k=0}^2(n-2-k) = (n-2-0)(n-2-1)(n-2-2) = (n-2)(n-3)(n-4).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve without involving hyperbolic function. How to solve this integral without involving hyperbolic functions?
$$\int \frac{1}{4-5\sin^2 x}dx$$
The answer is $\frac{1}{4}(\ln (\sin x+2 \cos x)-\ln(2\cos x-\sin x))+c$
| \begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{dx}{4 \cos^2 x + 4 \sin^2 x - 5 \sin^2 x}\\
&= \int \frac{dx}{4 \cos^2 x - \sin^2 x}\\
&= \int \frac{\sec^2 x}{4 - \tan^2 x} \, dx
\end{align*}
Let $u = \tan x, du = \sec^2 x \, dx$. So
\begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{du}{4 - u^2}\\
&= \frac{1}{4} \int \frac{du}{2 - u} + \frac{1}{4} \int \frac{du}{2 + u}\\
&= -\frac{1}{4} \ln |2 - u| + \frac{1}{4} \ln |2 + u| + \cal{C}\\
&= -\frac{1}{4} \ln |2 - \tan x| + \frac{1}{4} \ln |2 + \tan x| + \cal{C}\\
&= -\frac{1}{4} \ln \left |2 - \frac{\sin x}{\cos x} \right | + \frac{1}{4} \ln \left |2 + \frac{\sin x}{\cos x} \right | + \cal{C}\\
&= -\frac{1}{4} \ln \left |\frac{2 \cos x - \sin x}{\cos x} \right | + \frac{1}{4} \ln \left |\frac{2 \cos x + \sin x}{\cos x} \right | + \cal{C}\\
&= -\frac{1}{4} \ln |2 \cos x - \sin x| + \frac{1}{4} \ln |2 \cos x + \sin x| + \cal{C}
\end{align*}
as required to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?
| Opposed to all of these answers, I will simply go ahead and solve the cubic:
$$0=x^3+x^2-x-2$$
Applying the cubic formula, I get:
$$x=\frac16\left(-2+\sqrt[3]{4}(\sqrt[3]{42+3\sqrt{177}})+\sqrt[3]{42-3\sqrt{177}})\right)$$
According to Wolfram|Alpha.
Then I guess you would either do the floor function after getting it in decimal form or you could somehow take the floor function now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all values of $(1+i)^{1+i}$ $$z=(1+i)^{1+i}$$
I'm having trouble with this one. I got as far as the below, but then I got stuck. Could someone give me a hint??
$$\ln(z)=(1+i)\ln(1+i)$$
After reading the hints + suggested answer:
$$z=(1+i)^{1+i}$$
$$z=e^{(1+i)ln(1+i)}$$
$$z=e^{(1+i)ln|1+i|+(arg(1+i)+2\pi k)i}$$
$$z=e^{ln(\sqrt{2})+\pi i/4+2\pi ik+iln(\sqrt{2})-\pi /4-2\pi k}$$
Is it ok to leave my final answer like this?
| $\forall n\in \mathbb{Z}$,
\begin{align*}
e^{2n\pi i} &= 1 \\
1+i &=
\exp \left[ \frac{\ln 2}{2}+i\pi \left( 2n+\frac{1}{4} \right) \right] \\
(1+i)^{1+i} &=
\exp \left[
\frac{1+i}{2} \ln 2+i(1+i) \left( 2n+\frac{1}{4} \right) \pi
\right] \\
&=
\exp \left[
\frac{\ln 2}{2}-\left( 2n+\frac{1}{4} \right) \pi+
i\left( \frac{\ln 2}{2}+2n\pi+\frac{\pi}{4} \right)
\right] \\
&= \sqrt{2} \, \exp \left[
-\left( 2n+\frac{1}{4} \right) \pi+
i\left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)
\right] \\
&= \frac{\sqrt{2}}{e^{\left( 2n+\frac{1}{4} \right) \pi}}
\left[
\cos \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)+
i\sin \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)
\right]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1655886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Generating function with G'(1) and G''(1) reducing to 0/0 = undefined My question is about an analysis of an algorithm in D. E. Knuth's book The Art of Computer Programming, Vol. 1. More specifically, it is about section 1.2.10, equations 20 to 22.
First we have a generating function
$$G(z)=\frac{1}{n}z+\frac{1}{n}z^2+...\frac{1}{n}z^n=\frac{1}{n}\frac{z^{n+1}-z}{z-1}$$.
For the analysis of the algorithm $G'(1)$ and $G''(1)$ have to be calculated. These derivatives are as follows.
$$G'(z)=\frac{nz^{n+1}-(n+1)z^n+1}{n(z-1)^2}$$
and
$$G''(z)=\frac{n(n-1)z^{n+1}-2(n+1)(n-1)z^n+n(n+1)z^{n-1}-2}{n(z-1)^3}$$
By setting $z=1$ one realizes that $G'(1)=\frac{0}{0}=undefined$ and $G''(1)=\frac{0}{0}=undefined$. Therefore the two values needed cannot easily be determined.
To circumvent any too tedious calculations, Knuth creates a Taylor series
$$G(1+z)=G(1)+G'(1)z+\frac{G''(1)}{2!}z^2+...$$
by replacing $z$ by $z+1$ in the first equation
$$G(1+z)=\frac{1}{n}\frac{(1+z)^{n+1}-1-z}{z}$$.
In the last step Knuth transforms this term again to a Taylor series
$$G(1+z)=1+\frac{n+1}{2}z+\frac{(n+1)(n-1)}{6}z^2+...$$
and this was the point where I had to surrender. How does one get from the second last equation to the last equation? There probably is a simple explanation, but I have been thinking about it for hours and haven't come up with any reasonable idea.
| The essential tool is the binomial formula
$$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$
Expanding $(1+z)^{n+1}$ in the second last formula gives
$$(1+z)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} z^k = 1 + (n+1) z + \frac{(n+1) \cdot n}{2} z^2 + \frac{(n+1) \cdot n \cdot (n-1)}{6} z^3 + \ldots$$
Hence,
$$(1+z)^{n+1} - 1- z = n\cdot z + \frac{(n+1) \cdot n}{2} z^2 + \frac{(n+1) \cdot n \cdot (n-1)}{6} z^3 + \ldots$$
Dividing by $n \cdot z$ gives the desired formula
$$\frac{(1+z)^{n+1} - 1 - z}{n \cdot z} = 1 + \frac{n+1}{2} z + \frac{(n+1)(n-1)}{6} z^2 + \ldots$$
| {
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"url": "https://math.stackexchange.com/questions/1656540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem evaluating $ \int_{0}^{\pi/2}\frac{x}{\tan x}dx $ I am trying to evaluate $\int_0^{\pi/2}\frac{x}{\tan x} \, dx$. This is how I started:
$$ \int_0^{\pi/2} x\cot x \, dx$$
Integrating by parts we get:
$$x\ln\left|\sin x\right| - \int_0^{\pi/2}\ln\left|\sin x\right| \, dx $$
Since $x$ is between $0$ and $\pi/2$, we can drop the absolute value.
$$x\ln \sin x - \int_0^{\pi/2}\ln \sin x \, dx $$
Integrating the second integral again by parts we get:
$$x\ln \sin x - x\ln \sin x - \int_0^{\pi/2}\frac{x}{\tan x} \, dx$$
Now we take the last integral to the other side to get $$2\int_0^{\pi/2}\frac{x}{\tan x} \, dx = 0$$ and therefore $\int_0^{\pi/2} \frac{x}{\tan x} \, dx = 0$. However, when I checked the memo, it said the answer was $\frac{\pi}{2} \ln2$. What is wrong with my solution? Please help me find where I went wrong.
| Let $\displaystyle{I = \int^{\frac{\pi}{2}}_0 x \cot x \, dx}$. Now
\begin{align*}I &= \int^{\frac{\pi}{4}}_0 x \cot x \, dx + \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} x \cot x \, dx = \int^{\frac{\pi}{4}}_0 x \cot x \, dx + I_1\end{align*}
For the integral $I_1$, set $x = \frac{\pi}{2} - u, dx = - du$ while for the limits of integration at $x = \frac{\pi}{4}, u = \frac{\pi}{4}$ while at $x = \frac{\pi}{2}, u = 0$. Thus
\begin{align*}I_1 &= - \int^0_{\frac{\pi}{4}} \left (\frac{\pi}{2} - u \right ) \cot \left (\frac{\pi}{2} - u \right ) \, du\\
&= \int^{\frac{\pi}{4}}_0 \left (\frac{\pi}{2} - u \right ) \tan u \, du\\
&= \frac{\pi}{2} \int^{\frac{\pi}{4}}_0 \tan u \, du - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\
&= \frac{\pi}{2} \Big{[}-\ln |\cos u| \Big{]}^{\frac{\pi}{4}}_0 - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\
&= \frac{\pi}{4} \ln 2 - \int^{\frac{\pi}{4}}_0 u \tan u \, du
\end{align*}
So for integral $I$ we have
\begin{align*}I &= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \cot x \, dx - \int^{\frac{\pi}{4}}_0 x \tan x \, dx\\
&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\cot x - \tan x \right ) \, dx\\
&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right ) \, dx\\
&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \, dx\\
&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \frac{\cos 2x}{\sin 2x} \, dx\\
&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \cot 2x \, dx
\end{align*}
For the integral to the right, set $u = 2x, du = 2 dx$ while for the limits of integration, at $x = 0, u = 0$, while at $x = \frac{\pi}{4}, u = \frac{\pi}{2}$. Thus
\begin{align*}I &= \frac{\pi}{4} \ln 2 + \frac{1}{2} \int^{\frac{\pi}{2}}_0 u \cot u \, du\\
&= \frac{\pi}{4} \ln 2 + \frac{1}{2} I\\\Rightarrow \frac{1}{2} I
&= \frac{\pi}{4} \ln 2\\\Rightarrow I
&= \frac{\pi}{2} \ln 2.\end{align*}
Thus $\displaystyle{\int^{\frac{\pi}{2}}_0 x \cot x \, dx = \frac{\pi}{2} \ln 2}$ as indicated in one of the comments.
| {
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Find two square roots a 2 by 2 matrix filled with twos Given matrix $A = \begin{pmatrix} 2 && 2 \\ 2 && 2\end{pmatrix}$, I want to find two square roots of A.
I have to go about this with only very introductory-type tools, those covered in an introductory matrix operations chapter.
*
*My Approach
*
*Since I know that the square root matrix is a 2x2 matrix, let the square root matrix be $B = \begin{pmatrix} a && b \\ c && d\end{pmatrix}$.
*Now, for B to be a square root matrix of A, the following must hold true: BB = A. Evaluating BB I get $BB = \begin{pmatrix} a^2 + bc && b(a + d) \\ c(a+d) && d^2 + bc\end{pmatrix}$
*This leaves me with the following equations:
*
*$a^2 + bc=2$
*$d^2 + bc=2$
*$b(a+d)=2$
*$c(a+d)=2$
*From here on I've tried solving the equations but none of my attempts yielded the correct solution. I get the feeling I'm overlooking something very basic.
| Note that line 1 and 2 give $a^2 = 2-bc = d^2$ and line 3 and 4 give $b = 2/(a+d) = c$. Now since $b(a+d) =2\neq0$ we can't have $a=-d$, thus $a=d$. Therefore our four equations simplify to
*
*$a^2+b^2=2$
*$2ab=2$
Rewriting the second we get $b = 1/a$ and substituting in the first gives $a^2 +a^{-2} = 2$ or $a^4-2a^2+1=0$. The two real solutions to this polynomial are $a=\pm 1$. We conclude that the two square roots of $A$ are
$$
\begin{pmatrix}
1&1\\1&1
\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}
-1&-1\\-1&-1
\end{pmatrix}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| Let $a,b,c$ be roots of $$x^3+mx-n=0$$
From Vieta formula we get:
$$ 0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$
so $$\boxed{a^2+b^2+c^2 =-2m}$$
Further $$x^3 = n-mx\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n $$
so $$\boxed{a^3+b^3+c^3 =3n}$$
Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$
so $$\boxed{a^5+b^5+c^5 =-5mn}$$
Finally $$x^7 = x(n-mx)^2 = n^2x-2mnx^2+m^2x^3 $$ $$=n^2x-2mnx^2+m^2n-m^3x$$
so $$\boxed{a^7+b^7+c^7 = 7m^2n}$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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integration of $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}*\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{1+x}+2\sqrt{({1+x})*({1-x})}+1-x}{{1+x}-{1+x}}=\frac{{2}+2\sqrt{1-x^2}}{2x}=\frac{2(1+\sqrt{1-x^2})}{2x}=\frac{(1+\sqrt{1-x^2})}{x}$$
$$\int \frac{(1+\sqrt{1-x^2})}{x}=\int \frac{1}{x}+\int\frac{(\sqrt{1-x^2})}{x}=\ln|x|+\int\frac{(\sqrt{1-x^2})}{x}$$
$x=\sin t$
$dx=\cos t \; dt$
$$\int\frac{(\sqrt{1-\sin^2t})}{\sin t}\cos t\;dt=\int \frac{\cos^2t}{\sin t}\;dt =\int \frac{1-\sin^2t}{\sin t}\; dt=\int \frac{1}{\sin t}\;dt -\int \sin t \; dt$$ $$=\ln\left(\tan\frac{t}{2}\right)+\cos t+c$$
How do I substitute t back to x?
| Since $x= \sin t$, you have that $t=\arcsin x$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 \int 1 - 2\frac{1}{u} du = -2 (u - 2\ln u) + C$$
After substituting for $u$, I get $4 \ln(2+ \cos x) - 2\cos x - 4 + C$. According to Wolfram Alpha, the solution is $4 \ln(2+ \cos x) - 2\cos x + C$. Not sure where my mistake is, but any help is greatly appreciated.
| There is no difference between $4 \ln(2+ \cos x) - 2\cos x + C$ and $4 \ln(2+ \cos x) - 2\cos x -4 + C$. Either way it means $4 \ln(2+ \cos x) - 2\cos x$ plus some quantity not depending on $x$, which could be any number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Relationships between points lines and planes Develop the Cartesian equation of a plane with $x$-intercept $a$, $y$-intercept $b$ and $z$-intercept $c$.
Show that the distance $d$ from the origin to this plane is given by $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
In the picture below I have included what I have done so far. I was able to set all my variables and begin to define them.
| Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. The plane passes through the point $A$, so its equation is $(X-A)\cdot N =0$. Note that $A \cdot N = abc$, so this can also be written as:
$$
bcx + cay + abz = abc
$$
The line through the origin normal to the plane has equation $L(t) = tN$. Where this line intersects the plane we get a point $F$. This is the "foot" of the normal from the origin to the plane, so it's the point of the plane that's closest to the origin. At $F$, we have $(tN - A) \cdot A = 0$. This gives $t = (A \cdot N)/(N \cdot N) = abc/(N \cdot N)$, so the intersection point is
$$
F = \frac{abc}{N \cdot N}N = \frac{abc}{\| N \|^2}N
$$
The distance $d$ from the origin to $F$ is given by
$$
d = \|F\| = \frac{abc}{\| N \|^2}\|N\| = \frac{abc}{\| N \|}
= \frac{abc}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
This formula is correct even if one of $a$, $b$, $c$ is zero and the other two are non-zero, in which case it gives $d=0$, as you would expect.
If $a$, $b$, $c$ are all non-zero, this formula gives
$$
\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
$$
| {
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| You can write
$$2^8+2^{11}+2^n=(2^4)^2+2.2^4.2^6+2^n$$
now, note that if $n=12$, follows that
$$(2^4)^2+2.2^4.2^6+2^{12}=(2^4)^2+2.2^4.2^6+(2^6)^2=(2^4+2^6)^2$$
Thus, $n=12$ to solve the problem.
| {
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Proving that $1^n+2^n+3^n+4^n$ $(n\in \Bbb N)$ is divisible by 10 when $n$ is not divisible by 4 I was solving some math problems to prepare for math contests and came across this one:
Prove that $1^n+2^n+3^n+4^n$ $(n\in N)$ is divisible by 10 if and only if $n$ is not divisible by 4.
So, from what I understand, we have to prove that:
$1+16^n+81^n+256^n$ $(n\in \Bbb N)$ is never divisible by 10;
$1+2 \cdot 16^n+3 \cdot 81^n+4 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+4 \cdot 16^n+9 \cdot 81^n+16 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+8 \cdot 16^n+27 \cdot 81^n+64 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10.
The problem is, these equations seem even more complex than the starting one. How would you prove these? Am I on the right track or is there an easier and more elegant way to do this? Thanks.
| $1^n+2^n+3^n+4^n$ is divisible by $10$ if and only if it's divisible by $2$ and $5$.
