Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Numbers divisible by $11$ A number is divisible by $11$, when the difference between the sum of the digits in the odd positions counting from the left (the first, third, ....) and the sum of the remaining digits is either 0 or divisible by 11.
Why is that?
| \begin{align}
abcde_{10}
&= 10000a + 1000b + 100c + 10d + e \\
&= (9999a + a) + (1001b - b) + (99c + c) + (11d - d) + e \\
&= (9999a + 1001b + 99c + 11d) + (a-b+c-d+e) \\
&= 11(101a + 91b + 11c + d) + (a-b+c-d+e) \\
&\equiv a-b+c-d+e \pmod{11}
\end{align}
Another way to look at this problem is to note that
$10^{2n} \equiv (-1)^{2n} \equiv 1 \pmod{11}$
So $10^{2n}(10m + n) \equiv 1(-m + n) \equiv -m + n \pmod{11}$
So
\begin{align}
abcdef_{10}
&\equiv 10^4(10a + b) + 10^2(10c + d) + 10^0(10e + f) \\
&\equiv (-a+b) + (-c + d) + (-e + f) \\
&\equiv -a+b-c+d-e+f
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a-c = 9$ then find the value of $b-d$. If $a,b,c,d$ are positive integers such that $\log_a b=\frac{3}{2}$ and $\log_c d=\frac{5}{4}$, if $a-c = 9$ then find the value of $b-d$.
We get
$b=a^{3/2}$ and $d=c^{5/4}$
Hence
$b-d=a^{3/2}-c^{5/4}$
Now we need to use the fact that $a,b,c,d$ are integers, but is hit and trial the only way to go here?
| For $b$ and $d$ to be integers, we need some integers $p$ and $q$ such that $p^2=a$ and $q^4=c$.
So, $p^2-q^4=9$, and we want to find $p^3-q^5$.
$(p+q^2)(p-q^2)=9$. As $p$ and $q$ are integers, so must $(p+q^2)$ and $(p-q^2)$. Considering their product is 9, we have either: $p+q^2=9$ and $p-q^2=1$, or both are equal to $3$. The latter event locks $q=0$, meaning $c=0$, which means that $log_c d$ is undefined, so it must be the former.
$(p+q^2)-(p-q^2)=9-1$ reduces to $2q^2=8$, meaning $q=2$. From there, it is easy to see that $p=5$.
$b=p^3=125$; $d=q^5=32$. $b-d=125-32=93$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Summation of factorial. $$2(\frac{1}{3!\times7!}+\frac{1}{1\times9!})+\frac{1}{5!\times5!}=\frac{2^a}{b!}$$ find $a,b$ by some predictions I see $b=10$ but what about numerator. I think we have to $\sum {N\choose r}=2^N$ but how?. Thanks.
| Firstly write it as follows:
$$\frac{1}{1!\times9!}+\frac{1}{3!\times7!}+\frac{1}{5!\times5!}+\frac{1}{7!\times3!}+\frac{1}{9!\times1!}$$
$$=\sum_{n=0}^4\frac{1}{(2n+1)!(10-(2n+1))!}$$
$$=\frac{1}{2}\sum_{n=0}^{10}\frac{1}{n!(10-n)!}$$
$$=\frac{1}{2}\times\frac{2^{10}}{10!}$$
$$=\frac{2^9}{10!}$$
Hence $a=9$, $b=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does one find any polynomials common roots I was wondering if there was a way to find the common roots of 2 polynomials.
For example let $P_1(x) = x^5+x^3+2x^4-5x^2-7$ and $P_2(x) = 2x^7 +3x^3+4x^6+6x^2-14x^4-21$
Is there an algorithm or a method to find $P_1$ and $P_2$'s common roots?
| The general way to do this is to take the GCD of the two polynomials. In particular, one can apply the Euclidean algorithm to this, just like taking the GCD of two integers. In particular, the computation goes as follows: We start with
$$\gcd(x^5+2x^4+x^3-5x^2-7,2x^7+4x^6-14x^4+3x^3+6x^2-21)$$
Then, we take the second polynomial mod the first. That is, we subtract multiples of $P_1$ from $P_2$ until we get a polynomial of degree 4. In particular, $P_2-(2x^2-2)P_1=5x^3+10x^2-35$, so we reduce the above to
$$\gcd(5x^3+10x^2-35,x^5+2x^4+3x^3-5x^2-7)$$
Then, we take the polynomial remainder of the second polynomial mod the fist again, which results in $0$ since
$$x^5+2x^4+3x^3-5x^2-7-(-1/5x^2-1/5)(5x^3+10x^2-35) = 0$$
thus, we can reduce to
$$\gcd(0,5x^3+10x^2-35)$$
meaning that $5x^3+10x^2-35$ is the GCD of the two polynomials. This means that $x$ is a root of both $P_1$ and $P_2$ if and only if it is a root of $5x^3+10x^2-35$ or, equivalently, of $x^3+2x^2-7$.
This essentially works on the principle that if a polynomial has roots $r_1,\ldots,r_n$, then $(x-r_1)(x-r_2)\ldots (x-r_n)$ divides it. Thus, the product of all the shared roots must divide both polynomials and, in particular, must divide their GCD as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve this indefinite integral: $\int \frac{x+2}{x^2-1}dx$ I need to solve this indefinite integral: $$\int \frac{x+2}{x^2-1}dx$$
but the result is wrong.
Here my steps:
$D° > N° $ so I need to reduce the Denominator, which is a difference of two square: $(x^2-1) = (x-1)(x+1)$ so my new integral is:
$$ \int \frac{x+2}{x^2-1}dx = \frac{A}{x-1} + \frac{B}{x+1} = \frac{A(x+1)+B(x-1)}{(x-1)(x+1)}$$
$$
\begin{cases}
x = Ax+Bx \\
2 = A-B
\end{cases}
$$
and I get:
\begin{cases}
A= \frac{1}{2}\\
B = -\frac{1}{2}
\end{cases}
So:
$$\int \frac{x+2}{x^2-1}dx = \int \frac{1}{2(x-1)}dx -\int \frac{1}{2(x+1)}dx$$
I need to get the derivate of the denominator
$$\int \frac{x+2}{x^2-1}dx = \frac{1}{2}\int \frac{2}{2(x-1)}dx - \frac{1}{2}\int \frac{2}{2(x+1)}dx$$
So I get:
$$\frac{1}{2}ln|2x-2| - \frac{1}{2}ln|2x+2| + C$$
but the result is wrong: http://www.wolframalpha.com/input/?i=%5Cint+%5Cfrac%7Bx%2B2%7D%7Bx%5E2-1%7Ddx
| You made a mistake with your partial fractions:
$A$ should be $\frac{3}{2}$
$$ \frac{x+2}{x^2 -1} = \frac{x+2}{(x -1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$$
Eliminate A and B in turns by making $x+1=0$ or $x-1=0$
$$ x+2 \equiv A(x+1) + B(x-1)$$
let $x=1$
$$ A = \frac{3}{2}$$
let $x= -1$
$$ B = -\frac{1}{2}$$
$$ \frac{x+2}{x^2 -1} = \frac{3}{2(x-1)} - \frac{1}{2(x+1)}$$
You can then integrate the above which is very straight forward
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
I am trying to prove that, ${\sin{A}\cos{A}-\sin{B}\cos{B}\over \sin^2{A}-\sin^2{B}}=\tan(90^o-A-B)$ I am trying to prove that,
$${\sin{A}\cos{A}-\sin{B}\cos{B}\over \sin^2{A}-\sin^2{B}}=\tan(90^o-A-B)$$
Using: $\sin(2A)=2\sin{A}\cos{A}$, then we have
$${1\over 2}{\sin(2A)-\sin(2B)\over \sin^2{A}-\sin^2{B}}=\tan(90^0-A-B)$$
$${1\over 2}{\sin(2A)-\sin(2B)\over (\sin{A}-\sin{B})(\sin{A}+\sin{B})}=\tan(90^0-A-B)$$
I am out of ideas, I need some help, anyone's?
| $$
\frac{\sin A \cos A - \sin B \cos B}{\sin^2 A - \sin^2 B} = \frac{\sin (x-y) \cos (x+y)}{\sin (x-y) \sin(x+y)} =
$$
$$
= \frac{\cos (x+y) }{\sin (x+y)} = \cot (x+y) = \tan (90 - x-y)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
| Use Cauchy-Schwarz inequality :
\begin{align}|a\sin x+b \cos x| = |(a,b)\cdot (\sin x,\cos x)|
&\leq \sqrt{a^2+b^2}\sqrt{\sin^2 x+\cos^2 x}=\sqrt{a^2+b^2}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How to solve without solving by inspection? $\frac{x+5}{x+k}=\frac{-kx+5}{x-1}$ Background: This is from a test review on functions. The original problem was
Find the value of $k$ so that the function $f(x) = \frac{x+5}{x+k}$ will be its own inverse.
I found the answer by inspection, and then tried to solve it through more rigorous means. Continuing from the question title, I cross multiplied and used the quadratic formula on $k.$
$x^2+4x-5=5x+5k-kx^2-k^2x$
$k^2x+k(x^2-5)+(x^2-x-5)=0$
$k = \frac{-x+5\pm\sqrt{x^4-10x^2+25-4x(x^2-x-5)}}{2x}$
$k = \frac{-x+5\pm\sqrt{x^4-4x^3-6x^2+20x+25}}{2x}$
This is where I got stuck. Any ideas?
| Start by swapping the variables:
$$y=f(x)=\frac{x+5}{x+k}\implies x=\frac{y+5}{y+k}$$
and solving for $y$ from here:
$$\begin{align*}
x&=\frac{y+5}{y+k}\\[1ex]
xy+kx&=y+5\\[1ex]
xy-y&=5-kx\\[1ex]
y(x-1)&=5-kx\\[1ex]
y&=\frac{5-kx}{x-1}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$
All real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$
$\bf{My\; Try::}$ Here $\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x>0$
Now Let $f(x)=\sqrt{1+x}\;,$ Then equation convert into $f(f(f(x)))=x$
Now Here $f(x)=x$ be one function which satisfy above equation.
My question is how can we calculate other function which satisfy above functional equation.
Help required, Thanks
| Let $1+x=y^2$, where $y>0$ and $1+y=z^2$, where $z>0$.
Hence, $1+z=x^2$, where $x>0$.
Thus, $x-y=(y-z)(y+z)$ and $y-z=(z-x)(z+x)$.
*
*Let $x>y$. Hence, $y>z$, which says that $z>x$. It's contradiction.
*Let $x<y$. Hence, $y<z$, which says that $z<x$. It's contradiction again.
Id est, $x=y=z$ and $x=\frac{1+\sqrt5}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.
| $$1^2+2^2+\cdot\cdot\cdot+n^2=\dfrac{n(n+1)(2n+1)}{6}$$
This leads to the following:
$$1^2+\cdot\cdot\cdot+n^2+\cdot\cdot\cdot+(2n+1)^2=\dfrac{(2n+1)(2n+1+1)(2(2n+1)+1)}{6}$$
$$2^2+4^2+\cdot\cdot\cdot+(2n)^2=2^2(1^2+2^2+\cdot\cdot\cdot+n^2)=4\cdot\dfrac{n(n+1)(2n+1)}{6}$$
Thus,
$$1^2+3^2+\cdot\cdot\cdot+(2n+1)^2=\dfrac{(2n+1)(2n+1+1)(2(2n+1)+1)}{6}-\quad4\cdot\dfrac{n(n+1)(2n+1)}{6}=\dfrac{(n+1)(2n+1)(2n+3)}{3}=\sum_{i=0}^n (2i+1)^2$$
Note: You have an extra $n$ in your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the values of $x$ such that $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$.
Find the values of $x$ such that
$$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$.
Checking for $x\in [-1,1]$
In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ comes out to be $2\tan^{-1}x$ hence the taken function comes out to be equal to $4\tan^{-1}x$ hence the function is clearly dependent on $x$.
Now checking for $x\in (1,\infty)$
In the taken domain $2\tan^{-1}x$ comes out to be $\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes independent of $x$.
Now checking for $x\in (-\infty,-1)$
In the taken domain $2\tan^{-1}x$ comes out to be $-\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes $-\pi$ therefore becomes, independent of $x$.