Clearly $1^n+2^n+3^n+4^n$ is even for all $n\in\mathbb Z^+$, so let's work with mod $5$.
If $n=4k$ for some $k\in\mathbb Z^+$, then:
$$1^k+16^k+81^k+256^k\equiv 1^k+1^k+1^k+1^k\equiv 4\not\equiv 0\pmod{5}$$
If $n$ is odd, then:
$$1^n+2^n+3^n+4^n\equiv 1^n+2^n+(-2)^n+(-1)^n\pmod{5}$$
$$\equiv 1^n+2^n-2^n-1^n\equiv 0\pmod{5}$$
If $n=4k+2$ for some $k\in\mathbb Z_{\ge 0}$, then:
$$1^n+4\cdot 16^k+9\cdot 81^k+16\cdot 256^k\pmod{5}$$
$$\equiv 1+(-1)\cdot 1^k+(-1)\cdot 1^k+1\cdot 1^k\equiv 1-1-1+1\equiv 0\pmod{5}$$
| {
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How to prove the following binomial identity How to prove that
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
| Let be $$b_i=\frac{1}{2^{2i}}\binom{2i}{i}\quad \text{and}\quad a_i=\frac{i}{2^{2i-1}}\binom{2i}{i}.$$
Observing that $$a_i=2ib_i$$ and $$a_{i+1}=\frac{i+1}{2^{2i+1}}\binom{2i+2}{i+1}=\frac{i+1}{2}\frac{(2i+2)(2i+1)}{(i+1)(i+1)}\frac{1}{2^{2i}}\binom{2i}{i}=(2i+1)b_i$$
we have
$$
\begin{align*}
a_{i+1}-a_i&=(2i+1)b_i-2ib_i=b_i
\end{align*}
$$
and finally we have a telescoping sum:
$$
\sum_{i=0}^n b_i=\sum_{i=0}^n (a_{i+1}-a_i)=a_{n+1}-a_0=(2n+1)b_n
$$
that is
$$
\sum_{i=0}^n \frac{1}{2^{2i}}\binom{2i}{i}=\frac{1}{2^{2n}}\binom{2n}{n}(2n+1)
$$
| {
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
| Setting the AM $x+3=\dfrac1y$ to find
$$\lim_{y\to0}\dfrac{(1-5y^2+4y^4)^{1/5}-(1-3y)}y$$
$$=\lim_{y\to0}\dfrac{(1-5y^2+4y^4)-(1-3y)^5}y\cdot\dfrac1{\lim_{y\to0}\sum_{r=0}^4\{(1-5y^2+4y^4)^{1/5}\}^r(1-3y)^{4-r}}$$
Using Binomial expansion, this becomes
$$\lim_{y\to0}\dfrac{5\cdot3y+y^2\left(-5-\binom52\cdot3^2\right)+\cdots+(3y)^5}y\cdot\dfrac1{\lim_{y\to0}\sum_{r=0}^4\{1^{1/5}\}^r1^{4-r}}=?$$
Use $y\to0\implies y\ne0$
| {
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How to find a set of vectors spanning the solution space of $Ax=0$, where How to find a set of vectors spanning the solution space of $Ax=0$, where
Basically I have tried many times to solve it and my answer consistently comes in the following form:
$\pmatrix{1 \\ -1 \\-1\\0}$
While my book gives an answer of:
$\pmatrix{-1 \\-1\\1\\0}$
| Reducing:
$$\begin{pmatrix}1&0&1&0\\1&2&3&1\\2&1&3&1\\1&1&2&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&2&2&1\\0&1&1&1\\0&1&1&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&1&1&1\\0&0&0&\!\!-1\\0&0&0&0\end{pmatrix}$$
The general solution to the homogeneous system $\;A\vec x=\vec0\;$ is
$$x_4=0\;,\;\;x_1=x_2=-x_3\;\implies\;\left\{\;\begin{pmatrix}t\\t\\\!\!-t\\0\end{pmatrix}\;\;;\;\;\;t\in\Bbb F\right\}$$
whatever the field $\;\Bbb F\;$ is. Thus, a particular non-zero solution, and also a basis for the solution space, is for example
$$\begin{pmatrix}1\\1\\\!\!-1\\0\end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$
$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$$
$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}\cdot\frac{\cos 2x-1}{\cos 2x-1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{\cos^22x-1}dx$$
$$=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{-\sin^22x}dx$$
$u=\cos2x$
$du=-2\sin2x$
Is this substitution is ok? or do dx must be in the numerator?
| Remember the bisection/duplication formulas and
note that $$\cos2x=2\cos^2x-1=1-2\sin^2x,$$ so your integral is
$$
\int_{0}^{\pi/4}-\frac{\sin^2x}{\cos^2x}\,dx
=
\int_{0}^{\pi/4}\frac{\cos^2x-1}{\cos^2x}\,dx
=
\int_{0}^{\pi/4}\left(1-\frac{1}{\cos^2x}\right)\,dx
$$
Can you finish?
| {
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Issue with the Euclidean Algorithm in $\mathbb{F}_2[x]$ So I have two polynomials in $\mathbb{F}_2[x]$ that I am trying to find the GCD of: $f(x)=x^6+x^5+x^4+x^3+x+1$ and $g(x)=x^5+x^3+x^2+x$.
So I start the algorithm:
$f(x)=g(x)(x+1)+x^3+1$
$g(x)=(x^3+1)(x^2+1)+x+1$
$x^3+1=(x+1)(?)+...?$
This is where I am having an issue. To break down my work more:
$g(x)(x+1)=x^6+x^4+x^3+x^2+x^5+x^3+x^2+x=x^6+x^5+x^4+x$ so we have a remainder of $x^3+1$.
For the next one, $(x^3+1)(x^2+1)=x^5+x^3+x^2+1$ so we have a remainder of $x+1$.
However, for the last one, we need to find $h(x),r(x)$ such that $x^3+1=(x+1)h(x)+r(x)$. As $\deg(x+1)\gt \deg(r(x))$ this means that $r(x)=0$ or $1$.
Looking at $(x+1)h(x)$, to get the 3rd degree term we need $h(x)$ to have an $x^2$. I cannot seem to be able to find an $h(x)$ $r(x)$ combination that meets these requirements. Help would be appreciated.
| You want $(x+1)h(x)=x^3+1+r(x)$, where $\deg r(x)<1$. As you note, you see that $h(x)$ needs to be of the form $x^2+\dots$. But then you get an $x^2=1\cdot x^2$ term in $(x+1)h(x)=(x+1)(x^2+\dots)$ you need to cancel, so $h(x)$ needs to also have an $x$ term. You then have to cancel the $x=1\cdot x$ term in $(x+1)h(x)=(x+1)(x^2+x+\dots)$, so $h(x)$ needs to have a $1$ term. You now have $h(x)=x^2+x+1$, and $(x+1)h(x)=x^3+1$, so your remainder $r(x)$ is just $0$.
| {
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Chain rule for differentiation I've been given this problem:
$y= \sqrt{7+6x^3}$
Using the chain rule am I right in suggesting that
$$u = 7+6x^3$$
$$y = \sqrt{u}$$
| Yes, there are two ways to see it: $$\frac{{\rm d}y}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}u} \frac{{\rm d}u}{{\rm d}x} = \frac{1}{2\sqrt{u}}18x^2 = \frac{9x^2}{\sqrt{7+6x^3}},$$or calling $f(x) = \sqrt{x}$ and $g(x) = 7+6x^3$, we have that $y(x) = f(g(x))$, so: $$y'(x) = f'(g(x))g'(x) = \frac{1}{2\sqrt{g(x)}}g'(x) = \frac{1}{2\sqrt{7+6x^3}}18x^2 = \frac{9x^2}{\sqrt{7+6x^3}}. $$
| {
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Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-4\sqrt{3}i}=\sqrt{2^2+3i^2-4\sqrt{3}i}=\sqrt{(2-\sqrt{3}i)^2}$ which made calculation easier.
| $$\text{z}=\sqrt{\sqrt{1-4i\sqrt{3}}}=\left(\sqrt{1-4i\sqrt{3}}\right)^{\frac{1}{2}}=\left(\left(1-4i\sqrt{3}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$$
$$\left(1-4i\sqrt{3}\right)^{\frac{1}{4}}=\left(7e^{-\arctan\left(4\sqrt{3}\right)i}\right)^{\frac{1}{4}}=\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}$$
So:
$$\Re\left[\text{z}\right]=\Re\left[\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}\right]=\sqrt[4]{7}\cos\left(-\frac{\arctan\left(4\sqrt{3}\right)}{4}\right)$$
$$\Im\left[\text{z}\right]=\Im\left[\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}\right]=\sqrt[4]{7}\sin\left(-\frac{\arctan\left(4\sqrt{3}\right)}{4}\right)$$
| {
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$1=\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$ $$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$
this look amazingly, square root of zero is $1$.
$$s=\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$$
$$s=\sqrt{n+s}$$
$$s=\frac{ 1\pm \sqrt{1+4n^2}}{2}$$
$$ \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\frac{1\pm \sqrt{1+4n^2}}{2}$$
$$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\lim_{n \to 0}\frac{1\pm \sqrt{1+4n^2}}{2}$$
But there are two answer $0$ and $1$, which is correct?
| First, you must prove if the limit exist.
If $s=1$, then $s=\sqrt{n+s}$ means $1=\sqrt{n+1}$
If $s=0$, then $s=\sqrt{n+s}$ means $0=\sqrt{n+0}$
When n=0, both are correct, but the limit is unique, so the limit doesn't exist.
| {
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Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ I'm really at loss with this problem. I should prove that
$$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$
Only thing I managed to do by working the left side was
$$\sum_{n=1}^\infty \frac{1}{n^2} = 1+\sum_{n=2}^\infty \frac{1}{n^2}=1+ \sum_{n=1}^\infty \frac{1}{(n+1)^2}$$
How am I supposed to get to the right form?
| So if I got it right it goes something like this
$$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$
$$=1+\sum_{n=1}^\infty (\frac{1}{n^2} - \frac{1}{n(n+1)})$$
$$=1+\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty (\frac{1}{n(n+1)})$$
$$=1+\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+1})$$
and because of telescoping property
$$\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+1}) = 1 $$
so
$$\sum_{n=1}^\infty \frac{1}{n^2}=1 + \sum_{n=1}^\infty \frac{1}{n^2} - 1 = \sum_{n=1}^\infty \frac{1}{n^2}.$$
| {
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Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$
| The coefficient of $x^7$ will be ${n\choose 7}{2^{n-7}\over 3^7}$ and the coefficient of $x^8$ will be ${n\choose 8}{2^{n-8}\over 3^8}$ and not what you have got .you just missed out the factor of 3 that plays a role as a coefficient.So equating it you get
$${n\choose 7}{2^{n-7}\over 3^7}={n\choose 8}{2^{n-8}\over 3^8}$$
and get the answer as $55$
| {
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Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
| This equation reduces to:
$$ \frac{(3^a)^4 + (3^a)^2 + 1}{(3^a)^2 + 3^a + 1} = \frac{7}{9} $$
Let $3^a = x$
$$ \frac{x^4 + x^2 + 1}{x^2 + x +1} = \frac{7}{9}$$
$$ 9x^4 + 2x^2 - 7x + 2 = 0$$
Can you factorise the above equation and then find the values of a by substituting back in to $3^a = x$? Looks like a tricky factorisation.
Alternatively, you could use user236182's trick above to find the values of $x$.
| {
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A definite integral that surely needs contour integration: $\int_0^{\infty} \frac{1}{x^2 + a^2}\cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right)\, dx$ During my Master Thesis work I came up with an integral which I am going to consider as a hard challenge. I have been trying for days to crack it, but still nothing. The integral is the following
$$\int_0^{+\infty} \frac{1}{x^2 + a^2}\cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right)\ \text{d}x$$
Where $a, b, c$ are simplified real constants ("simplified" means that they are constants, well defined, which I wrote $a, b, c$ for simplicity and brevity).
I do believe (but I think it's obvious [?]) that some contour integration is required. However I didn't obtain anything but to extend the integration to the whole axis since it's an even function..
The solution does exist, and it has been confirmed by one of my professor's colleagues which used Mathematica I guess. The explicit solution is
$$\frac{\pi}{2a}\exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right)$$
Any hint to solve that integral?
| Assume that all the parameters are real and nonnegative.
Then the equation $$\int_{0}^{\infty} \frac{1}{x^2 + a^2} \, \cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right) \, dx = \frac{\pi}{2a} \, \exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right)$$ holds iff $a >0$ and $b \ge c$.
To see why this is the case, consider the function $$f(z) = \frac{1}{a^{2}+z^{2}} \, \exp \left(iz \, \frac{z^{2}-b^{2}}{z^{2}-c^{2}} \right) = \frac{1}{a^{2}+z^{2}} \, \exp(iz) \exp \left(-iz \, \frac{b^2-c^2}{z^2-c^2}\right). $$
(Daniel Fischer suggested expressing $f(z)$ in that alternative way to make the analysis a bit easier.)
In the upper half of the complex plane, both $|\exp(iz)|$ and $ \left| \exp\Bigl(-iz \, \frac{b^2-c^2}{z^2-c^2}\Bigr) \right|$ are bounded if $b \ge c$.
The latter is not particularly obvious. But by substituting $x+iy$ for $z$, one finds that the real part of $-iz \, \frac{b^2-c^2}{z^2-c^2} $ is $$-\frac{(b^{2}-c^{2})(c^{2}y+x^{2}y+y^{3})}{(x^2-y^2-c^{2})^{2}+4x^2y^{2}},$$ which is never positive if $y>0$ and $b \ge c$.
So if $b \ge c$, the magnitude of $\exp\Bigl(-iz \, \frac{b^2-c^2}{z^2-c^2}\Bigr)$, like the magnitude of $\exp(iz)$, never exceeds $1$ in the upper half-plane, which includes near the essential singularities at $z= \pm c$.
Therefore, if $b \ge c$, we can integrate around a closed semicircular contour in the upper half-plane that is indented at $z=\pm c$ and conclude after taking limits that $$ \begin{align} \text{PV}\int_{-\infty}^{\infty} \frac{1}{x^2 + a^2} \, \cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right) \, dx &= \text{Re} \, 2 \pi i \, \text{Res} \left[f(z), ia \right] \\ &= \frac{\pi}{a} \, \exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right). \end{align}$$
But since $\cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right)$ is bounded along the real axis, the integral coverges in the traditional sense we can drop the Cauchy principal value sign.
I don't know how to determine the value of the integral when $b < c$.
EDIT:
A solution was posted in The Gazette of the Royal Spanish Mathematical Society, but I don't think it's correct. It makes no mention of any restriction on the parameters.
I rechecked with Wolfram Alpha to make sure the equation doesn't hold if $b<c$, and indeed it doesn't appear to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Cauchy-Shwarz inequality in vector analysis Vectors $x$ and $y$ are related as follows $$\mathbf{x}+\mathbf{y(x \cdot y)}=\mathbf{a}.$$
Show $$\mathbf{(x \cdot y)}^2=\mathbf{\frac{|a|^2-|x|^2}{2+|y|^2}}$$
I think we need to proceed using Cauchy-Shwarz inequality.
$\mathbf{y(x \cdot y)}=\mathbf{a}-\mathbf{x}$
$\mathbf{y(y \cdot x)(y \cdot x)}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$
$\mathbf{y(y \cdot x)^2}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$
Then, I am lost.
| We have $a = x + \def\<#1>{\left<#1\right>}\<x,y>y$, hence
\begin{align*}
|a|^2 &= \<a,a>\\
&= \<x + {\<x,y>y, x+ \<x,y>y}>\\
&= \<x,x> + 2\<x,y>^2 + \<x,y>^2\<y,y>\\
&= |x|^2 + (2 + |y|^2)\<x,y>^2\\
\iff |a|^2 - |x|^2 &=(2 + |y|^2)\<x,y>^2\\
\iff \<x,y>^2 &= \frac{ |a|^2 - |x|^2 }{2 + |y|^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding vertical asymptotes of $\frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$ I'm trying to find the vertical asymptotes for $$f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$$
If I understand correctly, the vertical asymptote exists at $x=a$ when a value $a$ is found such that $f(a)$ increases to $∞$.
So, we must find a number that is infinitely small on the denominator, or in other words, $0$.
Setting $$x^4 - 25^2 + 144 = 0$$
$$(x^2 - 9)(x^2 - 16) = 0$$
$$(x+3)(x-3)(x+4)(x-4) = 0$$
So based on my understanding, $[-4, -3, 3, 4]$ should be values of $a$ where $x = a$ is a vertical asymptote.
Graphing this function, I can clearly see that there are only vertical asymptotes at $x = 4$ and $x = -3$.
Where did I go wrong in my theory?
| Hint:
By factoring the numerator we have
\begin{align}
3x^4+3x^3-36x^2&=3x^2(x^2+x-12)\\
&=3x^2(x+4)(x-3)
\end{align}
Then
$$\frac{3x^4+3x^3-36x^2}{x^4 - 25x^2 + 144}=\frac{3x^2(x+4)(x-3)}{(x+3)(x-3)(x+4)(x-4)}=\frac{3x^2}{(x+3)(x-4)}\qquad x\neq -4, x\neq 3$$
So, the limit of the function when $x\to -4$ and $x\to 3$ are finite, hence no asymptotes are at these points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I simplify this rational expression? I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$
I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$
$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
then I am stuck.
| $$\begin{align}
& \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(5^{k+1} - 1) + 3(4 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(1\cdot5^{k+1} + 4 \;\cdot\;5^{k+1} - 1)}{4}\\
= & \frac{3(5\cdot5^{k+1} - 1)}{4}\\
= & \frac{3(5^{k+2} - 1)}{4}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
Question:
calculate:
$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$
Attempt at a solution:
This can be written as:
$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}}$$
Here we can clearly see that if x would go to $+\infty$ the limit would converge towards $\frac{3}{2}$. But what happens when x goes to $-\infty$.
From the expression above it would seem that the answer would still be $\frac{3}{2}$. My textbook says it would be $- \frac{3}{2}$ and I can't understand why.
I am not supposed to use l'Hospital's rule for this exercise.
| $$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$
$$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$
$$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x^2+3x-(x^2+1)}{\sqrt{x^2\left(1+\frac{3}{x}\right)}+\sqrt{x^2\left(1+\frac{1}{x}\right)}}$$
$$=\lim_{x\to-\infty}\frac{3x-1}{2x}=-\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, otherwise:
*If $\frac{m+1}{n} \in \mathbb{Z}$ we use substitution $(a+bx^n)^p=t^s$
where $s$ is denominator of $p$;
*Finally, if $\frac{m+1}{n}+p \in \mathbb{Z}$ then we use substitution
$(a+bx^{-n})^p=t^s$ where $s$ is denominator of $p$.
If we look at this example:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx,$$
we can see that $m=\frac{1}{2}$, $n=1$, and $p=\frac{-1}{2}$ which means that we have to use third substitution since $\frac{m+1}{n}+p = \frac{3}{2}-\frac{1}{2}=1$ but when I use that substitution I get even more complicated integral with square root. But, when I tried second substitution I have this:
$$2-x=t^2 \Rightarrow 2-t^2=x \Rightarrow dx=-2tdt,$$
so when I implement this substitution I have:
$$\int \sqrt{2-t^2}\frac{1}{t}(-2tdt)=-2\int \sqrt{2-t^2}dt.$$
This means that we should do substitution once more, this time:
$$t=\sqrt{2}\sin y \Rightarrow y=\arcsin\frac{t}{\sqrt{2}} \Rightarrow dt=\sqrt{2}\cos ydy.$$
So now we have:
\begin{align*}
-2\int \sqrt{2-2\sin^2y}\sqrt{2}\cos ydy={}&-4\int\cos^2ydy = -4\int \frac{1+\cos2y}{2}dy={} \\
{}={}& -2\int dy -2\int \cos2ydy = -2y -\sin2y.
\end{align*}
Now, we have to return to variable $x$:
\begin{align*}
-2\arcsin\frac{t}{2} -2\sin y\cos y ={}& -2\arcsin\frac{t}{2} -2\frac{t}{\sqrt{2}}\sqrt\frac{2-t^2}{2}={} \\
{}={}& -2\arcsin\frac{t}{2} -\sqrt{t^2(2-t^2)}.
\end{align*}
Now to $x$:
$$-2\arcsin\sqrt{\frac{2-x}{2}} - \sqrt{2x-x^2},$$
which would be just fine if I haven't checked the solution to this in workbook where the right answer is:
$$2\arcsin\sqrt\frac{x}{2} - \sqrt{2x-x^2},$$
and when I found the derivative of this, it turns out that the solution in workbook is correct, so I made a mistake and I don't know where, so I would appreciate some help, and I have a question, why the second substitution works better in this example despite the theorem i mentioned above which says that I should use third substitution for this example?
| Let $u=\sqrt{2-x}$ then we simply want
$-2\int \sqrt{2-u^2}du$ which is simple after $u=\sqrt{2}\sin{v}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How can I find this sum? I'm doing some examples related to convolution (digital signal processing). I post my problem here because it is actually mathematics problem.
I have to calculate this sum:
$$\sum_{k\ = \ n-5}^{n+5} e^{-|k|}$$
Any suggestion?
|
We show the following is valid:
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}=
\begin{cases}
{\displaystyle \frac{e^{-5}-e^{6}}{1-e}e^{-n}}&\qquad\qquad |n|>5\\
\\
{\displaystyle \frac{e^{-n-5}-e^{n-5}-1-e}{1-e}}&\qquad\qquad |n|\leq 5
\end{cases}
\end{align*}
Since
\begin{align*}
e^{-|k|}=
\begin{cases}
e^{-k}&\qquad k\geq 0\\
e^k&\qquad k<0
\end{cases}
\end{align*}
we distinguish three different cases:
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}=
\begin{cases}
{\displaystyle\sum_{k=n-5}^{n+5}e^{k}}&\qquad n<- 5\qquad \qquad\text{(i)}\\
{\displaystyle\sum_{k=n-5}^{-1}e^{k}+\sum_{k=0}^{n+5}e^{-k}}&\qquad -5\leq n\leq 5\qquad \text{(ii)}\\
{\displaystyle\sum_{k=n-5}^{n+5}e^{-k}}&\qquad n> 5\ \ \qquad \qquad\text{(iii)}\\
\end{cases}
\end{align*}
Note that cases (i) and (iii) coincide, since substituting $k$ with $-k$ and changing the order of summation in (iii) gives
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-k}=\sum_{-k=n-5}^{n+5}e^{k}=\sum_{k=-n-5}^{-n+5}e^{k}
\end{align*}
which is the same as (i) when $n$ is replaced is with $-n$.
We recall the formula of finite geometric series and do some index transformation
\begin{align*}
\sum_{k=0}^mq^k&=\frac{1-q^{m+1}}{1-q}\\
\sum_{k=a}^bq^k&=\sum_{k=0}^{b-a}q^{k+a}
=q^a \sum_{k=0}^{b-a}q^{k}\\
&=q^a\frac{1-q^{b-a+1}}{1-q}
\end{align*}
Now we are well prepared to calculate the cases (i) - (iii)
Case (i): $n<-5$
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{n+5}e^{k}\\
&=\frac{1-e^{[n+5-(n-5)+1]}}{1-e^{-1}}e^{n-5}\\
&=\frac{1-e^{11}}{1-e}e^{n-5}\\
&=\frac{e^{-5}-e^{6}}{1-e}e^{n}\\
\end{align*}
Case (ii): $-5\leq n\leq 5$
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{-1}e^{k}+\sum_{k=0}^{n+5}e^{-k}\\
&=\sum_{k=0}^{-n+4}e^{k+n-5}+\frac{1-e^{-[(n+5)+1]}}{1-e^{-1}}\\
&=e^{n-5}\frac{1-e^{[(-n+4)+1]}}{1-e}+\frac{1-e^{-n-6}}{1-e^{-1}}\\
&=e^{n-5}\frac{1-e^{-n+5}}{1-e}+\frac{1-e^{-n-6}}{1-e^{-1}}\\
&=\frac{e^{-n-5}-e^{n-5}-1-e}{1-e}
\end{align*}
Case (iii): $n>5$
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{n+5}e^{-k}\\
&=\frac{1-e^{-[n+5-(n-5)+1]}}{1-e^{-1}}e^{-(n-5)}\\
&=\frac{1-e^{-11}}{1-e^{-1}}e^{-n+5}\\
&=\frac{e^5-e^{-6}}{1-e^{-1}}e^{-n}\\
&=\frac{e^{-5}-e^{6}}{1-e}e^{-n}\\
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the asymptotes of the graph $y=\frac{x^2-5x-4}{x^2-5x+4}$? I just came across this question asking me to find the asymptotes of
$$y=\frac{x^2-5x-4}{x^2-5x+4}$$
I typed this into my graphics calculator but I've never seen this type of graph before so this is something new. Currently studying functions.
Would someone be able to explain to me thoroughly how to solve this problem and how to go about this type of questions in general?
Thank you!
| Hint.
$$\begin{aligned}
f(x) &=\frac{x^2-5x-4}{x^2-5x+4}\\
&= \frac{x^2-5x+4 - 8}{x^2-5x+4}=1-\frac{8}{x^2-5x+4}\\
&=1-\frac{8}{(x-1)(x-4)}\\
&=1+\frac{8}{3} \left( \frac{1}{x-1} -\frac{1}{x-4} \right)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?
It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
| \begin{align}
\sin^3(x)+\cos^3(x) & = (\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
& \{\sin(x)+\cos(x) = \sqrt{2}\sin(x+\pi/4)\} \\
& = \sqrt{2}\sin(x+\pi/4)(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
& \{\sin(2x) = 2\sin(x)\cos(x)\} \\
& =\sqrt{2}\sin(x+\pi /4)(\sin^2(x) + \cos^2(x) -\sin(2x)/2)) \\
& =\sqrt{2}\sin(x+\pi /4)(1-\sin(2x)/2)) \\
& \{u = x +\pi/4, \: \sin(2x)=-\cos(2u)\} \\
& =(\sqrt{2}/2) \sin(u)(2+\cos(2u)) \\
& \{\cos^2(u) = \frac{1}{2}(1 + \cos(2u)\} \\
& =(\sqrt{2}/2) \sin(u)(1+2 \cos^2(u)) \\
\end{align}
\begin{align}
\ \int \frac{1}{\sin^3 (x) + \cos^3 (x)}dx & = \int{\frac{\sqrt{2}}{\sin(u)(1+2\cos^2(u))}du} \\
& =\int{\frac{\sqrt{2}\sin(u)}{\sin^2(u)(1+2\cos^2(u))}du} \\
& \{\cos(u)=t, \:-\sin(u) du = dt\} \\
& =-\int{\frac{\sqrt{2}}{(1-t^2)(1+2t^2)}dt} \\
& =-\sqrt{2} \int{\frac{1/6}{1-t}+\frac{1/6}{1+t}+\frac{2/3}{1+2t^2}dt} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Demidovich problems 556 and 557 Exist two problems:
$$
№556: \lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}};
$$
$$
№557: \lim_{x\rightarrow0}(\frac{a^{x+1}+b^{x+1}+c^{x+1}}{a+b+c})^{\frac{1}{x}}
$$
And $(a>0,b>0,c>0)$. I sure that have one solution for both.
| $$\lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}}$$
On substituting $x=0$, the limit is of the form $1^{\infty}$.
Hence,
$$\lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}}=\lim_{x\rightarrow0}e^{\left(\frac{a^x+b^x+c^x}{3}-1\right)\frac1x}=\lim_{x\rightarrow0}e^{\left(\frac{a^x-1}{x}+\frac{b^x-1}{x}+\frac{c^x-1}{x}\right)\frac13}=e^{\frac{\ln a+\ln b+\ln c}3}=e^{\ln ({abc}^{\frac13})}=(abc)^{\frac13}$$
Similarly, the other problem can be solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cute Diophantine equation (simplify the expression) Find the largest integer $n$ less than $1000$ of the form
$n=(x+\sqrt{x^2-1})^{\frac{4}{3}}+(x+\sqrt{x^2-1})^{\frac{-4}{3}}$
for some positive integer $x$.
| Write $$u:=x+\sqrt{x^2-1}=e^{3t}\quad(t\geq0)\ .$$
Then
$$x={1\over2}\left(u+{1\over u}\right)=\cosh(3t)$$
and
$${n\over2}={1\over2}(u^{4/3}+u^{-4/3})=\cosh(4t)\ .$$
It follows that
$$8x^4-8x^2+1=\cosh(12t)=4\left({n\over2}\right)^3-3\left({n\over2}\right)\ $$
(see the comment by san below), so that $x$ and $n$ are related by
$$4(2x^2-1)^2=n^3-3n+2=(n-1)^2(n+2)\ .\tag{1}$$
This implies that $n+2$ has to be a square: $n+2=m^2\geq4$. Introducing this into $(1)$ and taking the square root we obtain
$$2(2x^2-1)=(m^2-3)m\ ,$$
which then leads to
$$x^2={1\over4}(m^2-3m+2)={1\over4}(m-1)^2(m+2)\ .$$
This shows that $m+2$ has to be a square as well: $m+2=p^2\geq4$. This leads to
$$x={p(p^2-3)\over2},\qquad n=m^2-2=p^4-4p^2+2\ .$$
In this way we obtain for $p\geq2$ the pairs
$$(x,n)=\quad(1,2),\quad (9,47),\quad(26,194),\quad(55,527),\quad(99, 1154),\quad\ldots\ .$$
The answer to the original question therefore is $527$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is it true that for any $x>2$ and $k,j \in N$ with $k>j$, $p_k(x)$ is larger than $p_j(x)$? Let $p_k(x)=(x+\sqrt{x^2-4})^k-(x-\sqrt{x^2-4})^k$.
Is it true that for any $x>2$ and $k,j \in N$ $k>j$, $p_k(x)$ is larger than $p_j(x)$?
| Yes, it is true.
We can have
$$\begin{align}\frac{d}{dk}&p_k(x)\gt 0\\&\iff \left(x+\sqrt{x^2-4}\right)^k\ln \left(x+\sqrt{x^2-4}\right)-\left(x-\sqrt{x^2-4}\right)^k\ln\left(x-\sqrt{x^2-4}\right)\gt 0\\&\iff \left(\frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\right)^k\gt\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}\\&\iff k\gt\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\tag1\end{align}$$
where $$X=\frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\gt 1$$
Note here that
$$\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\lt 1\tag2$$
because
$$\begin{align}&\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\lt 1=\log_XX\\&\iff \frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}\lt \frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\\&\iff \left(x+\sqrt{x^2-4}\right)\ln\left(x+\sqrt{x^2-4}\right)\gt \left(x-\sqrt{x^2-4}\right)\ln\left(x-\sqrt{x^2-4}\right)\end{align}$$
which is true.
It follows from $(1)(2)$ that $p_k(x)$ is increasing for $k\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
The case $x < - 3$ in the absolute value equation $|x + 3| + |x - 2| = 5$ In the absolute value equation $|x + 3| + |x - 2| = 5$, why do we replace $|x + 3|$ by $-x - 3$ rather than $3 - x$ when $-\infty < x < -3$?
$$|x+3|+|x-2|=5$$
What is the result set?
$$\begin{array}{c|c|c|c}
& \hphantom{xxx}-3 & \hphantom{xxx-2} & 2\hphantom{-xxx} \\\hline
x+3 & - & + & + \\\hline
x-2 & - & - & +
\end{array}$$
My problem is in the case $-\infty<x<3$.
$-\infty<x<3$ $\implies$ $-x-3-x-2=5$ $\implies$ $-2x-1=5$ $\implies$ $-2x=6$ $\implies$ $x=-3$
As a basic rule in absolute value we can change $|x+3|$ to $3-x$ when $x<0$ and we want to remove the absolute value.
So that why we didn't solve it like:
$3-x+2-x=-5$ $\implies$ $-2x=0$ $\implies$ $x=0$ and $x$ is inacceptable
| Reminders about absolute value:
$$|x|:=\begin{cases} x&\text{when}~x\geq 0\\ -x&\text{when}~x<0\end{cases}$$
For your specific question, finding the values of $x$ such that $|x+3|+|x-2|=5$, it behooves us to break into cases so that we may remove the absolute value signs.
$|x+3|+|x-2| = \begin{cases} x+3+x-2&\text{when}~x\geq 2\\
x+3-x+2&\text{when}~-3<x<2\\
-x-3-x+2&\text{when}~x\leq -3\end{cases}$
Setting the expression equal to five gives us for the first case, $x\geq 2$ and $2x+1=5\Rightarrow 2x=4\Rightarrow x=2$. Checking, $x=2$ does agree with $x\geq 2$ so it is a solution. Indeed $|x+3|+|x-2|=|2+3|+|2-2|=|5|+|0|=5$
For the second case, we get $x+3-x+2$ which is equal to five always regardless the value of $x$, so all values in the second case will work. For example, $x=0.1$, we have $|0.1+3|+|0.1-2|=|3.1|+|-1.9|=3.1+1.9=5$
For the third case, we get $x\leq -3$ and $-x-3-x+2=5\Rightarrow -2x-1=5\Rightarrow -2x=6\Rightarrow x=-3$. Checking, $x=-3$ does agree with $x\leq -3$ so it is also a solution.