But the answer has been mentioned as just $x\in [1,\infty)$ Can anybody tell me why the second set has not been included.
| Let $\tan^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2\iff-\pi\le2y\le\pi$
$$\implies\dfrac{2x}{1+x^2}=\sin2y$$
$$\sin^{-1}\dfrac{2x}{1+x^2}=\begin{cases}-2y-\pi &\mbox{if }2y<-\dfrac\pi2\iff x<-1 \\
2y & \mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\\ \pi-2y & \mbox{if }2y>\dfrac\pi2\end{cases}$$
Can you take it home from here?
See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove that $\int_{0}^{1}\ln{(x/(1-x))}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$
$$\int_{0}^{1}\ln{\big(\frac{x}{1-x}\big)}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$$
Put $$\frac{x}{1-x}=y$$
$$I=\int_{0}^{\infty}\ln{y}\ln{(1+3y+y^2)}\frac{dy}{y(y+1)}=\frac{8}{5}\zeta{(3)}$$
Simple integral at first sight, however I cannot prove that. I would appreciate your help.
| Since $x\left(1-x\right)<1$ if $x\in\left(0,1\right)
$ we have $$\begin{align}
\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx & =\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}\left(1-x\right)^{k}dx \\
& -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(1-x\right)x^{k-1}\left(1-x\right)^{k}dx.
\end{align}
$$ Now by definition of Beta function we have $$
\int_{0}^{1}x^{a}\left(1-x\right)^{k}dx=B\left(a+1,k+1\right)
$$ and so $$\begin{align}
\int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(x\right)dx = &\frac{\partial}{\partial a}\left(B\left(a+1,k+1\right)\right)_{a=k-1} \\
= & B\left(k,k+1\right)\left(\psi\left(k\right)-\psi\left(2k+1\right)\right)
\end{align}
$$ and in a similar way we have $$\begin{align}
\int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(1-x\right)dx= & \frac{\partial}{\partial b}\left(B\left(k,b\right)\right)_{b=k+1} \\
= & B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(2k+1\right)\right)
\end{align}
$$ then $$\begin{align}
\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx= & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(k\right)\right) \\
= & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{2}}B\left(k,k+1\right) \\ = & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}}
\end{align}
$$ and the last series has a well known closed form $$\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}}=\frac{2}{5}\zeta\left(3\right).\tag{1}$$
For the other integral note that $$I=\int_{0}^{\infty}\frac{\log\left(y\right)\log\left(1+3y+y^{2}\right)}{y\left(y+1\right)}dy\overset{y=\frac{x}{1-x}}{=}\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(\frac{1+x-x^{2}}{\left(1-x\right)^{2}}\right)}{x}dx
$$ $$=\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1+x-x^{2}\right)}{x}dx-2\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1-x\right)}{x}dx
$$ and it is sufficient to observe that $$-2\int_{0}^{1}\frac{\log\left(x\right)\log\left(1-x\right)}{x}dx=2\sum_{k\geq1}\frac{1}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}dx
$$ $$=-2\sum_{k\geq1}\frac{1}{k^{3}}=-2\zeta\left(3\right)
$$ and $$ 2\int_{0}^{1}\frac{\log^{2}\left(1-x\right)}{x}dx=2\int_{0}^{1}\frac{\log^{2}\left(x\right)}{1-x}dx=2\sum_{k\geq0}\int_{0}^{1}\log^{2}\left(x\right)x^{k}dx
$$ $$=4\sum_{k\geq1}\frac{1}{k^{3}}=4\zeta\left(3\right)
$$ so using the previous result we have $$I=\zeta\left(3\right)\left(2-\frac{2}{5}\right)=\frac{8}{5}\zeta\left(3\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim\limits_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$
$$\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$$
$$y=\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$$
$$\ln y=\ln \left(\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}} \right)$$
$$\ln y=\left(\lim_{x\to -2} \frac{\ln(3x+7)}{x^2-4}\right)$$
Applying L'hopital
$$\ln y= \left(\lim_{x\to -2} \frac{\frac{3}{3x+7}}{2x}\right)$$
$$\ln y=-\frac{3}{4}$$
$$y=e^{-\frac{3}{4}}$$
Is there a way to solve it using something like $(3x+7)^{\frac{1}{3x+7} \cdot (3x+7) \cdot\frac{1}{x^2-4}}$ ?
| Let assume $$L=\lim_{x \rightarrow -2}(3x+7)^{\frac1{x^2-4}}$$
We can see $$\ln L=\lim_{x \rightarrow -2}\frac{\ln(3x+7)}{{x^2-4}}$$
$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+3(x+2))}{{(x-2)(x+2)}}$$
If we use Taylor expansion at x+2, we get
$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+3(x+2))}{{(x-2)(x+2)}}=\frac{3(x+2))}{{(x-2)(x+2)}}-\frac{(3(x+2)))^2}{2{(x-2)(x+2)}}+O((x+2)^2) \rightarrow -\frac3{4}$$
This gives $L=e^{-\frac34}$
Or if not using Taylor expansion, let's use substitution $x=\frac13y-2$
$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+y)}{{y}}\frac1{(\frac19y-\frac43)}$$
Keep this in mind, $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}$$
The limit $L=e^{-\frac34}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the divisors of $5040$ in the Plato's dialogue "Theaetetus" In the Plato's dialogue "Theaetetus", at a certain point, we have the following "problem"
\begin{align*}
5040 &= 7! \\
&= 1\times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \\
&= 2 \times 3 \times 2 \times 2 \times 5 \times 2 \times 3 \times 7 \\
&= 2^4 \times 3^2 \times 5 \times 7 \\
&= 2^4 \times 3^2 \times 5 \times 7
\end{align*}
Then we should find the number of divisors of $5040$ as $$5 \times 3 \times 2 \times 2 - 1 = 59$$
where the numbers $5, 3, 2$ and $2$ should come apparently respectively from the exponents ($+1$) of the terms in $ 2^4 \times 3^2 \times 5 \times 7$.
My questions is
Where does this formula come from exactly?
For example, I understood we can divide $5040$ by $2$, $2^2$, $2^3$, $2^4$, $3$, $3^2$, $5$ and $7$, and by any combination of products of these numbers, but I don't understand exactly what's the reasoning that has been done to obtain that formula.