Putting all of this together, we have the full solution set is the set of all numbers $x$ such that $-3\leq x\leq 2$ (or using interval notation, $[-3,2]$)
In your work, you say something along the lines of "When $x<0$ we can change $|3+x|$ to $|3-x|$." That is not true. Take $x=-1$ for counterexample. $|3+(-1)|=|2|=2$ whereas $|3-(-1)|=|3+1|=|4|=4$. Refer again to the definition of absolute value given above. It says that if whatever is in the absolute value as a whole is negative, then change its sign. You do not change the sign of single pieces of what is in it.
$$|3+x|=\begin{cases}3+x&\text{when}~3+x\geq 0\\ -(3+x)&\text{when}~3+x<0\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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the system of diophantine equations: $x+y=a^3$; $xy=\dfrac{a^6-b^3}{3}$ has only trivial solutions. Without using Fermat's Last Theorem, how can one prove that the following system of diophantine equations has only trivial solutions:
$$x+y=a^3$$
$$xy=\dfrac{a^6-b^3}{3}$$
We suppose of course that $\gcd(x,y)=\gcd(a,x)=\gcd(a,y)=\gcd(a,b)=1$
| If $a$, $b$, $x$ and $y$ are integers with $\gcd(x,y)=\gcd(a,b)=1$ such that
$$x+y=a^3\qquad\text{ and }\qquad xy=\frac{a^6-b^3}{3},$$
then $b^3=a^6-3xy$ and $a^6=(x+y)^2=x^2+xy+y^2$, so
$$b^3=(x^2+xy+y^2)-3xy=x^2-2xy+y^2=(x-y)^2.$$
It follows that for some integer $c$ we have $b=c^2$ and $x-y=c^3$. Then
$$x=\frac{a^3+c^3}{2}\qquad\text{ and }\qquad y=\frac{a^3-c^3}{2},$$
from which it follows that
$$\frac{a^6-c^6}{4}=\frac{a^3+c^3}{2}\frac{a^3-c^3}{2}=xy=\frac{a^6-b^3}{3}=\frac{a^6-c^6}{3}.$$
This shows that $a^6=c^6$ and hence that $xy=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) &\geq
&\left( 1+1\right) \left( 1+\frac{1}{\cos x}\right) \\
&=&2\left( 1+\frac{1}{\cos x}\right) \\
&\geq &2\left( 1+1\right) =4,
\end{eqnarray*}
but $4\leq 3+\sqrt{2}$.
| We have $$\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}=\dfrac{\sin x \cos x+\sin x +\cos x +1}{\sin x \cos x}=\dfrac{1/2 \sin 2x+\sqrt{2}\sin(\frac{\pi}{4}+x)+1}{1/2\sin 2x}$$
$=1+\dfrac{2\sqrt{2}\sin(\frac{\pi}{4}+x)+2}{\sin 2x}$ is at its minimum when $x=\pi/4$ since we want the denominator closest to 1.
At $x=\pi/4$ the expression becomes $1+2\sqrt{2}+2=3+2\sqrt{2}$
Thus, we have shown the inequality $\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}\geq 3+2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Writing complex numbers in form $a+bi$ Can $\sqrt{i+\sqrt{2}}$ be expressed as $a+bi$ with $a,b \in \mathbb{R}$? In general, what kinds of expressions can be rewritten in that form?
| An efficient approach relies on use of Cartesian-to-polar coordinate transformation. To that end, let $z=x+iy=\sqrt{x^2+y^2} e^{i\arctan2(x,y)+i2\ell \pi}$. Then, the square root of $z$ is given by
$$\begin{align}\sqrt{z}&=\sqrt{x+iy}\\\\&=\sqrt{\sqrt{x^2+y^2}\,e^{i\arctan2(x,y)+i2\ell \pi}}\\\\
&=(-1)^{\ell}(x^2+y^2)^{1/4}e^{i\frac12\text{arctan2}\,(x,y)}\\\\
&=\bbox[5px,border:2px solid #C0A000]{(-1)^{\ell}(x^2+y^2)^{1/4}\left(\sqrt{\frac{1+\frac{|x|}{\sqrt{x^2+y^2}}}{2}} +i\sqrt{\frac{1-\frac{|x|}{\sqrt{x^2+y^2}}}{2}} \right)} \tag 1\\\\
\end{align}$$
where $\ell $ is any integer and $\arctan2(x,y)$ is the arctangent function with $2$ arguments.
Note that $(-1)^\ell = \pm 1$. Then, for $x=\sqrt 2$ and $y=1$, $(1)$ becomes
$$\begin{align}\sqrt{\sqrt 2+i}&=\pm 3^{1/4}\left(\sqrt{\frac{1+\frac{\sqrt 2}{\sqrt{3}}}{2}} +i\sqrt{\frac{1-\frac{\sqrt 2}{\sqrt{3}}}{2}} \right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\pm \left(\sqrt{\frac{\sqrt 3+\sqrt 2}{2}} +i\sqrt{\frac{\sqrt 3-\sqrt 2}{2}} \right)}
\end{align}$$
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
| HINT :
Dividing the both sides by $\cos^2\theta\ (\not=0)$ gives
$$6\tan^2(x)+\tan(x)-1=\frac{5}{\cos^2(x)}=5(1+\tan^2(x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
Integrating $\sqrt{r^2 - x^2 - z^2}$ I need to integrate the following piece:
$\int \sqrt{R^2 - x^2 - z^2} \, dx$.
This is what I tried:
*
*$\int \sqrt{R^2 - x^2 - z^2} \, dx = \int (R^2 - x^2 - z^2)^\frac{1}{2} \, dx$
*Substitute $u = (R^2 - x^2 - z^2)^\frac{1}{2}, du = \frac{du}{dx}dx = \frac{1}{2}(R^2 - x^2 - z^2)^\frac{-1}{2} * 2x dx, dx = \frac{(R^2 - x^2 - z^2)^\frac{1}{2}}{x} du = \frac{u}{x}du$
*But I'm stuck.
The expected result is: $\frac{R^2- z^2}{2} * \arcsin(\frac{y}{\sqrt{R^2 - z^2}}) + \frac{y}{2}*\sqrt{R^2-z^2-y^2}$
| Hint:
Let $R^2-z^2=a^2$. Then, substitute $x=a\sin\theta$.
$$I=\int\sqrt{a^2-x^2}dx=\int a^2\cos^2\theta d\theta$$
Use $\cos^2\theta=\frac{1+\cos2\theta}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
= \mathrm{e}^{-1}
$$
I know that $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ and so my attempt was to write
$$
\left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)\\
= \underbrace{\left(1-\frac{1}{n} \right)^n}_{\to \text{e}^{-1}} \cdot \underbrace{\frac{1-\frac{n}{n^2-1}}{1-\frac{1}{n}}}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\frac{1-\frac{n}{n^2-n+1}}{1-\frac{1}{n}}}_{\rightarrow 1}
$$
which I thought would solve my problem. But after a second look I see that splitting the limits in the product cannot be allowed. This would be the same nonsense as
$$
\underbrace{\left(1-\frac{n}{n^2} \right)}_{\rightarrow 1} \underbrace{\left(1-\frac{n}{n^2-1} \right)}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\left(1-\frac{n}{n^2-n+1} \right)}_{\rightarrow 1}.$$
Maybe anybody can make the situation clear to me.
| $$ \left(1-\frac{n}{n^2 - n + 1} \right)^n \leq \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) \leq \left(1-\frac{n}{n^2} \right)^n
$$
$$\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right)^n = \frac{1}{e}$$
$$\begin{align*} \lim_{n \to \infty} \left(1-\frac{n}{n^2 - n + 1} \right)^n
&= \lim_{n \to \infty} \left(1-\frac{1}{n - 1 + 1/n} \right)^n \\
&= \lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \frac{1}{e}
\end{align*}$$
so $\lim_{n \to \infty} \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) = \frac{1}{e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}} $$
$$ \frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}$$
$$ \frac{(\sin(x)-\cos(x))}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
$$ \frac{1}{\sin(x)+\cos(x)} $$
Now this is where I am stuck , I thought of multiplying the numerator and denominator by $$ \frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}} $$ but that did not work out well..
| Multiply the LHS by product of the reciprocal of RHS, and the RHS.
$$\begin{array}{lll}
\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}&=&\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}\cdot\frac{\sec x + \csc x}{\tan x + \cot x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{\sec^2 x - \csc^2 x}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{(\tan^2 x + 1) - (\cot^2 x+1)}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{\tan x + \cot x}{\sec x + \csc x}\\
\end{array}$$
The trick: if LHS=RHS then $LHS\times\frac{1}{RHS}$ will always equal 1, just as $\frac{1}{RHS}\times RHS$ will always equal 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to simplify this surd: $\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$ $$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$
We have to find the value of $x$.
Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.
I think the answer lies in simplification, but can't do it. Also I have tried taking $\sqrt{2}$ common, but it doesn't help.
| try converting expr inside surd into a square.
$$x = \sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$$
$$ = \sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}} $$
$$ = \sqrt{\frac{1+3+2\sqrt{3}}{2^2}}+\sqrt{\frac{1+3-2\sqrt{3}}{2^2 }} $$
$$ = \sqrt{\left(\frac{1+\sqrt{3}}{2}\right)^2}+\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^2} $$
$$ = \frac{|1+\sqrt{3}|}{2} + \frac{|1-\sqrt{3}|}{2}$$
$$ = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}$$
$$ = \sqrt{3}$$
Edit
My earlier answer was incorrect since i foolishly took square root and forgot about modulus. I have ammended it. It is important to note that:
$$\sqrt{x^2} = |x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Finding the value of a trigonometric function given the value of $\tan(\alpha)\tan(\beta)$ Given:
$$\tan{\alpha} \tan{\beta} = -\frac{b^2}{a^2}$$
where $a$ and $b$ are constants, find:
$$\cos^2(\frac{\alpha - \beta}{2})$$
in terms of $a$ and $b$.
Here is my attempt:
$$\frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}} = -\frac{b^2}{a^2}$$
$$\frac{\sin{\alpha}\sin{\beta} + \cos{\alpha}\cos{\beta}}{\sin{\alpha}\sin{\beta} - \cos{\alpha}\cos{\beta}} = \frac{a^2 - b^2}{-a^2 - b^2}$$
$$\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{b^2 - a^2}{a^2 + b^2}$$
|
actually this is related to ellipse, the question was "for the ellipse x^2/a^2+y^2/b^2, find the locus of the centroid of the triangle formed by the center and the points of intersection of chord of the ellipse which subtend right angle at the origin". while solving this problem i encountered this trigonometry problem
I understand how you faced the trigonometry problem.
Let $(a\cos\alpha,b\sin\alpha),(a\cos\beta,b\sin\beta)$ be the points on the ellipse. Then, we have
$$\frac{b\sin\alpha}{a\cos\alpha}\cdot\frac{b\sin\beta}{a\cos\beta}=-1\iff \tan\alpha\tan\beta=-\frac{a^2}{b^2}\tag1$$
Let $(x,y)$ be the coordinate of the centroid. Then,
$$x=\frac{0+a\cos\alpha+a\cos\beta}{3},\quad y=\frac{0+b\sin\alpha+b\sin\beta}{3}\tag2$$
and so
$$\begin{align}\left(\frac{3x}{a}\right)^2+\left(\frac{3y}{b}\right)^2&=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2\\&=1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\\&=2+2\cos(\alpha-\beta)\\&=4\cos^2\left(\frac{\alpha-\beta}{2}\right)\end{align}$$
Now, I think that we cannot represent $\cos^2(\frac{\alpha-\beta}{2})$ only by $a,b$.
So, let us use another approach.
We have
$$\begin{align}\left(\frac{3x}{a}\right)^2+\left(\frac{3y}{b}\right)^2&=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2\\&=1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\\&=2+2\cos\alpha\cos\beta(1+\tan\alpha\tan\beta)\\&=2+2\cos\alpha\cos\beta\left(1-\frac{a^2}{b^2}\right)\tag3\end{align}$$
From $(1)$, letting $\cos\alpha\cos\beta=P$,
$$\begin{align}\sin\alpha\sin\beta=-\frac{a^2}{b^2}P&\Rightarrow \sin^2\alpha\sin^2\beta=\frac{a^4}{b^4}P^2\\&\Rightarrow (1-\cos^2\alpha)(1-\cos^2\beta)=\frac{a^4}{b^4}P^2\\&\Rightarrow 1-(\cos^2\alpha+\cos^2\beta)+P^2=\frac{a^4}{b^4}P^2\\&\Rightarrow 1-\left(\left(\frac{3x}{a}\right)^2-2P\right)+P^2=\frac{a^4}{b^4}P^2\\&\Rightarrow a^2(b^4-a^4)P^2+2a^2b^4P+b^4(a^2-9x^2)=0\\&\Rightarrow P=\frac{-ab^4\pm b^2\sqrt{a^6+9x^2(b^4-a^4)}}{a(b^4-a^4)}\end{align}$$
Hence, from $(3)$,
$$\left(\frac{3x}{a}\right)^2+\left(\frac{3y}{b}\right)^2=2+2\left(1-\frac{a^2}{b^2}\right)\cdot \frac{-ab^4\pm b^2\sqrt{a^6+9x^2(b^4-a^4)}}{a(b^4-a^4)},$$
i.e.
$$\left(\frac{3x}{a}\right)^2+\left(\frac{3y}{b}\right)^2=\frac{2a^3\pm 2\sqrt{a^6+9x^2(b^4-a^4)}}{a(a^2+b^2)}$$
| {
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"url": "https://math.stackexchange.com/questions/1708334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| $$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
$$\sin^6 x+\cos^6 x=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)=1-3\sin^2x\cos^2x$$
Thus your expression simplifies to
$$2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$$
Which is $0$, for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.
| A calculus approach. Let $$f(x) = \frac{x}{2}+\frac{1}{x}.$$ So $$f(x)-x = -\frac{x}{2}+\frac{1}{x}$$ is the sum of two decreasing functions on $]0,+\infty]$. Hence $f(p_{n})-p_n<0$, the sequence is decreasing.
$f(x)$ is strictly increasing on $]\sqrt{2},2]$ (easy with the derivative), thus if $x \in ]\sqrt{2},2] $, $f(x) \in ]\sqrt{2},\frac{3}{2}] $. You begin to feel that $f$ could be a contracting map on this interval, and indeed $0<f'(x)\le \frac{1}{4}$.
By the Mean Value Theorem For Derivatives, there exist $c_x\in[\sqrt{2},x]$ such that
$$ f'(c_x) = \frac{f(x)-\sqrt{2}}{x-\sqrt{2}}\,.$$
So:
$$ 0< p_{n+1}-\sqrt{2}\le \max_{[\sqrt{2},x]} f' \times (p_n-\sqrt{2})\,$$
and
$$ 0< p_{n+1}-\sqrt{2}\le \frac{2-\sqrt{2}}{4^n}\,$$
which provides you with some rate of convergence:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Resolve $ \frac{120}{x+y} + \frac{60}{x-y} = 6;\,\frac{80}{x+y} + \frac{100}{x-y} = 7$ I want to resolve this system of equations:
$$\begin{cases} \frac{120}{x+y} + \frac{60}{x-y} = 6 \\\frac{80}{x+y} + \frac{100}{x-y} = 7\end{cases}$$
I came to equations like
$$x - \frac{10x}{x-y} + y - \frac{10y}{x-y} = 20$$
and
$$-2xy - y^2 - 10y = 20 - x^2 -10x$$
I need to leave $x$ or $y$ alone and didn't succeed. Any help?
| You could also solve by eliminating one of the variables:
Multiply the first equation by 100 and the second by 60:
$$ \frac{12000}{x+y} + \frac{6000}{x-y}= 600$$
$$ \frac{4800}{x+y} + \frac{6000}{x-y}= 420$$
Subtract them from each other and get rid of $\frac{6000}{x-y}$:
$$ x+y = \frac{720}{180} = 40$$
Substitute $x+y$ into either of your original equations:
$$ \frac{120}{x+y} + \frac{60}{x-y}= 6 \implies \frac{120}{40} + \frac{60}{x-y}= 6 $$
$$ x-y = 20$$
Now solve these two simultaneous equations:
$$ x+y = 40, \ x-y = 20$$
Trust you can finish this off
| {
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Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + \frac{3yx}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 3x + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 2.5x = 3y$$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?
| Solve to get
$$t = \frac{5x}{2y-2x} = \frac{3(x+y)}{y}$$
This gives the quadratic equation:
$$ 6x^2 - 5xy - 6y^2 = 0 $$
Factorize to get:
$$(3x + 2y) (2x - 3y) = 0$$
yielding
$$\frac{y}{x} = \frac{-3}{2} \,\,\,\text{or} \,\,\,\frac{2}{3}$$
Thus unless some sort of restriction is placed on $x$ or $y$
(for e.g both > $0$), $\frac yx$ can assume $2$ distinct values.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$? For a prime number $p \neq 2$, $3$, does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$?
A case where this is true as to suggest plausibility: $13 = 2^2 + 9 \times 1^2$.
| Finish:
For $p=12k+1$, we have $p=x^2+y^2$, and, if $gcd(xy, 3)=1$, then $p\equiv x^2+y^2\equiv 2 \pmod 3 $, a contradiction.
Conversely, $x^2+9y^2\equiv 0$ or $1 \pmod 3$, and $x^2+9y^2\equiv 0, 1, 2\pmod 4$. If this number is prime, then we have only one possibility:
$$x^2+9y^2\equiv 1 \pmod {12}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Linear Algebra Identity Proof Prove that the following identity holds for all $x,y$ are in $\mathbb{R}^n$:
$x\cdot{}y = \frac{1}{4}(\left\|x+y\right\|^2 - \left\|x-y\right\|^2)$
| We have
$$
\begin{align}
\frac14 \left(\left\|x + y\right\| ^2 - \left\|x - y\right\|^2\right) &= \frac14 \left((x + y)\cdot(x+y) - (x - y)\cdot(x-y)\right)\\
& = \frac14(x \cdot x + 2x \cdot y + y \cdot y - (x \cdot x - 2 x \cdot y + y \cdot y))\\
&= \frac14(4x \cdot y)\\
&= x \cdot y
\end{align}$$
and so we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse function of $x^2-4$ The function $h$ is defined by $$h(x)=x^2-4$$. for $$x\leq0$$
Find an expression for $h^{-1}(x)$
My attempt,
Let $h^{-1}(x)=a$
$x=h(a)$
$x=a^2-4$
$a=\sqrt{x+4}$
$h^{-1}(x)=\sqrt{x+4}$
Am I wrong? Is the answer $-\sqrt{x+4}$?
| There is a misstep in your procedure.
$y = x^2 - 4 \implies x^2 = y + 4\implies \sqrt{x^2} = \sqrt{y+4}$
Note that $\sqrt x^2 = x$ but $\sqrt{x^2} \neq x$ in general. Actually, $\sqrt{x^2} = |x|$, and combined with $|x| = -x$ for $x\leq 0$ we have
$-x = \sqrt{y+4}\implies x=-\sqrt{y+4}\implies h^{-1}(x) = -\sqrt{x+4}$
Quick check gives us
$h(h^{-1}(x)) = h(-\sqrt{x+4})= (-\sqrt{x+4})^2 - 4 = x + 4 - 4 = x$
$h^{-1}(h(x)) = h^{-1}(x^2 - 4) = -\sqrt{x^2-4 + 4} = -\sqrt{x^2} = - |x| = x\quad (x\leq 0)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.
| Have a look at the scene to come up with an idea:
(Large Version)
The green surfaces belong to the cone, the red and blue surfaces are spheres around $P$.
E.g you could try to find the intersection between cone and sphere and fiddle with the radius to shrink it to a point.
We can describe the intersection by the system
$$
x^2 + y^2 - z^2 = 0 \\
(x-1)^2 + (y - 4)^2 + z^2 = r^2
$$
This gives
$$
r^2 = (x-1)^2 + (y-4)^2 + x^2 + y^2 = 2x^2 + 2y^2 - 2x - 8y + 17 \iff \\
r^2/2 = x^2 + y^2 - x - 4y + 17/2
= (x - 1/2)^2 + (y-2)^2 + 17/2 - 1/4 - 4 \iff \\
\left(\frac{\sqrt{2r^2-17}}{2}\right)^2
= \left(x - \frac{1}{2}\right)^2 + \left(y - 2\right)^2
$$
If I made no error, this indicates a circle around $(1/2, 2, 0)$ as projection of the intersection surface onto the $x$-$y$-plane.
Then $z = \pm \sqrt{1/4 + 4} = \pm \sqrt{17}/2$.
That would suggest $Q=(1/2,2,\pm\sqrt{17}/2)$ as closest points, at distance $r = \sqrt{17/2}$.
(Large Version)
(Large Version)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$? How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$ ? Explain your answer.
I started by factoring $2016$. I found the factors to be $36$, but I couldn't go further.
| We know that $2016 = 2^5\cdot 3^2\cdot 7^1$ and so $2016^{2016} = 2^{2016\cdot 5}\cdot 3^{2016\cdot 2}\cdot 7^{2016}$
We want $x^y$ to be equal to $2016^{2016}$. It is clear then by the prime factorization of $2016$ and the fundamental theorem of arithmetic that $x$ must be of them form $x=2^{a}\cdot 3^b\cdot 7^c$, implying that $x^y=2^{ay}\cdot 3^{by}\cdot 7^{cy}$
Further, to have $2016^{2016}=x^y$ we will require that: $\begin{cases}ay=5\cdot 2016\\ by=2\cdot 2016\\ cy=2016\end{cases}$
Thus, we may simplify, $x^y = 2^{5cy}\cdot 3^{2cy}\cdot 7^{cy}$
Since we want $x$ and $y$ to be positive integers, this implies further that $c$ must be a positive integer. We see that the number of solutions $(x,y)$ will be in bijection with the solutions to $cy=2016$ which are in bijection with the divisors of $2016$.
As there are $(5+1)\cdot (2+1)\cdot (1+1)=6\cdot 3\cdot 2 = 36$ divisors of $2016$, each of these divisors may play the role of $y$ and the corresponding value for $x$ will be $x=2^{2016\cdot 5/y}\cdot 3^{2016\cdot 2/y}\cdot 7^{2016/y}$ (which is clearly an integer since $y\mid 2016$)
| {
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Differential equation with nasty coefficients $ x^2(1-x)^2 y'' + (Ax + b)y = 0 $ I have encountered a differential equation on the form
$$ x^2(1-x)^2 y'' + (Ax + b)y = 0 $$
My math background is too limited to even know where to begin, so any help of solving the equation (if a solution exist?) would be greatly appreciated!
| In fact it just belongs to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=262.
Let $y=x^p(1-x)^qu$ ,
Then $y'=x^p(1-x)^qu'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u$
$y''=x^p(1-x)^qu''+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u=x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u$
$\therefore x^2(1-x)^2\left(x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u\right)+(Ax+b)x^p(1-x)^qu=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\dfrac{Ax+b}{x^2(1-x)^2}u=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\left(\dfrac{A+2b}{x}+\dfrac{b}{x^2}+\dfrac{A+2b}{1-x}+\dfrac{A+b}{(1-x)^2}\right)u=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)+b}{x^2}+\dfrac{A+2b-2pq}{x(1-x)}+\dfrac{q(q-1)+A+b}{(1-x)^2}\right)u=0$
Choose $p(p-1)+b=0$ and $q(q-1)+A+b=0$ , the ODE reduces to the Gaussian hypergeometric equation.
Note that the special case of $A=0$ the ODE reduces to the form of http://eqworld.ipmnet.ru/en/solutions/ode/ode0224.pdf.
| {
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Roots of a quadratic equation If $ a $ , $b$ are the roots of $ x^2+x+1=0 $ , then the equation whose roots are $a^k $ and $b ^k$ where k is a positive integer not divisible by 3 is
$a)$ $x^2 - x + 1 = 0$
$b)$ $x^2 + x+1 = 0$
$c)$ $ x^2 -x -1 =0$
$d)$ None of the above
My attempt : I was able to solve the question , only after I used the options given below . I put k = 2 and checked the options for which the above statement was coming to be true.Using the formula for $(a+b)^2$ .
However I am looking for a more genera approach , preferably one which doesn't apply checking against the options
| Since $x^3-1=(x-1)(x^2+x+1)$, you have $a^3=1$ and $b^3=1$.
If $k=3h+1$, then $a^k=a(a^3)^h=a$ and $b^k=b(b^3)h=b$.
On the other hand, $a^2=b$ and $b^2=a$, so if $k=3h+2$, you have $a^k=a^2(a^3)^h=b$ and $b^k=b^2(b^3)^h=a$.
Why is $a^2=b$? It is clear that $(a^2)^3=(a^3)^2=1$, so $a^2$ is a root of $x^3-1$. However, $a^2\ne1$ because neither $1$ nor $-1$ are roots of $x^2+x+1$; it cannot be $a^2=a$, because otherwise $a=0$ or $a=1$. So the only remaining possibility is $a^2=b$. For the same reason, $b^2=a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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finding recursive formula and show it converges to a limit Suppose we are playing cards and we start with $1000$ dollars. Every hour we lose $\frac{1}{2}$ of our money and then we buy another $100$ dollars. I am trying to find $x_n$ for the amount of money the player has after $n$ hours.
I think we can just take $x_n = \frac{x_{n-1}}{2} + 100 $
An so, let $L = \lim x_n$. Then $L = \frac{L}{2} + 100 $ and so $L = 200$
Is this correct?
| $x_n = \frac{x_{n-1}}{2} + 100 = (\frac{\frac{x_{n-2}}{2} + 100}{2}) + 100 =\frac{x_{n-2}}{4} + 100.(\frac{1}{2} + 1)= \frac{\frac{x_{n-3}}{2} + 100}{4} + 100.(\frac{1}{2} + 1)=...$
$\implies x_n = \frac{x_0}{2^n} + 100.(\frac{1}{2^{n-1}} + \frac{1}{2^{n-2}} + ... + \frac{1}{2} + 1 ) $
where $x_0 = 1000$.
| {
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Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
| Using Henry W.'s answer $$A_n=\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = 2^n\frac{(n!)^2 }{(2n)!}$$ $$\log(A_n)= 2\log(n!) + n \log( 2) - \log \big((2n)!\big)$$ Now, using Stirling approximation
$$\log(p!)\approx p\log(p)-p+\frac 12\log(2\pi p)$$ $$\log(A_n)\approx \frac{1}{2} \log (\pi n)-n \log (2)$$ $n$ varies faster than $\log(n)$; so $\log(A_n)\to -\infty$ and $A_n\to 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding area bounded by two graphs of functions I know that someone posted this same question before here, because i found it, but i couldn't find the answer about this example that i was looking for, so i will post everything i did so far:
I have to find area bounded by graphics of functions
$y_1=x\sqrt{4x-x^2}$
$y_2=\sqrt{4x-x^2}$
First, i found domains, the are the same for both of these functions $[0,4]$
Now that i have the domains i need to find intersection points, and there's one $x=1$ now , it means that graph of one function is above the graph of the other one up to on interval $[0,1]$ and after that they switch their roles on $[1,4]$
It turns out that $y_1>y_2 (\forall x \in [0,1])$
and $y_2>y_1 (\forall x \in [1,4])$
So i have to find the following integral:
$P=\int_0^1 (\sqrt{4x-x^2} - x\sqrt{4x-x^2})dx + \int_1^4(x\sqrt{4x-x^2}-\sqrt{4x-x^2})dx$
This is equal to
$P=\int_0^1 (1-x)\sqrt{4x-x^2}dx - \int_1^4 (1-x)\sqrt{4x-x^2}dx$
so i actually have to solve
$\int (1-x)\sqrt{4x-x^2}dx = \int \sqrt{4x-x^2} - \int x\sqrt{4x-x^2}dx $
the first one can be easily solved using substitution $x=2sint$ and the second one can be solved using the same substitution except that we will have i little bit different solution.
It turns out that the solution of the first integral is
$2arcsin\frac{x-2}{2} + \frac{(x-2)\sqrt{4x-x^2}}{2}$
and for the second one:
$4arcsin\frac{x-2}{2} + (x-2)\sqrt{4x-x^2} + \frac{\sqrt{(4x-x^2)^3}}{3}$
Now when i insert the values i have:
$P=\int_0^1 \sqrt{4x-x^2}dx - \int_0^1 x\sqrt{4x-x^2}dx =[ 2arcsin\frac{1-2}{2} + \frac{(1-2)\sqrt{4*1-1^2}}{2} ] - [2arcsin\frac{0-2}{2} + \frac{(1-2)\sqrt{4*0-0^2}}{2}] - [2arcsin\frac{4-2}{2} + \frac{(4-2)\sqrt{4*4-4^2}}{2}] + [2arcsin\frac{1-2}{2} + \frac{(1-2)\sqrt{4*1-1^2}}{2}]$
In this large expression i have $arcsin(\frac{-1}{2})$ Now, i know that $arcsin(\frac{-1}{2}) = \frac{7\pi}{6} and \frac{11\pi}{6} $ my question is, which one i should use? In my solution, i used $\frac{7\pi}{6}$ And $\frac{11\pi}{6} = \frac{-\pi}{6}$ what about that?
And at the end, the most important thing, i got solution $P=\frac{-2\pi}{3}$ which is the negative value, but area can't be negative, what's wrong with this?
|
$P=\int_0^1 \sqrt{4x-x^2}dx - \int_0^1 x\sqrt{4x-x^2}dx $
You should have
$$P=\int_{0}^{1}\sqrt{4x-x^2}\ dx-\int_{0}^{1}x\sqrt{4x-x^2}\ dx\color{red}{-\left(\int_{1}^{4}\sqrt{4x-x^2}\ dx-\int_{1}^{4}x\sqrt{4x-x^2}\ dx\right)}$$
Also, $\arcsin(-1/2)=-\pi/6$ because the range of usual principal value of arcsine is $[-\pi/2,\pi/2]$.
I think you can continue from here.
| {
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Find the sum of the series $\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$ I need to test the convergence and find the sum of the following series:
$\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$
But i am not really sure what kind of series is this?
Since
$2\sqrt{n+1}>\sqrt{n}+\sqrt{n+2} \forall n \in \mathbb{N}$
It means that this is the series with all negative terms, what can i do with that kind of series? Can i use the same tests as for series with all positive terms (non-negative to be precise)? If that is true, then how can i use them?
| Since
\begin{align*}
\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}}
\end{align*}
Then, the sum is telescopic:
\begin{align*}
\sum_{n=1}^N(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})&=\frac{1}{\sqrt{N+2}+\sqrt{N+1}}-\frac{1}{\sqrt{2}+1}\\
\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})&=-\frac{1}{\sqrt{2}+1}\\
&=-(\sqrt{2}-1)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Show that $2^{15}-2^3$ divides $a^{15}-a^3$ for all $a$ Show that for all $a$, $2^{15}-2^3$ divides $a^{15}-a^3$.
I was able to prove that this is true for all $a$, such that $\gcd(a,2^{15}-2^3)=1$, by using Euler's theorem, where I concluded that $a^{12}\equiv1 \pmod{2^{15}-2^3}$, since $a^{12}\equiv1$, mod all of $2^{15}-2^3$ divisors.
For $a$, such that $\gcd(a, 2^{15}-2^3)\neq1$, I wasn't able to solve this. I thought of going over all of the cases of $\gcd(a,2^{15}-2^3)$, but this seems tedious.
| Similar to Ian's answer san the brute force: since
$$2^{15}-2^3=2^3\times 3^2 \times 5 \times 7 \times 13$$
we only need to show $a^{15}-a^3=a^3(a^{12}-1)$ is divisible by each of the above factors.
By Fermat's theorem
*
*$a(a^{12}-1)$ is divisible by $13$.
*$a(a^6-1)$ is divisible by $7$.
*$a(a^4-1)$ is divisible by $5$.
On the other hand, it is well-known that $a^2-1$ is divisible by $8$ if $a$ is odd. So $a^3(a^2-1)$ is divisible by $8$.