I'm very bad in facing problems related to combinations and permutations. Any advice on how to improve this defect? Any specific book or set of exercises that you advise me?
| There is a basic counting principle called the product rule, which says the following: in an experiment consisting of two steps, if the number of ways to do the first step is $n_1$, and the number of ways to do the second step is $n_2$ for each of the $n_1$ outcomes in the first step, then the total number of possible outcomes for the experiment is the product $n_1 n_2$. The product rule generalizes to experiments with more than $2$ steps. For example, the number of vehicle license plates consisting of three lower case letters followed by three digits is $26^3 \cdot 10^3$ because the first blank can be filled in 26 ways, the second blank can be filled in 26 ways regardless of how the first blank is filled, and so on.
How many divisors does $2^4 \cdot 3^2 \cdot 5 \cdot 7$ have? If $d$ is a divisor of this number, then $d$ is of the form $2^a 3^b 5^c 7^d$, where $a \in \{0,1,2,3,4\}, b \in \{0,1,2\}$ and $c, d \in \{0,1\}$. The number of ways to choose $a$ is $5$, the number of ways to choose $b$ is $3$, and the number of ways to choose $c$ and $d$ are $2$ each. By the product rule, we get that the number of divisors is $5 \cdot 3 \cdot 2 \cdot 2$. Of course, this includes the special case $a=b=c=d=0$ corresponding to the divisor $1$ and the special case $a=4, b=2, c=d=1$ corresponding to the divisor $5040$. If you are counting only the number of proper divisors, then $5040$ should not be counted and so we subtract $1$ from the product.
The product rule is discussed in Kenneth Rosen's text Discrete Mathematics and its Applications.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{x^2+y^2}{x+y}$ is a divisor of $1978$
Two nonzero integers $x,y$ (not necessarily positive) are such that $x+y$ is a divisor of $x^2+y^2$, and the quotient $\dfrac{x^2+y^2}{x+y}$ is a divisor of $1978$. Prove that $x = y$.
Let $A = \dfrac{x^2+y^2}{x+y}$ where $A$ is an integer divisor of $1978$. Then $A \mid x^2+y^2$ and if $p$ is a prime divisor of $A$ we must have $x^2+y^2 \equiv 0 \pmod{p}$.
We first deal with the case $x,y \equiv 0 \pmod{q}$. Otherwise $x,y \not \equiv 0 \pmod{q}$ where $q \equiv 3 \pmod{4}$ and is a prime divisor of $A$.
I didn't find an easy way of dealing with the case that $x,y \equiv 0 \pmod{q}$. Is there an easier way of solving this?
| For the equation.
$$A=\frac{x^2+y^2}{x+y}$$
It may be better to write a decision?
$$x=(p-s)(k-s)$$
$$y=(p-s)(p-k)$$
$$A=p^2+s^2+2k(k-p-s)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Positive integers $a$ and $b$ are such that $a+b=a/b + b/a$. Find $a^2+ b^2$. This is PRE RMO 2015 question. Positive integers $a$ and $b$ are such that $$a+b=a/b + b/a$$ Find $$a^2+ b^2$$
My try:-
Given that
$$a+b=a/b + b/a$$On simplification we get
$$a^2 b+ b^2 a= a^2 + b^2$$
But in my book the given answer is 2. Please tell me how it is possible.
| This is not that elegant but provides any way of treating this question.
WLOG, we can assume a is larger than b by k (i.e. a = b + k; where k is an integer).
Positive integer $= a + b = \dfrac {a}{b} + \dfrac {b}{a} = \dfrac { b + k }{ b } + \dfrac { b }{ b + k } = 1 + \dfrac { k }{ b } + \dfrac { b }{ b + k }= 1 + \dfrac {bk + k^2 + b^2}{b(b + k)}$
Note that the last fraction must be reduced to an integer but the numerator cannot be further factorized and hence there is no way to cancel the factors in the denominator. The only way that cancelling can be done is when $k = 0$ and this leads to $a = b$.
Result follows after applying the finding to the given equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int \frac{x^{13}\ dx}{x^5 + 1}$ We get this problem from our teacher today. I only wish that it was $x^{14}$ in the numerator, so we can use substitution method:
$$\int \dfrac{x^{13}\ dx}{x^5 + 1}$$
I can't find way to integrate this. Please help me out to solve this seeming simple integral.
| E.g. with $\frac{x^{13}}{x^5+1}=\sum\limits_{k=0}^\infty (-1)^k x^{5k+13}$ you can integrate easily for $|x|<1$.
You can simplify with $x^{13}=(x^8-x^3)(x^5+1)+x^3$.
The useful link of user 170039 gives you the result for $\int\frac{x^{3}}{x^5+1}$ and shows how complicated it becomes.
If you like to use the proposal of boaz, then you need $\int\frac{ex+f}{ax^2+bx+c}=$
$$=\frac{e}{2a}\ln|ax^2+bx+c|+\frac{2af-be}{a\sqrt{4ac-b^2}}\cdot\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}+C$$
for $4ac-b^2>0$ and
$$=\frac{e}{2a}\ln|ax^2+bx+c|+\frac{2af-be}{2a\sqrt{b^2-4ac}}\cdot\ln\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}+C$$
for $4ac-b^2<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
For which $n$ can $(a, nb, c)$ and $(b, c, d)$ be Pythagorean triples? Fermat proved that if $(a, b, c)$ is a Pythagorean triple, then $(b, c, d)$ cannot be a Pythagorean triple.
Suppose $(a, nb, c)$ form a Pythagorean triple. Can $(b, c, d)$ be a Pythagorean triple? For which $n$ is this possible?
Another formulation is the following Diophatine system of equations:
$$a^2 + n^2b^2 = c^2$$
$$c^2 + b^2 = d^2$$
Fermat's theorem is then that $n$ cannot equal 1.
Searching with a computer for $d < 50,000$ and $n < 100$, I have found solutions for $n = 6, 16, 30, 48, 84$. For example, $n = 6$ has the solution $(a, b, c, d) = (99, 28, 195, 197)$ because $(99, 6*28, 195)$ and $(28, 195, 197)$ are both triples.
I am especially interested if there are any $n \neq 1$ that can be ruled out.
| An Attempt
You have $a^2+\left(n^2+1\right)b^2=d^2$. All integral solutions $(a,b,d)$ to this equation satisfies (1) $|a|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right)$ for some $r\in\mathbb{Q}$.