Lastly, to show divisibility by $3^2$, note that if $a$ is divisible by $3$, then we are done. If $a$ is not divisible by $3$, then $a^4=3k+1$, so $a^{12}-1=(3k+1)^3-1$ is divisible by $3^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Evaluating $ \int_{a}^{b} \frac{\mathrm{d}{x}}{x^{2}} $ using Riemann sums only. How does one evaluate the following Riemann integral by using Riemann sums only?
$$
\int_{a}^{b} \frac{\mathrm{d}{x}}{x^{2}}.
$$
| Compute the integral as the limit of a Riemann sum:
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k-1)\right)^{-1},$$
and decomposing into partial fractions,
$$\sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}k\right)^{-1}-\left(a + \frac{b-a}{n}(k+1)\right)^{-1}\right\} \leqslant S_n \\\leqslant \sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}(k-1)\right)^{-1}-\left(a + \frac{b-a}{n}k\right)^{-1}\right\}. $$
The sums are telescoping. Hence,
$$\left(a + \frac{b-a}{n}\right)^{-1}-\left(a + \frac{b-a}{n}(n+1)\right)^{-1} \leqslant S_n \leqslant a^{-1} - b^{-1}.$$
By the squeeze theorem, we get the value of the integral as
$$\lim_{n \to \infty}S_n = a^{-1} - b^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ x+iy = \sqrt{\frac{a+ib}{c+id} } ,$Show that$ (x^2+y^2)^2 = \frac {a^2+b^2}{c^2+d^2} $ $ x+iy = \sqrt{\frac{a+ib}{c+id} } , $Show that $ ({x^2+y^2})^2 = \frac {a^2+b^2}{c^2+d^2} $
How do i do this ?I tried squaring both sides but x+iy expansion becomes difficult when squaring the next time .I also tried conjugating the R.H.S
| $$x^2+y^2=|x+iy|^2=\left|\sqrt{\frac{a+ib}{c+id}}\right|^2=\left|\frac{a+ib}{c+id}\right|$$
Therefore,
$$(x^2+y^2)^2=\frac{|a+ib|^2}{|c+id|^2}=\frac{a^2+b^2}{c^2+d^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume of ice cream cone using cylindrical/spherical coordinates I'm stuck on what the boundaries are for the volume bounded by the cone $z=-\sqrt{(x^2+y^2)}$ and the surface $z=-\sqrt{(9-x^2-y^2)}$ $\,\,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-\sqrt{(9-r^2)}\le z \le -r$
$0\le r \le 3$
$0\le \theta \le 2\pi$
And for spherical coordinates the triple integral boundaries would be
$0\le r \le 3$
$\pi/2\le \phi \le \pi$
$0\le \theta \le 2\pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18\pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$\int_0^{2\pi}\int_0^3\int_{-\sqrt{9-r^2}}^{-r} r\,\, dzdrd\theta$
$\int_0^{2\pi}\int_0^3\ [rz]_{-\sqrt{9-r^2}}^{-r} \,\, dzdrd\theta$
$\int_0^{2\pi}\int_0^3\ -r^2+{r\sqrt{9-r^2}} \,\, dzdrd\theta$
$\int_0^{2\pi}[(\int_0^3\ -r^2 \,dr)+(\int_0^3\ {r\sqrt{9-r^2} \,dr)}]d\theta$
Let $u=9-r^2$
$du=-2r\,dr$
$\int_0^{2\pi}[[\frac{-r^3}{3}]_0^3\,-\frac{1}{2}\int_0^3\ u^{\frac{1}{2}} \,du)]\,d\theta$
$\int_0^{2\pi}[-\frac{27}{3}\,-\frac{1}{2}[\frac{2}{3}u^{\frac{3}{2}}]_{r=0}^{r=3} ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-r^2)^{\frac{3}{2}}]_{0}^{3} ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-3^2)^{\frac{3}{2}}\,-{\frac{2}{3}}(9-0^2)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}{\frac{2}{3}}[(9-9)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[(0)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[-27] ]\,d\theta$
$\int_0^{2\pi}[-9\,+9]\,d\theta$
$\int_0^{2\pi}0\,d\theta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
| By cylindrical coordinates the set up of the integral should be
$$\int_0^{2\pi}\int_{-3}^{-\frac3{\sqrt2}}\int_{0}^{\sqrt{9-z^2}} r\,\, dzdrd\theta+\int_0^{2\pi}\int_{-\frac3{\sqrt2}}^0\int_{0}^{-z} r\,\, dzdrd\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the general solution to $\sin(4x)-\cos(x)=0$
Question: Find the general solution to $\sin(4x)-\cos(x)=0$
My attempt:
$$ \sin(4x)-\cos(x)=0$$
$$ \Leftrightarrow \sin(4x) = \cos(x) $$
$$ \Leftrightarrow \sin(4x) = \sin( \frac{\pi}{2} -x)$$
From another post I learnt that you can equate $\sin(x) = \sin(y)$ on 2 conditions so applying it here:
$$ \Leftrightarrow 4x = \frac{\pi}{2} -x + 2 \pi n$$
and
$$ \Leftrightarrow4x = \pi - (\frac{\pi}{2} -x)+ 2 \pi n$$
Solving both for $x$
$$ x = \frac{\pi}{10} + \frac{2\pi n }{5}$$
$$ x = \frac{\pi}{6} + \frac{2\pi n }{3}$$
However I checked Wolfram alpha and they have different solutions?
Am I correct or not?
| $\sin(4x)−\cos(x)=0$
$2\sin(2x)\cos(2x)-\cos(x)=0$
$4\sin(x)\cos(x)(1-2\sin^2(x))-\cos(x)=0$
One possible solution is $\cos(x)=0$
$4\sin(x)(1-2\sin^2(x))=1$
$8\sin^3(x)-4\sin(x)+1=0$
Now, let $\sin(x)=m$ and solve the resulting cubic...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Number of subsets from an ordered set where adjacent elements may or may not be tied together Assume we have an ordered set $S$ with a finite number of elements $S=\{1,2,3,\ldots,N\}$. I need to know the number of subsets where adjacent elements from the original set may either be tied together as one "unit" shown with a "-" between them or separate elements shown as "," as normally in a subset.
For instance, with 2 elements, if $S=\{1,2\}$ this number is 5 where the 5 subsets are: $\{\},\{1\},\{2\},\{1,2\}$ and $\{1-2\}$.
And with 3 elements, if $S=\{1,2,3\}$ there are 13 subsets of this kind: $\{\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1-2\}, \{1,3\}, \{2,3\}, \{2-3\}, \{1,2,3\}, \{1-2,3\}, \{1,2-3\}$ and $\{1-2-3\}$.
With 4 elements I have counted this number to be 34. What is this number in the general case of $N$ elements, where $S=\{1,2,3,\ldots,N\}$ and can a formula be given?
| Let $a_n$ be the count of these listings for given $n$.
Then $a_n$ is obtained as sum of those listings not ending in $n$ (there are $a_{n-1}$ of these), those ending in "$,n$" (there are $a_{n-1}$ of these), and those anding in "${-}n$" (there are $a_{n-1}-a_{n-2}$ of these).
Hence we have the recursion
$$a_n=3a_{n-1}-a_{n-2}. $$
The general solution to this is
$$a_n=\alpha_1 \lambda_1^n+\alpha_2\lambda_2^n $$
where $\lambda_{1,2}$ are the roots of $x^2-3x+1=0$. So $\lambda_{1,2}=\frac{3\pm\sqrt{5}}{2}$. We determine $\alpha_{1,2}$ so that the result matches $a_0=1$, $a_1=2$. This leads to $\alpha_{1,2}=\frac{5\pm\sqrt 5}{10}$ so that
$$ a_n=\frac{5+\sqrt 5}{10}\cdot\left(\frac{3+\sqrt{5}}{2}\right)^n+\frac{5-\sqrt 5}{10}\cdot\left(\frac{3-\sqrt{5}}{2}\right)^n.$$
As the second summand is always between $0$ and $1$, we might as well say
$$ a_n=\left\lceil\frac{5+\sqrt 5}{10}\cdot\left(\frac{3+\sqrt{5}}{2}\right)^n\right\rceil.$$
Remark. In particular, the formualas above lead to $a_4=34$, not $30$. Indeed, here's the list:
$$\begin{matrix}\{\}&
\{1\}&
\{2\}&
\{1,2\}&
\{1-2\}\\
\{3\}&
\{1,3\}&
\{2,3\}&
\{2-3\}&
\{1,2,3\}\\
\{1-2,3\}&
\{1,2-3\}&
\{1-2-3\}&
\{4\}&
\{1,4\}\\
\{2,4\}&
\{1,2,4\}&
\{1-2,4\}&
\{3,4\}&
\{3-4\}\\
\{1,3,4\}&
\{1,3-4\}&
\{2,3,4\}&
\{2-3,4\}&
\{2,3-4\}\\
\{2-3-4\}&
\{1,2,3,4\}&
\{1-2,3,4\}&
\{1,2-3,4\}&
\{1-2-3,4\}\\
\{1,2,3-4\}&
\{1-2,3-4\}&
\{1,2-3-4\}&
\{1-2-3-4\}&
\end{matrix}$$
| {
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"url": "https://math.stackexchange.com/questions/1735881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}.......(1)$
By Heron's formula,$\Delta=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}}$
$\Delta^2=\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}$
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=16\Delta^2$
Putting in $(1)$
$16\Delta^2=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$
By formula $R=\frac{abc}{4\Delta}$,we get
$a^2+b^2+c^2=8R^2$
By sine rule,$a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\sin^2A+\sin^2B+\sin^2C=2$
I am stuck here.
| $\sin^2A+\sin^2B+\sin^2C =1-\cos^2A+ \dfrac{1-\cos(2B)}{2} + \dfrac{1-\cos(2C)}{2}=1-\cos^2A+ 1 - \cos(B+C)\cos(B-C)=2-\cos^2A+\cos A\cos(B-C)=2$. From this you can see $\cos A = 0$,or $A = B-C \Rightarrow B = \dfrac{\pi}{2}$. Thus $B)$ is the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is given by
$(A)5<x<10(B)x<\frac{5}{2}(C)\frac{5}{3}<x<10(D)\frac{5}{2}<x<10$
$\cos\angle A=\frac{AB^2+AC^2-BC^2}{2\times AB\times AC}=\frac{AB^2+100-x^2}{20\times AB}$
$\frac{\sqrt3}{2}=\frac{AB^2+100-x^2}{20\times AB}$
I am stuck here.
|
$$x^2=c^2+100-20c\frac {\sqrt 3}2$$
$$c^2-10\sqrt3 c+100-x^2=0$$
Two triangles exist -- equation has two roots:
$$\frac D4=75-(100-x^2)>0$$
$$x^2> 25$$
$$x > 5$$
$$x<b+c=10+c_1 \Rightarrow x<10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$, then equation becomes
$$y^2 + z^2=2xy$$
This implies that $(y-z)^2=0$.
So this implies that the required value is $$\frac{y^3 + z^3}{(y+z)(y^2 + z^2)}=\frac{1}{2}$$
But this wrong as the correct answer is $\frac{11}{18}$.
What is wrong with my method?
| Suppose one of the variables is zero. Then you have shown the maximum is $\frac12$.
So let us consider the case when none of the variables are zero. Then as both the objective and constraint are symmetric & homogeneous, we may set WLOG $z=1$ and $1\ge x\ge y>0$. Thus in this case we have the constraint $x^2+y^2+1 = 2(x+y+xy) \iff (x-y)^2+1=2(x+y)$ and need to maximize
$$F=\frac{x^3+y^3+1}{(x+y+1)(x^2+y^2+1)} $$
Now let $u= x+y, v = x-y \implies v^2+1=2u$. Now we can express everything in terms of $v \in [0, 1)$,
$$F = \frac{(v^2+3)^3+6(v^2-1)^2}{2(v^2+3)^3}=\frac12+\frac3{v^2+3}\left(\frac4{v^2+3}-1\right)^2\le \frac12+\frac3{3}\left(\frac43-1\right)^2=\frac{11}{18}$$
clearly this is a higher maximum than the earlier case, and equality is possible when $(u, v) = (\frac12, 0) \iff x=y=\frac14$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A conditional inequality which itself implies a sharper version of it Problem: Given that $m, n$ are positive integers such that $\sqrt{7} -\frac{m}{n} > 0$. Then show that $\sqrt{7}-\frac{m}{n} > \frac{1}{mn}$.
I have failed to do this fascinating problem.
My efforts: I tried to approach by contradiction.
Assume that we have $\sqrt{7} -\frac{m}{n} > 0$ and $\sqrt{7} -\frac{m}{n} > \frac{1}{mn}$ both hold simultaneously.
Now, $\sqrt{7} -\frac{m}{n} > \frac{1}{mn}$ implies $(m-\frac{n\sqrt{7}+\sqrt{7n^2-4}}{2})(m-\frac{n\sqrt{7}-\sqrt{7n^2-4}}{2}) >0$.
Since, $m \geq 1$ and $\frac{n\sqrt{7}-\sqrt{7n^2-4}}{2} < 1 $.
We end up with, $\frac{n\sqrt{7}+\sqrt{7n^2-4}}{2} < m < n\sqrt{7}$.
So, now I need to show that the last inequality cannot hold for positive integers $m, n$. But I am unable to do that.
So, someone please help me. Please..
| A more direct approach.
We write:
$$1\leq 7n^2-m^2 = (n\sqrt{7}+m)n \left(\sqrt{7}-\frac{m}{n}\right)$$
If $\sqrt{7}-\frac{m}{n}\lt \frac{1}{mn}$ (equality is not possible) then $n\sqrt{7}< m+\frac{1}{m}$ so $m+n\sqrt{7}\lt 2m+\frac{1}{m}$ and $7n^2-m^2\lt (2m+\frac{1}{m})n\frac{1}{mn}=2+\frac{1}{m^2}$.
So $7n^2-m^2=1\text{ or }2$. But $-1$ and $-2$ are not squares modulo $7$.
I doubt you will be able to get this without looking at congruences, because, for example, $m^2-Dn^2=-1$, then $$0<\sqrt{D}-\frac{m}{n}=\frac{1}{n(m+n\sqrt{D})}<\frac{1}{mn}$$ for these values, but when, say prime $p\equiv 7\pmod{8}$ then you never have $0<\sqrt{p}-\frac{m}{n}<\frac{1}{mn}$. So the general result is at least partly about these equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?
| Hint: except $-1^2$ make pairs of other terms: $$-1^2+(-2^2+3^2)+(4^2-5^2)+(-6^2+7^2)+....$$ and then proceed further to get $-1^2+2+3+4+5+6+...$ and than solve it to get the result
| {
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"timestamp": "2023-03-29T00:00:00",
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Numbers expressible as sum of 2 squares in 2 distinct ways I was trying this question here which goes like:
Find numbers which are squares and can be expressed as $x^2y^2-x^2-y^2+2$ for non-consecutive positive integers only.
Let the number be $a$
\begin{align*}
a^2 &= x^2y^2-x^2-y^2+2 \\
a^2-1 &= (x^2-1)(y^2-1) \\
&=(x-1)(x+1)(y-1)(y+1) \\
\text{Rearranging, }\\
&= \underbrace{(x-1)(y+1)}_{\text{First factor}} \text{ }\underbrace{(x+1)(y-1)}_{\text{Second factor}}\\
&=(\color{red}{xy-1}+\color{blue}{x-y})(\color{red}{xy-1}-\color{blue}{(x-y)}) \\
&= \color{red}{(xy-1)^2}-\color{blue}{(x-y)^2} \\
\text{So, } a^2-1&=(xy-1)^2-(x-y)^2 \\
\text{When $x-y= \pm1$, (consecutive) } \\
a = |xy-1| &= |(x(x\pm 1)-1)|=|x^2\pm x-1| \\
\text{Or else, }\\
a^2+(x-y)^2 &= 1+(xy-1)^2\\
\text{Let $x-y=t$ and $xy-1=u$, then } \\
a^2 + t^2&= 1^2 +u^2 = p \text{ (say)} \\
\end{align*}
So, I think we need to look at numbers which can be expressed as sum of 2 squares in 2 distinct ways, with 1 being one of the squares in one case.
For example, $$5^2 + 5^2 =1^2 + 7^2 = 50$$
Now, is there some general form for such numbers? Or some way in which I could generate them? How do I proceed? Also, is there some flaw in my work so far?
(I'm NOT looking for solutions to the problem I linked. I want some ideas to proceed with MY ATTEMPT)
| The condition $(x^2-1)(y^2-1)+1=a^2$ implies that the set $\{1,x^2-1,y^2-1\}$ is a Diophantine triple see (e.g. here), i.e. the product of any two of its distinct elements increased by $1$ is a perfect square. For fixed $x$, finding $y$ leads to Pellian equation $a^2 - (x^2-1)y^2 =-x^2$. It has infinitely many solutions: $y=1$, $y=x\pm 1$, $y=2x^2\pm 2x-1$, $y=4x^3 \pm 4x^2-3x \mp 1$, ... . E.g. for $y=2x^2+2x-1$ we get $a=2x^3 +2x^2 −2x−1$, and $x=2$ gives $a=19$ and the triple $\{1,3,120\}$, which is subtriple of the famous Fermat's quadruple $\{1,3,8,120\}$.
| {
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Find $\sqrt 7 \pmod {2579}$ Find $\sqrt 7 \pmod {2579}$.