Now, if $b^2+c^2=d^2$, then (1) $|c|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{b}{d},\frac{c}{d}\right)=\left(\frac{2s}{s^2+1},\frac{s^2-1}{s^2+1}\right)$ for some $s\in\mathbb{Q}$. Thus, we have to find $r,s\in\mathbb{Q}$ such that $$\frac{r}{\left(n^2+1\right)r^2+1}=\frac{s}{s^2+1}\,.$$
Consequently, $$rs^2-\big(\left(n^2+1\right)r^2+1\big)s+r=0\,.$$
This means
$$\big(\left(n^2+1\right)r^2+1\big)^2-4r^2=q^2$$
for some $q\in\mathbb{Q}$. Hence,
$$\left(n^2+1\right)^2r^4+2\left(n^2-1\right)r^2+1=q^2\,.$$
Therefore, we have
$$\left(\left(n^2+1\right)r^2+\frac{n^2-1}{n^2+1}\right)^2+\left(\frac{2n}{n^2+1}\right)^2=q^2\,.$$
We have again run into a Pythagorean problem. Using a similar argument, the problem boils down to finding $(r,u)\in\mathbb{Q}\times\mathbb{Q}$ satisfying $u\neq 0$ and $$\frac{\left(n^2+1\right)^2r^2+\left(n^2-1\right)}{n}=\frac{u^2-1}{u}\,.$$
For example, $(n,r,u)=\left(6,\frac27,-\frac2{49}\right)$ produces $(n,a,b,c,d)=(6,99,28,195,197)$. I do not have any idea how to proceed further than this. (To be honest, I haven't reduced the number of variables at all. Being rationals, $r$ and $u$ each account for $2$ integer variables.)
Will Jagy's infinite family is produced by $$(n,r,s,q,u)=\left(2\left(t^2-1\right),\frac{t}{2t^2-1},t\left(2t^2-1\right),\frac{4t^6-4t^4+t^2-1}{\left(2t^2-1\right)^2},-\frac{2}{\left(2t^2-1\right)^2}\right)$$
for $t=2,3,\ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How to solve this inequality with absolute value: $ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $ Good morning to everyone. I have an inequality that I don't know how to solve: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $$ I tried to solve it in this way: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 \rightarrow \frac{\left|x-3\right|}{\left|x+2\right|} - 3 \le 0 \rightarrow \frac{\left|x-3\right|-3\left|x+2\right|}{\left|x+2\right|}\le 0 \rightarrow $$
Case 1:
$$ \left|x+2\right| > 0 \wedge \left|x-3\right|-3\left|x+2\right| \le 0 $$
Case 1 a) $$ x+2 >0 \rightarrow x>-2$$
Case 1 b) $$ -x-2 <0 \rightarrow x>-2$$
Case 1 c) $$ \left|x-3\right|-3\left|x+2\right| \le 0 \rightarrow \left|x-3\right| \le 3\left|x+2\right| \rightarrow x-3 \le x+2 \rightarrow $$
The solution are all real numbers
Case 1 d)$$ -x+3 < x+2 \rightarrow x>\frac{1}{2} $$
Case 1 e) $$ x-3 \le -x-2 \rightarrow x \le \frac {1}{2} $$
Case 1 f) $$ x-3 < x+2 $$
The solution: all real numbers
Case 2: $$ \left|x+2\right| \le 0 \wedge \left|x-3\right|-3\left|x+2\right| > 0 $$
It'll have the same solutions because are the same inequalities. Therefore $x$ belongs to $(\frac{1}{2}, \infty)$ But my answer sheet shows that it belongs to $(-\infty, -\frac{9}{4}) \wedge (-\frac{3}{4},\infty)$
| Simply write the homographic function in canonical form:
$$\biggl\lvert\frac{x-3}{x+2}\biggr\rvert=\biggl\lvert\frac{(x+2)-5}{x+2}\biggr\rvert=\biggl\lvert1-\frac{5}{x+2}\biggr\rvert.$$
The inequation means the distance between $1$ and $\dfrac{5}{x+2}$ is at most $3$, so it's equivalent to
$$1-3=-2\le \frac{5}{x+2}\le 1+3=4.$$
As $\dfrac{5}{x+2}$ can't be $0$, we have two cases
*
*$-2\le \dfrac{5}{x+2}<0\iff x+2\le-\dfrac52\iff x\le-\dfrac92$,
*$0<\dfrac{5}{x+2}\le 4\iff x+2\ge\dfrac54\iff x\ge-\dfrac34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Given $A+B+C=180^{\circ}$, find value of $\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C$
Given $A+B+C=180^{\circ}$, find value of $$\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C$$
I know about some basic conditional identities but don't know how to use them here.
| Observe that
$$\cos{A+B+C}=\cos{A}\cos{B}\cos{C}-\sin{A}\sin{B}\cos{C}-\sin{A}\cos{B}\sin{C}-\cos{A}\sin{B}\sin{C}\tag{1}$$
Let $k=\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C$
Divide $(1)$ by $\cos^2{A}\cos^2{B}\cos^2{C}$, to get the $-k=\cos{\pi}=-1 \Rightarrow k=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integer solutions to $x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)$
Find all pairs of integer solutions $(x,y)$:
$$x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1).$$
Let $s = x+y$ and $p = xy$. We then reduce the given equation to $2p(s-4) = s^3-8s^2-8$. Thus, $s-4 \mid s^3-4s^2-4(s^2+2) \implies s-4 \mid 4$ or $s-4 \mid s^2+2$.
What do I do from here?
| Hint: If $f(x)\in\mathbb{Z}[x]$ and $n\in\mathbb{N}$, then $(n-k)\mid f(n)$ implies $(n-k)\mid f(k)$. This shows that $s-4\mid 72$. Since $s$ must be even, $s-4$ can be one of the $18$ possible values.
Furthermore, we need $s^2-4p=(x-y)^2\geq 0$. Therefore, $$0\leq s^2-4p=s^2-4\left(\frac{s^3-8s^2-8}{2(s-4)}\right)=-\left(\frac{s^3-12s^2-16}{s-4}\right)\,.$$ This shows that $4< s<13$. There are now only $4$ cases. All solutions are $(x,y)=(2,8)$ and $(x,y)=(8,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the complete (polynomial) factorization of $\sigma(p^k)$, where $p$ is prime with $p \equiv k \equiv 1 \pmod 4$? The title says it all.
What is the complete (polynomial) factorization of $\sigma(p^k)$, where $p$ is prime with $p \equiv k \equiv 1 \pmod 4$?
Here, $\sigma = \sigma_{1}$ is the classical sum-of-divisors function.
My Attempt
$$\sigma(p^k) = 1 + p + \ldots + p^{k-1} + p^k = \left(1+p\right)\left(1 + p^2 + \ldots + p^{k-1}\right) = \sigma(p)\left(1 + p^2 + \ldots + p^{k-1}\right)$$
Note that $\sigma(p^k) = \sigma(p)$ if $k=1$.
Now, the finite geometric series
$$1 + p^2 + \ldots + p^{k-1}$$
has $n=\dfrac{k-1}{2}+1=\dfrac{k+1}{2}$ terms, and common ratio $r=p^2$. Thus, we have
$$1 + p^2 + \ldots + p^{k-1} = \dfrac{\left(p^2\right)^{(k+1)/2} - 1}{p^2 - 1} = \dfrac{p^{k+1}-1}{p^2 - 1} = \dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}.$$
Consequently, we obtain
$$\sigma(p^k) = \sigma(p)\left(\dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}\right).$$
Question
Is $\sigma(p^k)$ when given as
$$\sigma(p^k) = \sigma(p)\left(\dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}\right)$$
considered completely factored?
By complete factorization, I mean:
$$\bf{e.g.} \hspace{0.5in} x^4 - 1 = (x-1)(x+1)(x^2 + 1).$$
| If $k=1$ then we just have the irreducible polynomial $\sigma(p) = 1 + p$, and your second factor cancels out to just $1$. So we will assume that $k > 1$.
Have a look at cyclotomic polynomials: in general, the polynomial $\sigma(p^k) = 1 + p + \cdots + p^k = (p^{k+1}-1)/(p-1)$ will factor over $\mathbb Q[p]$ exactly as $\prod_{d\mid k+1, d>1} \Phi_d(p)$, with the $d=1$ term having been cancelled by the $p-1$ denominator. In the particular case that $k\equiv 1 \pmod 4$, you will always get the factor $\Phi_2(p) = 1+p = \sigma(p)$, and if $(k+1)/2$ happens to be prime then your other factors correspond roughly to the irreducible factors
$$\Phi_{k+1}(p) = 1 - p + p^2 - \cdots + p^{(k-1)/2},$$
$$\Phi_{(k+1)/2}(p) = 1 + p + p^2 + \cdots + p^{(k-1)/2}.$$
However in the typical case where $(k+1)/2$ is not prime, then the polynomial is certainly not completely factored by your attempt. The first time this happens (subject to the modular restriction on $k$) is when $k=17$:
$$\sigma(p^{17}) = \Phi_2(p) \Phi_3(p) \Phi_6(p) \Phi_9(p) \Phi_{18}(p) \\ = (p+1)(p^2+p+1)(p^2-p+1)(p^6+p^3+1)(p^6-p^3+1).$$
By the way, there seems to be no point whatsoever in specifying that $p\equiv 1\pmod 4$, since you are only treating $p$ as an indeterminate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Factorial Proof by Induction Question: $ \frac1{2!} + \frac2{3!} + \dots+ \frac{n}{(n+1)!} = 1 - \frac1{(n+1)!} $? $\text{Use the PMI to prove the following for all natural numbers n.}$
$ \frac{1}{2!} + \frac{2}{3!} + \cdot \cdot \cdot + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!} $
So for this question I get stuck because I cannot seem to make the left side and right side be equivalent.
$\color{Green}{Proof} :$
$\mathbf (I) \; Basis \; Step :$ Show $p(1).$ For $n=1$
$\frac{1}{(1+1)!} = 1- \frac{1}{(1+1)!} \Rightarrow \frac12 = \frac12$
$\mathbf (II) \; Inductive \; Step : $
Assume $p(k)$ for a PAC $k \in ℕ ,\; \; \;\; n=k$
$ \frac{1}{2!} + \frac{2}{3!} +\cdot\cdot\cdot + \frac{k}{(k+1)!} = 1-\frac{1}{(k+1)!} \; \; \text{Induction Hypothesis}$
Left Hand Side : We need to show : $p(k+1)$ i.e. $n = k+1$
$\frac{k}{(k+1)!} + \frac{k+1}{(k+1+1)!} \Rightarrow \ $
$\frac{(k+1)}{(k+1+1)!} +1 - \frac{1}{(k+1)!} $
Right Hand Side: $ n =k+1$
$\left( 1- \frac{1}{(k+1+1)!}\right) = 1- \frac{1}{(k+2)!}$
This is where I have made my mistake because I think I have made an arithmetic mistake on the left side because I cannot get them to be verifiable. Any hints on how to correct the error would be appreciated. Have a good one.
| By the inductive hypothesis, we know that \begin{align*} \frac{1}{2!} + \cdots + \frac{k}{(k+1)!} + \frac{k+1}{(k+2)!} &= 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{1}{(k+2)!}.\end{align*} Hence, if the statement is true for $n=k$, then it true for $n = k+1$, so by induction, it is true for all $n \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
find a vector function that represents the curve of intersection of the two surfaces The cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$
I parameterized the plane and put it into vector form:
$t=1+y \to y=t-1$
$z= 1+t-1 \to z = t$
$y= t-1$, $z=t$
Since I'm finding the intersection, I can just plug the parameters from the plane into the cone and get x:
$t = \sqrt{x^2+(t-1)^2} \to t^2 = x^2 + t^2 - 2t +1 \to x^2 = -2t+1 \to x = \sqrt{-2t+1}$
for the vector equation I have:
$\Big(\sqrt{-2t+1}i+t-1j+tk\Big)$
| I would approach it this way.