I think I understand how I would solve a very basic equation like this:
$x^2 = 1 \pmod 5$
make a table of all the possible solutions like this
$x=0 \implies x^2=0 \\
x=1 \implies x^2=1 \\
x=2 \implies x^2=4 \\
x=3 \implies x^2=4 \\
x=4 \implies x^2=1 \\
x=5 \implies x^2=0$
then take the ones that work. In this case $\pm 1$ and $\pm 4$ are the roots, but how do I do this with a huge prime that I can't list out all the possibilities for?
| First, check that there are solutions for $x^2 \equiv 7 \bmod 2579 $ using Euler's criterion:
If $p$ is a prime that does not divide $a$,
then
$\qquad x^2 \equiv a \bmod p$ has a solution
$\iff a^{\tfrac{p-1}{2}} \equiv 1 \bmod p$.
We compute and see that $7^{\frac{2579-1}{2}} \equiv 1 \bmod 2579$.
Next, we find the solutions using a theorem of Lagrange, which is easily checked:
If $p \equiv 3 \bmod 4$ and $a$ is a quadratic residue mod $p$, then
$\qquad$ the solutions of $x^2 \equiv a \bmod p$ are $x \equiv \pm a^{\frac{p+1}{4}} \bmod p$.
This theorem applies because $2579 \equiv 3 \bmod 4$ and $7$ is a quadratic residue mod $2579$, as above.
So, we compute $7^{645} \equiv 88 \bmod 2579$ and so the solutions are $\pm 88 \bmod 2579$.
| {
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Volume of the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0$ I want to find volume the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0.$
I set it up as $\displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{-\sqrt{\frac{1}{2} - x^2}}^{\sqrt{\frac{1}{2} - x^2}}\int_0^{\frac{1}{\sqrt{2}}} \, dz \, dy \, dx$ - how come that's not the concerned volume?
| $x^2+y^2+z^2 \leq 1$ is clearly a ball. And for $z^2 \geq x^2 + y^2$, you can think in this way, let $r^2 = x^2+y^2$, then $x^2+y^2 = r^2 \leq z^2$, which is a cone. The overlap of a cone and a ball is still a cone. So try to use spherical coordinates.
| {
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The possible values of $\gcd(a^2,b^2)+\gcd(a,bc)+\gcd(b,ca)+\gcd(c,ab)$
Let $A=\gcd(a^2,b^2)+\gcd(a,bc)+\gcd(b,ca)+\gcd(c,ab)$, where $a$, $b$ and $c$ are positive integers. What are the values that $A$ can take, when $a$, $b$ and $c$ range over all positive integers?
I have no clue how to start. Any kind of help will be appreciated.
Addition:
It is understood that $A\ge4$.
Let $\gcd(a,b)=\gcd(a,c)=\gcd(c,b)=1$; then $A=4$.
But $A\ne5$ for any $a$, $b$ and $c$.
If $a=c=2$ and $b=1$, then $A=\gcd\left(2^2,1^2\right)+\gcd(2,1\times2)+\gcd(1,2\times2)+\gcd(2,2\times1)=1+2+1+2=6$.
But $A\ne7$ for any $a$, $b$ and $c$.
| The values that $ A $ can take are exactly the composite positive integers. I first noticed this by writing a computer program checking the value of $ A $ when $ a $, $ b $ and $ c $ are less than $ 30 $. The result looked amazing to me, as the defining expression of $ A $ didn't indicate it in a clear way. So, before I start presenting a proof, I'd like to ask about the source of the problem, and whether there was any motivation behind designing such a problem given somewhere. I want to add that the computer program also helped me find out how to show that every composite positive integer can appear as the value of $ A $ for some $ a $, $ b $ and $ c $.
First, let's show that $ A $ must be composite. To see this, let $ d = \gcd ( a , b ) $, $ a ' = \frac a d $ and $ b ' = \frac b d $, and note that $ \gcd \left ( a ^ 2 , b ^ 2 \right ) = d ^ 2 $, $ \gcd ( a , b c ) = d \gcd ( a ' , c ) $ and $ \gcd ( b , c a ) = d \gcd ( b ' , c ) $. Since $ \gcd ( c , d ) $ divides both $ d $ and $ \gcd ( c , a b ) $, in case $ c $ and $ d $ are not coprime, $ A $ is composite. In case $ c $ and $ d $ are coprime, we get $ \gcd ( c , a b ) = \gcd ( c , a ' b ' ) $, and since $ a ' $ and $ b ' $ are coprime by their definitions, $ \gcd ( c , a b ) = \gcd ( c , a ' ) \gcd ( c , b ' ) $. Then, we will have $ A = \bigl ( d + \gcd ( a ' , c ) \bigr ) \bigl ( d + \gcd ( b ' , c ) \bigr ) $, which shows that $ A $ is composite, as both factors in the last product are greater than $ 1 $.
Now, let's show that for any composite positive integer $ n $, we can choose positive integers $ a $, $ b $ and $ c $ such that $ A = n $. Given $ n $, there are $ m $ and $ k $, both greater than $ 1 $, such that $ n = m k $. Without loss of generality, assume that we have $ m \le k $. Set $ a = m - 1 $, $ b = ( m - 1 ) ( k - m + 1 ) $ and $ c = k - m + 1 $. Then, We have $ \gcd \left ( a ^ 2 , b ^ 2 \right ) = ( m - 1 ) ^ 2 $, $ \gcd ( a , b c ) = m - 1 $, $ \gcd ( b , c a ) = ( m - 1 ) ( k - m + 1 ) $ and $ \gcd ( c , a b ) = k - m + 1 $, which implies $ A = n $.
| {
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Find all the parameters and such that the line $y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles in at least one point . Problem:
Find all the parameters and such that the line
$y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles
in at least one point.
My work:
We try to find tangent to hyperbola such that tangent line intersects given line at rigth angle. We know that two line $y_1=ax+b$ and $y_2=cx+d$ intersects at right angle iff $a=-\frac{1}{c}$.
So we need to find tangent to hyperbola. Let $t...y=kx+l$ be our tangent line. To find that first we will need $y'$ and we will find it by implicit differentiation.
$$xy=1$$
$$y+xy'=0 \iff y'=\frac{-y}{x}$$
$$k=-\frac{y}{x}$$
But I haven't idea how to finish my work.
| Let $P=(x_P,y_P)$ the intersection point between the line and the hyperbola.
The hyperbola has equation $y=\frac{1}{x}$ so the derivative (without implicit differentiation) is $y'=\frac{-1}{x^2}$ and, at $P$ it is $y'(x_P)=\frac{-1}{x_P^2}$.
This is the slope of the tangent to the hyperbola at $P$, and if we want that it's orthogonal to the straight line $y=ax+\frac{1}{2}a-2$ we must have:
$\frac{-1}{x^2}=\frac{-1}{a} \Rightarrow a=x_P^2 $
Now, we want also that $P$ is a point of the straight line, se we must also have:
$\frac{1}{x_P}=ax_P+\frac{1}{2}a-2=x_P^3+\frac{1}{2}x_P^2-2 \Rightarrow x_P^4+\frac{1}{2}x_P^3-4x_P-2=0 \Rightarrow (2x_P-2)(x_P^3-2)=0$
Solving this equation you find the possible values for $x_P$ and for $a$.
| {
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Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question though, so if someone can explain it to me in a highschool math language, it will be appreciated. Now here is how I try to do it.
Let $X$ stand for any natural number and let $X+1$ and $X+2$ stand for the two consecutive numbers. I will be cubing, expanding and simplifying them
\begin{align*}
& (x)^{3}+(x+1)^{3}+(x+2)^{3}\\
&= x^3+x^3+3 x^2+3 x+1+x^3+6 x^2+12 x+8\\
&=3x^{3}+9x^{2}+15x+9 \\
&= 3\left ( x^{3}+3x^{2}+5x+3 \right )\\
\end{align*}
This can be used to deductively prove that the sum of cube of $3$ consecutive numbers is divisible by $3$ but I can't prove it is divisible by $9$
| Need to show that
$$x^3+3x^2+5x+3 =0 \mod 3$$
That is
$$x(x^2+5)=0 \mod 3$$
It's true when $x=0 \mod 3$.
For $$x=\pm 1 \mod 3$$,
$$(3k + 1)((3k + 1)^2+5)=(3k + 1)(9k^2+6k+6)=0 \mod 3$$,
$$(3k - 1)((3k - 1)^2+5)=(3k - 1)(9k^2-6k+6)=0 \mod 3$$.
| {
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Eigenvalue of 3x3 matrix with real unknown constants how do i find the eigenvectors of this matrix which has unknown real constants a and b ?
$$
\begin{bmatrix}
a & -\sqrt{2}b & 0 \\
-\sqrt{2}b & 0 & -\sqrt{2}b \\
0 & -\sqrt{2}b & a \\
\end{bmatrix}
$$
Eigenvalues of the matrix are $\lambda_{1}=a, \lambda_{2}=\frac{a-\sqrt{a^2+16b^2}}{2}, \lambda_{3}=\frac{a+\sqrt{a^2+16b^2}}{2}$
When I tried to solve I am getting $\Big[ 1,0,-1 \Big], \Big[1,\tfrac{{4}\sqrt{2}b}{-a+\sqrt{a^2+16b^2}},1\Big], \Big[ 1,\tfrac{4\sqrt{2}b}{-a-\sqrt{a^2+16b^2}},1 \Big]$.
But in mathematica it gives $\Big[ 1,0,-1 \Big], \Big[1,\tfrac{a+\sqrt{a^2+16b^2}}{{2}\sqrt{2}b},1\Big], \Big[ 1,\tfrac{a-\sqrt{a^2+16b^2}}{2\sqrt{2}b},1 \Big]$
I do not understand what is wrong with the calculation
| You have that $\frac{4\sqrt2 b}{-a+\sqrt{a^2+16b^2}}=\frac{a+\sqrt{a^2+16b^2}}{2\sqrt2 b}$ and $\frac{4\sqrt2 b}{-a-\sqrt{a^2+16b^2}}=\frac{a-\sqrt{a^2+16b^2}}{2\sqrt2 b}$. So your answer and the output from Mathematica are the same.
| {
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Expand the function $f(z)=\frac{1}{(z-a)(z-b)}$ where $0 < |a| < |b|$ in a Laurent series in different annuli I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)$ and $(c)$, which I'll share below; however, I was not sure what to do about the $0<|z|$ part in (a). I'm kind of teaching these to myself, and the only examples I've seen with a $0$ in them have involved something like $0< |z+2| <|a|$, or something like that where you'd have to make a substitution $w$... Please let me know how to take care of examples like (a).
Also, here are my attempts at parts (b) and (c). Please let me know if they're correct, and if not, let me know what I need to do in order to fix them:
(b) $\mathbf{|a|<|z|<|b|}$. Using partial fractions, I wrote $\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left( \frac{1}{z-a}\right) - \frac{1}{a-b}\left( \frac{1}{z-b}\right)$.
Now, if $|z|> |a|$, then $\left\vert \frac{a}{z}\right\vert< 1$, so we write $\frac{1}{(a-b)(z-a)} = \frac{1}{(a-b)z(1-a/z)} = \frac{1}{a-b}\cdot \frac{1}{z} \cdot \sum_{n=0}^{\infty} \left( \frac{a}{z}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|z|<|b|$, then $\left\vert \frac{z}{b}\right\vert < 1$, so we write $\displaystyle \frac{-1}{(a-b)}\cdot \frac{1}{(z-b)} = \frac{-1}{(a-b)}\cdot \frac{1}{b\left( \frac{z}{b} - 1 \right)} = \frac{1}{b(a-b)}\frac{1}{1-\frac{z}{b}} = \frac{1}{b(a-b)}\sum_{n=0}^{\infty}\left( \frac{z}{b}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}$
So, the whole expansion is $\displaystyle \frac{1}{z-b} \left( \sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}} + \sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}\right)\\ \displaystyle = \frac{1}{a-b} \left( \frac{1}{z} + \frac{a}{z^{2}} + \frac{a^{2}}{z^{3}} + \cdots + \frac{1}{b} + \frac{z}{b^{2}}+\frac{z^{2}}{b^{3}}+\cdots\right) \\ \displaystyle = \frac{1}{z(a-b)} + \frac{a}{z^{2}(a-b)} + \frac{a^{2}}{z^{3}(a-b)} + \cdots + \frac{1}{b(a-b)} + \frac{z}{b^{2}(a-b)}+\frac{z^{2}}{b^{3}(a-b)}+\cdots$
(c) $\mathbf{|b|<|z|}$. If $|a|<|z|$, we have, as in part (b) $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-a)} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|b|<|z|$, $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-b)} = \frac{1}{(a-b)} \frac{1}{z \left( 1 - \frac{b}{z}\right)} \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z} \sum_{n=0}^{\infty} \left( \frac{b}{z}\right)^{n} \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z}\left( 1 + \frac{b}{z} + \frac{b^{2}}{z^{2}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{1}{z} + \frac{b}{z^{2}} + \frac{b^{2}}{z^{3}}+ \cdots\right)$
Then, the required Laurent expansion valid for both $|z|>|a|$ and $|z|>|b|$ is found by subtraction:
$ \displaystyle \frac{1}{(a-b)} \left( \frac{1}{z} - \frac{1}{z} + \frac{a}{z^{2}} - \frac{b}{z^{2}} + \frac{a^{2}}{z^{3}} - \frac{b^{2}}{z^{3}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{a-b}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}} + \frac{a^{3}-b^{3}}{z^{4}} + \cdots \right) \\ \displaystyle = \left( \frac{1}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}(a-b)} + \frac{a^{3}-b^{3}}{z^{4}(a-b)} + \cdots \right)$
Again, please let me know how to do part (a), as well as if my parts (b) and (c) are ok. Thanks.
| Hint:
For $\left|z\right|\lt\left|c\right|$,
$$
\frac1{z-c}=-\frac1c\left(1+\frac zc+\frac{z^2}{c^2}+\frac{z^3}{c^3}+\dots\right)
$$
For $\left|z\right|\gt\left|c\right|$,
$$
\frac1{z-c}=\frac1z\left(1+\frac cz+\frac{c^2}{z^2}+\frac{c^3}{z^3}+\dots\right)
$$
Apply these to
$$
\begin{align}
\frac1{(z-a)(z-b)}
&=\frac1{b-a}\left(\frac1{z-b}-\frac1{z-a}\right)\\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $x^3-6x-6=0$
Evaluate the roots of
$$x^3-6x-6=0$$
I solved it using Cardano's method, but I'm looking for other elementary approaches through substitutions and properties of polynomials.
Thanks.
| Hint:
$x^3-6x-6=x^3-3bcx+b^3+c^3$
$$bc=-2, b^3+c^3=-6$$
$$b=-\sqrt[3]{2}, c=-\sqrt[3]{4}$$
So $(b+c)^3=b^3+c^3+3bc(b+c)$, then number $b+c$ - solution of $x^3-3bcx+b^3+c^3=0$
| {
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Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ I am being asked to find the sum of the following convergent series :
$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$
Attempting to generalize from partial sums yields nothing of interest:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43}{36}$
$s_3 = \frac{43}{36} + \frac{35}{216} = \frac{293}{216}$
$s_4 = \frac{293}{216} + \frac{97}{1296} = \frac{1855}{1296} $
I do not see a pattern here...
How must I proceed to find the sum of this series?
| $\frac{2^n}{6^n}=(\frac{1}{3})^n$ and $\frac{3^n}{6^n}=(\frac{1}{2})^n$
Now, use the well known result that for $|x|<1$,
$$\sum \limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
| Hint: we have: $\dfrac{1+a\cos x}{x^2} - \dfrac{b}{\sin^2 x} \to 1$, and rewrite $\dfrac{b}{\sin^2 x} = \dfrac{b}{x^2}\cdot \dfrac{x^2}{\sin^2 x}$ then the limit on the left equals $\dfrac{1+a\cos x - b}{x^2} = 1$, this means you can assume L'hopitale rule meaning: $1 + a - b = 0$, and differentiate both numerator and denominator to get another equation and solve for $a, b$.
| {
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Prove that $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$ Prove that
$$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$$
I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}=\tan^{-1}\frac{1}{2}$ and so on but cannot observe any pattern. Could someone suggest something?
| Hint:
$$\arctan\frac {1}{1+k+k^2}=\arctan\frac{(k+1)-k}{1+(k+1)k}=\arctan(k+1)-\arctan k$$
| {
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Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this is the part I don't understand
$\rightarrow \frac{6}{32}$
How did the numerator simplify to $6n^3$?
| You have a little mistake with the binomial coefficient:
$$\frac{n^4}{\binom{4n}4}=\frac{4!n^4(4n-4)!}{(4n)!}=\frac{24n^4}{4n(4n-1)(4n-2)(4n-3)}\xrightarrow[n\to\infty]{}\frac{24}{256}=\frac3{32}$$
| {
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Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{2x}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{\frac1x \cdot 2x}{\sqrt{\frac{x^2}{x^2} + \frac{3x}{x^2}} + \sqrt{\frac{x^2}{x^2} + \frac{x}{x^2}}} = \\
&= \lim_{x \to -\infty} \frac{2}{\sqrt{1 + \frac3x} + \sqrt{1 + \frac1x}} = 1
\end{align}$$
Wolfram result
| Well very obviously Wolfram's answer is incoherent, since the constant term in the expansion at infinity and the limit should be equal. Your reasoning, however, fails on the last line, since $x\sqrt{a}$ does not equal $\sqrt{x^{2}a}$ as $x$ goes to minus infinity, but $-\sqrt{x^{2}a}$ instead.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another smarties of a different color?