$1+y = \sqrt{x^2 + y^2}\\
1+2y + y^2 = x^2+ y^2\\
1+2y = x^2\\
y = \frac 12 x^2 - \frac 12$
now I would say:
$x = t\\
y = \frac 12 t^2 - \frac 12\\
z = \frac 12 t^2 + \frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a nicer way to show that the series is convergent? I'd like to show that for a fixed $z\in\mathbb C\setminus\mathbb Z$ the series $$\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right|$$ is convergent.
I think, one can do it as follows. Fix some $n_0> |z|$. Then
\begin{align*}
\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| & = \underbrace{\sum_{n=1}^{n_0-1} \left| \frac{1}{z-n} + \frac{1}{n} \right|}_{=:C} + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{z}{(z-n)n} \right| \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{|z-n|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{||z|-|n||n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{(n-|z|)n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n^2} \\
& \leq C + \frac{|z|}{1-|z|} \underbrace{\sum_{n=n_0}^\infty \frac{1}{n^2}}_{<\infty} \\
\end{align*}
That strikes me as somewhat cumbersome. Is there a nicer way? E.g. without separating the series into before and after $n_0$?
| I don't know about elegance, but you can also rewrite $z$ as $x+iy$ to get
$\sum_{n=1}^\infty \sqrt{[\frac{x-n}{(x-n)^2+y^2}+\frac{1}{n}]^2+O(\frac{1}{n^4})} $
Now $\frac{x-n}{(x-n)^2+y^2}+\frac{1}{n}$ can be simplified as $\frac{x^2-xn+y^2}{n[(x-n)^2+y^2]}=O(\frac{1}{n^2})$ and the rest is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do you write a polynomial as a linear combination of binomial coefficients? So here Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 in a proof, the needed polynomial is written as a linear combination of binomial coefficients, and I just can't figure out how. If some of you could give me a step by step explanation on how do you exactly turn a polynomial into a form like that, I would be thankful.
| Note the powers $n^k$ can also be expressed with binomial coefficients $\binom{n}{k}$ by means of the so-called Stirling numbers of the second kind denoted with
$\begin{Bmatrix}n\\k\end{Bmatrix}$.In the referred Wiki-page the relationship is given using falling powers
\begin{align*}
(n)_k=n(n-1)(n-2)\cdots(n-k+1)
\end{align*}
as
\begin{align*}
n^k=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}(n)_j\tag{1}
\end{align*}
Since binomial coefficients $\binom{n}{j}=\frac{(n)_j}{j!}$ we can write (1) as
\begin{align*}\
n^k=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n}{j}j!\tag{2}
\end{align*}
In order to represent $n^5-5n^3+4n$ with binomial coefficients we calculate according to (2)
\begin{align*}
n^5&=\sum_{j=0}^5\begin{Bmatrix}5\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!+15\binom{n}{2}2!+25\binom{n}{3}3!+10\binom{n}{4}4!+\binom{n}{5}5!\\
n^3&=\sum_{j=0}^3\begin{Bmatrix}3\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!+3\binom{n}{2}2!+1\binom{n}{3}3!\\
n^1&=\sum_{j=0}^1\begin{Bmatrix}1\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!\\
\end{align*}
Putting all together gives
\begin{align*}
n^5-5n^3+4n&=120\binom{n}{5}+240\binom{n}{4}+140\binom{n}{3}+30\binom{n}{2}+\binom{n}{1}\\
&\qquad-5\left(6\binom{n}{3}+6\binom{n}{2}+\binom{n}{1}\right)\\
&\qquad+4\binom{n}{1}\\
&=120\left(\binom{n}{5}+2\binom{n}{4}+\binom{n}{3}\right)\tag{3}
\end{align*}
Applying the induction step $n\rightarrow n+1$ to (3) it is obvious that the expression is a multiple of $120$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{x\to \infty} \frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x+2}+\sqrt{x}-\sqrt{4x+12}}$ Using the equivalency, I got the answer equal to $1$, however the answer is given as $\frac34$.
WolframAlpha also gives $\frac34$.
Could someone tell me that the given answer is wrong or I made a mistake!
| Substitute $x=1/t$; then, with a simple massage, the limit becomes
$$
\lim_{t\to0^+}
\frac{\sqrt{1+t}+\sqrt{1+2t}-2\sqrt{1+3t}}
{\sqrt{1+2t}+1-2\sqrt{1+3t}}
=\lim_{t\to0^+}
\frac{(1+\frac{1}{2}t)+(1+t)-2(1+\frac{3}{2}t)+o(t)}
{(1+t)+1-2(1+\frac{3}{2}t)+o(t)}=
\frac{-3/2}{-2}=\frac{3}{4}
$$
using the first order Taylor expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the smallest number $\alpha$, such that for all $x,y,z$ $\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$
Find a smallest number $\alpha$, such that for all $x,y,z$ (not all of which are positive) inequality
$$\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$$
My work so far:
Let $f(t)=t^2-t+1$. Then $f(t) \ge \frac34$.
If $x=0, y=z=\frac12$, then $$\alpha\ge \frac{16}9$$
| From your
$a(x^2-x+1)^3\ge x^6-x^3+1
$,
since
$(x+1)(x^2-x+1)
=x^3+1
$
and
$(x^3+1)(x^6-x^3+1)
=x^8+1
$,
$a(\frac{x^3+1}{x+1})^3\ge \frac{x^9+1}{x^3+1}
$
or
$a\ge \frac{(x^9+1)(x+1)^3}{(x^3+1)^4}
$.
According to Wolfy,
this has a maximum of
$2.1547$
at $x \approx 0.435421
$
and
$2.29663
$.
These are the roots of
$0
= x^6-2 x^5-x^4+x^2+2 x-1
$.
The exact roots are
$-1, 1,
\frac12 (1\pm\sqrt{2} 3^{1/4}+\sqrt{3}),
\frac12 (1\pm i\sqrt{2} 3^{1/4}-\sqrt{3})
$.
So $a$ must be at least
$2.1547$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Giving a basis for the column space of A
Let $A = \begin{bmatrix}3&3&3\\3&5&1\\-2&4&-8\\-2&-4&0\\4&9&-1\end{bmatrix}$
Give a basis for the column space of A
So what I've done so far is put it in RREF (which was a task itself) and got
$\begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}$, but I'm not sure what to do next to give the "basis" of the column space of A
|
Fact. Let $A$ be a matrix. The nonzero rows of $\DeclareMathOperator{rref}{rref}\rref(A^\top)$ form a basis of the column space of $A$.