Round your answer to the nearest integer.
Answer given is $67$. I don't get it. Is it not:
$$\left(\frac{16}{40} \times \frac{15}{39} \times\frac{24}{38}\right) + \left(\frac{14}{40} \times\frac{13}{39} \times\frac{26}{38}\right) +\left(\frac{10}{40} \times\frac{9}{39} \times\frac{30}{38}\right)= 22?$$
| We imagine taking out the candies one at a time.
Your $\frac{16}{40}\cdot\frac{15}{39}\cdot \frac{24}{38}$ calculates the provability of getting Yellow, Yellow, Other in that order. However, two Yellow and one Other can happen in two additional orders, Yellow, Other, Yellow or Other, Yellow, Yellow. Each of these turns out to have the same probability as Yellow, Yellow, Other.
So the probability of $2$ Yellow and $1$ Other is $3\cdot \frac{16}{40}\cdot\frac{15}{39}\cdot \frac{24}{38}$.
Similar adjustments need to be made in your other two terms.
Another way: There are $\binom{40}{3}$ equally likely ways to choose $3$ candies. We now count the favourables, where we have $2$ candies of one colour and $1$ of another. For example the number of ways to have $2$ Yellow and $1$ other is $\binom{16}{2}\binom{24}{1}$, and we have similar expressions for the other favourables. Add up, and divide by $\binom{40}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines
Suppose we have some segment $AB$ of constant length that slides in such a way that its endpoints are moving along orthogonal lines. Let $P$ be a point in the segment so that $|AP| = a$ and $|PB| = b$. How can we find the curve along which $P$ moves?
I was trying to write lines $ax + by = c$ since this is orthogonal to the other line, we know normal to the orthogonal one should be $N = (-b,-a)$ and so the equation of the other line is of the form $-b x - a y + d = 0$. Now, point $A$ lie in one of them and $B$ lies on the other one line. should I solve the equation $d(P,A) + d(P,B) = a+ b $? Am I on the right track?
| WLOG let the orthogonal lines be the $x$-axis and $y$-axis respectively.
Let $P=(x,y), A=(0,y+k), B=(x+h,0), AP=a, PB=b$.
By Pythagoras' theorem,
$$\begin{align}
(y+k)^2+(x+h)^2&=(a+b)^2\\
(y+\sqrt{a^2-x^2})^2+(x+\sqrt{b^2-y^2})^2&=(a+b)^2\\
y\sqrt{a^2-x^2}+x\sqrt{b^2-y^2}&=ab\\
y^2(a^2-x^2)&=a^2b^2-2abx\sqrt{b^2-y^2}+x^2(b^2-y^2)\\
2abx\sqrt{b^2-y^2}&=b^2x^2-a^2y^2+a^2b^2\\
4a^2b^2x^2(b^2-y^2)&=b^4x^4+a^4y^4+a^4b^4+2(-a^4b^2x^2y^2+a^2b^4x^2-a^4b^2y^2)\\
2a^2b^2x^2(b^2-y^2)&=b^4x^4+a^4y^4+a^4b^4-2a^4b^2y^2\\
2a^2b^2(b^2x^2+a^2y^2-x^2y^2)&=b^4x^4+a^4y^4+a^4b^4\\
(a^2y^2+b^2x^2-a^2b^2)^2&=0\\
a^2y^2+b^2x^2&=a^2b^2\\
\frac {x^2}{a^2}+\frac{y^2}{b^2}&=1
\end{align}$$
which is an ellipse.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that
$$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$
Could you give me some hints on this?
I thought that Jensen's inequality might be of use in this exercise, but I haven't managed to solve this on my own.
| Another way:
By AM-GM we obtain:
$$\sum_{cyc}a\sqrt{b+c}=\sqrt{\sum_{cyc}(a^2b+a^2c+2ab\sqrt{(b+c)(a+c)}}\leq$$
$$\leq\sqrt{\sum_{cyc}(a^2b+a^2c+ab(b+c+a+c)}=\sqrt{2(a+b+c)(ab+ac+bc)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the sum of this series: $1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 +\cdots$? Say I have a series like the following;
$$1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 + \frac{1 \times 6 \times 11}{5 \times 10 \times 15}x^3 + \cdots.$$
How do I find the sum of this?
I'm trying to find an expression for the nth term but struggling!
| The term for $x^n$ seems to be :
$$\frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the limit $\lim\limits_{n\rightarrow \infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n})$ $$\displaystyle\lim_{n\rightarrow \infty} \left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n}\right)$$
I am not able to find any technique to proceed. It might be simple but I am not able to understand. If the hint and the concept used is explained it will be very helpful.
Thanks in advance
| To complement Ákos' answer:
$\begin{align}&\sum_{n=2}^\infty\frac 1 {n^3-n}=\sum_{n=2}^\infty \frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\\
&=\frac 1 2 \left[ \left(\frac 1 2 +\frac 1 6 + \frac 1 {12} + \cdots \right)-\left(\frac 1 6 +\frac 1 {12} + \cdots \right)\right]\\
&=\frac 1 2 \times \frac 1 2=\frac 1 4
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding minimum distance between a circle and curve what is the minimum distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
in Question there is a circle and a curve and we have to find the least distance between them
| Hint: These are the two ellipses:
Solution:
The red ellipse seems to be a rotated ellipse, centered around the origin. The minimal distance then is the length of the major semi-axis.
So we try to find the rotation:
We can write the second the second equation as
$u^t Q u = 1$ with $u = (x, y)^t$ and
$$
Q =
\begin{pmatrix}
2 & 3 \\
3 & 10
\end{pmatrix}
$$
The symmetric matrix $Q$ has the characteristic polynomial
$$
\chi(\lambda) = (2-\lambda)(10-\lambda) - 9
= \lambda^2 - 12 \lambda + 11
= (\lambda - 6)^2 - 25
$$
with roots $\lambda = 6 \pm 5$.
For the eigenvalue $\lambda = 1$ we solve $Q_1 u = 0$ for
$$
Q_1 =
\begin{pmatrix}
2-1 & 3 \\
3 & 10-1
\end{pmatrix}
=
\begin{pmatrix}
1 & 3 \\
3 & 9
\end{pmatrix}
$$
and get $u = \alpha (3,-1)^t$. We pick the normed vector $u_1 = (1/\sqrt{10}) (3,-1)^t$ as eigenvector.
For the eigenvalue $\lambda = 11$ we solve $Q_2 u = 0$ for
$$
Q_2 =
\begin{pmatrix}
2-11 & 3 \\
3 & 10-11
\end{pmatrix}
=
\begin{pmatrix}
-9 & 3 \\
3 & -1
\end{pmatrix}
$$
and get $u = \alpha (1,3)^t$. We pick the normed vector $u_2 = (1/\sqrt{10}) (1,3)^t$ as eigenvector.
Transforming to the base of eigenvectors via
$$
T
= (u_1, u_2)
=
\frac{1}{\sqrt{10}}
\begin{pmatrix}
3 & 1 \\
-1 & 3
\end{pmatrix}
$$
we get
$$
\DeclareMathOperator{diag}{diag}
T^{-1} Q T = T^{-1} (1 \cdot u_1, 11 \cdot u_2) = 1 \cdot e_1 + 11 e_2 = \diag(1, 11) \iff \\
Q = T \diag(1, 11) T^{-1}
$$
It turns out $T^{-1} = T^t$, so $T$ is orthogonal, and $\det(T) = 1$, so $T$ is the looked for rotation. This was expected because $Q$ is a symmetric ($Q = Q^t$) real valued matrix.
In the transformed system, $u' = T u$, we have the equation
\begin{align}
1
&= (u')^t Q u' \\
&= (T u)^t T \diag(1, 11) T^{-1} T u \\
&= u^t \diag(1, 11) u \\
&= x^2 + 11 y^2 \\
&= \left( \frac{x}{1} \right)^2 + \left( \frac{y}{1/\sqrt{11}} \right)^2
\end{align}
from which we infer the ellipse parameters $a = 1$ and $b=1/\sqrt{11}$.
So the inner (red) ellipse has a circumscribed circle with $r = 1$, the sought distance is $d = 2$.
About the above image: The tiny arrows are the images of $e_1$ and $e_2$ under $T$, thus $u_1$ and $u_2$. The blue ellipse is the transformed ellipse $x^2 + 11 y^2 = 1$. The yellow circle is the unit circle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in \mathbb{Z}$ such that $x^2 + y^2 = z^2$, and $x$ and $y$ are odd. Let $x = 2m + 1$ and $y = 2n + 1$. Hence, $x^2 + y^2$ = $(2m + 1)^2 + (2n + 1)^2$
$$= 4m^2 + 4m + 1 + 4n^2 + 4n + 1$$
$$= 4(m^2 + n^2) + 4(m + n) + 2$$
$$= 2[2(m^2 + n^2) + 2(m + n) + 1]$$
Since $2(m^2 + n^2) + 2(m + n) + 1$ is odd it shows that the sum of the squares of two odd integers cannot be the square of an integer.
This is what I have so far but I think it needs some work.
| Here's a quick method, not unrelated to your approach or to the other answers here.
The squares mod $4$ are $0$ and $1$ (can be verified easily by checking all four). Odd numbers are congruent to $1$ or $3$ mod $4$ and these each have square congruent to $1$ mod $4$. Hence the sum of two odd squares is congruent to $2$ mod $4$ which isn't a square.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to show $\frac{19}{7}How can I show $\dfrac{19}{7}<e$ without using a calculator and without knowing any digits of $e$?
Using a calculator, it is easy to see that $\frac{19}{7}=2.7142857...$ and $e=2.71828...$
However, how could this be shown in a testing environment where one does not have access to a calculator?
My only thought is to use the Taylor series for $e^x$ with $x=1$ to calculate $\displaystyle e\approx\sum \limits_{n=0}^{7}\frac{1}{n!}=\frac{685}{252}=2.7182...$
However, this method seems very time consuming and tedious, finding common denominators and performing long division. Does there exist a quicker, more elegant way?
| Since you know what the result should be, try remove terms one by one by computing residues $r_i$. That will reduce the size of integers, as long as yoou factorize while you can, which is easy because of the factorials (highly composite numbers).
First, notice that the first $3$ terms for $e$ give $\frac{5}{2}$.
Now, find residues:
$$r_1 = \frac{19}{7} - \frac{5}{2} = \frac{3}{2.7}\,, $$
$$r_2 = \frac{3}{2.7} -\frac{1}{6} = \frac{1}{2}\left(\frac{3}{7}-\frac{1}{3}\right) =\frac{1}{3.7}\,,$$
$$r_3 =\frac{1}{3.7} -\frac{1}{3.8} = \frac{1}{3}\frac{1}{7.8}\,,$$
$$r_4 =\frac{1}{3.7.8} -\frac{1}{120} = \frac{1}{3.8}\left(\frac{1}{7}-\frac{1}{5}\right)\,.$$
Since $5<7$, the last residue $r_4$ is negative, you can stop here, never having to do long multiplications, the hardest being $19\times 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Student is to answer 7 out of 10 questions in an examination
Student is to answer 7 out of 10 questions in
an examination. How
many if she must answer at least 3 of the first 5
questions?
Credit: A FIRST COURSE IN PROBABILITY - Sheldon Ross
University of Southern California
The answer is:
$$\binom{5}{3}\binom{5}{4}+\binom{5}{4}\binom{5}{3}+\binom{5}{5}\binom{5}{2}=110$$
Why my idea is not good?
$$\binom{5}{3}\binom{7}{4}=350$$
I take 3 of 5 of the first questions, and then I have another 7 questions that I can answer and I choose 4. Thank you!
| You are counting those cases in which she answers four or five of the first five questions more than once. You are counting them both among the $\binom{5}{3}$ selections of three of the first five questions and among the $\binom{7}{4}$ selections of four additional questions.
Suppose she answers four of the first five questions. You have counted that case four times, once for each of the $\binom{4}{3}$ ways you could have selected three of those four questions as a member of the three questions that you selected from the first five.
Similarly, if she answers all five of the first five questions, you have counted that case $10$ times, once for each of the $\binom{5}{3}$ ways you could have selected three of those five questions as a member of the three questions that you selected from the first five.
Note that
$$\binom{3}{3}\binom{5}{3}\binom{5}{4} + \binom{4}{3}\binom{5}{4}\binom{5}{3} + \binom{5}{3}\binom{5}{5}\binom{5}{2} = 350 = \binom{5}{3}\binom{7}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Deriving formula for asymptotes of a hyperbola I'm trying to find a precalculus-level derivation of the formula for the asymptotes of a hyperbola. My book says:
Solving $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
for $y$, we obtain
$y = \pm \frac ba \sqrt{x^2 - a^2}$
$ = \pm \frac ba \sqrt{x^2(1 - \frac {a^2}{x^2})}$
$ = \pm \frac bax \sqrt{(1 - \frac {a^2}{x^2})}$
then goes on to say $\frac {a^2}{x^2}$ approaches 0, and therefore the asymptotes are at $y = \pm \frac ba x$
However, in my attempt to derive the formula, I have been unable to get to the first equation. I got as far as
$y^2 = -b^2(1 - \frac {x^2}{a^2})$
| $$y^2=-b^2\left(1-\frac{x^2}{a^2}\right)$$
$$y^2=\frac{b^2}{a^2}(x^2-a^2)\tag{factor $1/a^2$ out}$$
$$y=\pm\sqrt{\frac{b^2}{a^2}(x^2-a^2)}$$
$$y=\pm\frac{b}{a}\sqrt{x^2-a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean triple (3,4,5)
It is true that this continued fraction is only valid for Pythagoraean Triple only?
I try other numbers and it seem that the only numbers that are valid are the Pythagoraean Triples. Can anyone verify this? Or show some examples where is also work for other numbers.
| Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then
\begin{align*}
\sqrt{2}+\frac{b}{x} &= x \\
x\sqrt{2}+b &= x^{2} \\
x^{2}-\sqrt{2} \, x-b &= 0 \\
x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\
&= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\
&= \sqrt{b+1+\sqrt{2b+1}} \\
\end{align*}
Take $a=\sqrt{2b+1}$ and $c=b+1$, it follows that $a^{2}+b^{2}=c^{2}$.
Note that the choices of $a$ and $c$ still have one degree of freedom if $(a,b,c)$ is not restricted to be a Pythagorean triple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Relation between $I_{k}$ and $I_{k+2}$
If $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,n\in \mathbb{N}$ Then Relation between $I_{k}$ and $I_{k+2}$
$\bf{My\; Try::}$ Given $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,$ Then $\displaystyle I_{k+2}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^{n+2}dx\;,$
So $$I_{k+2}-I_{k} = \int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^n\cdot \sin 2xdx$$
Now How can I solve it after that, Help me
Thanks
|
Note:
$$\int_0^\frac{\pi}{2} x(\sin x+\cos x)^n dx=\frac{\pi}{4}\int_0^\frac{\pi}{2}(\sin x+\cos x)^ndx=I_n$$
(Put $x=\frac{\pi}{2}-x$ to prove this)
We have $$I_{n+2}-I_n=\frac{\pi}{4}\int_0^\frac{\pi}{2} (\sin x+\cos x)^n \sin 2xdx$$
Applying by parts, differentiating $(\sin x+\cos x)^n$ and integrating $\sin 2x$, this becomes:
$$I_{n+2}-I_n=\frac{\pi}{4}\left (-\frac 12 \cos 2x(\sin x+\cos x)^n|_0^{\frac{\pi}{2}}+\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx-\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx\right)$$
Now use $$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx=\frac{4I_n}{\pi}$$
and
$$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx=\frac{4(I_{n+2}-I_n)}{\pi}$$
to get:
$$\color{green} {I_{n+2}=I_n+\frac{\pi}{2(n+2)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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