In our case we have
$$
A=
\left[\begin{array}{rrr}
3 & 3 & 3 \\
3 & 5 & 1 \\
-2 & 4 & -8 \\
-2 & -4 & 0 \\
4 & 9 & -1
\end{array}\right]
$$
Row-reducing $A^\top$ gives
$$
\rref(A^\top)=
\left[\begin{array}{rrrrr}
1 & 0 & -\frac{11}{3} & \frac{1}{3} & -\frac{7}{6} \\
0 & 1 & 3 & -1 & \frac{5}{2} \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$
The fact above implies that
\begin{align*}
\langle1, 0, -{11}/{3}, {1}/{3}, -{7}/{6} \rangle
&&
\langle0,1, 3, -1, 5/2 \rangle
\end{align*}
forms a basis of the column space of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Prove that: $\sum\limits_{n=1}^\infty \frac{n^2}{2^n}=6$. Prove that: $$\sum\limits_{n=1}^\infty \dfrac{n^2}{2^n}=6.$$ I am trying to find the sum of this infinite series. Got stuck. Any help will be highly appreciated.
| We have, $$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{n^2}{2^n} $$
$$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{(n+1)^2}{2^{n+1}} $$
Since $S = 2S - S $,
\begin{align}
S&=\sum_{n=0}^\infty \frac{(n+1)^2}{2^n} - \sum_{n=0}^\infty \frac{n^2}{2^n} \\
&=\sum_{n=0}^\infty \frac{2n+1}{2^n}\\ &=
\sum_{n=0}^\infty \frac{2n}{2^n} + \sum_{n=0}^\infty \frac{1}{2^n} \\&=
4+2 = 6
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$
My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?
| $\cos{5a} = \cos{(3a+2a)}$
$=\cos 3a \cos 2a -\sin 3a \sin 2a$
$=\cos{(2a+a)}\cos 2a -\sin{(2a+a)} \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=((\cos^2a-\sin^2a)\cos a-(2\sin a \cos a)\sin a)(\cos^2a-\sin^2a) -((2\sin a \cos a) \cos a + (\cos^2a-\sin^2a) \sin a) (2\sin a\cos a)$
$=(\cos^3a-\sin^2 a\cos a -2\sin^2a\cos a)(\cos^2a-\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\cos^2a)-(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)-(\cos^3a \sin^2a - \sin^4 a \cos a - 2 \sin^4a \cos a) - (4\sin^2 a \cos^3a + 2\cos^3a \sin^2 a - 2\sin^4 a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)+(-\cos^3a \sin^2a + \sin^4 a \cos a + 2 \sin^4a \cos a) + (-4 \sin^2 a \cos^3a - 2\cos^3a \sin^2 a + 2\sin^4 a \cos a)$
$\bf =\cos^5a - 10 \sin^2a \cos^3 a + 5 \sin^4a \cos a$ // Another proof
$=\cos^5a - 10 (1 - \cos^2a) \cos^3 a + 5 (1- \cos^2a)^2 \cos a$
$=\cos^5a - 10 \cos^3a + 10 \cos^5a + (5 \cos a- 10 \cos^3a + 5 \cos^5a)$
$\cos{5a}=16\cos^5a - 20 \cos^3a + 5 \cos a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Algebraic or Analytic Proof of a Polynomial Identity
Let $m$, $n$, and $r$ be integers with $0\leq r \leq \min\{m,n\}$. Define
$$f_{m,n,r}(q):=\left(\prod_{j=1}^r\,\left(q^m-q^{j-1}\right)\right)\,\left(\sum_{\substack{{j_1,\ldots,j_r\in\mathbb{Z}_{\geq 0}}\\{j_1+j_2+\ldots+j_r\leq n-r}}}\,q^{\sum_{i=1}^r\,i\,j_i}\right)\,,$$
so that
$$f_{n,m,r}(q)=\left(\prod_{j=1}^r\,\left(q^n-q^{j-1}\right)\right)\,\left(\sum_{\substack{{j_1,\ldots,j_r\in\mathbb{Z}_{\geq 0}}\\{j_1+j_2+\ldots+j_r\leq m-r}}}\,q^{\sum_{i=1}^r\,i\,j_i}\right)$$
as polynomials over $\mathbb{Z}$ in the variable $q$. Prove that $$f_{m,n,r}(q)=f_{n,m,r}(q)\,.$$
Here is a combinatorial proof of this identity. The polynomial $f_{m,n,r}(q)$ counts the number of $m$-by-$n$ matrices over $\mathbb{F}_q$ of rank $r$, when $q$ is a power of a prime natural number. Since the transpose map from $\text{Mat}_{m\times n}\left(\mathbb{F}_q\right)\to\text{Mat}_{n\times m}\left(\mathbb{F}_q\right)$ is a bijection that preserves rank, we conclude that the number of $n$-by-$m$ matrices over $\mathbb{F}_q$ of rank $r$ is also $f_{m,n,r}(q)$. Ergo, $f_{m,n,r}(q)=f_{n,m,r}(q)$ whenever $q$ is a prime power, whence the equality $f_{m,n,r}(q)=f_{n,m,r}(q)$ is indeed an identity in $\mathbb{Z}[q]$. See here.
| We conjecture that
$$f_{m,n,r}(z) =
\left(\prod_{j=1}^r (z^m-z^{j-1})\right)
\left(\prod_{j=1}^r (z^n-z^{j-1})\right)
\frac{(-1)^r}{z^{r(r-1)/2}} \prod_{j=1}^r \frac{1}{1-z^j}.$$
If we can prove this we have the symmetry. We must show that
$$\left(\prod_{j=1}^r (z^n-z^{j-1})\right)
\frac{(-1)^r}{z^{r(r-1)/2}} \prod_{j=1}^r \frac{1}{1-z^j}
= [w^{n-r}] \prod_{j=0}^r \frac{1}{1-wz^j}.$$
The LHS is
$$\begin{align}
&\left(\prod_{j=1}^r z^{j-1} \prod_{j=1}^r (z^{n-(j-1)}-1)\right)
\frac{(-1)^r}{z^{r(r-1)/2}} \prod_{j=1}^r \frac{1}{1-z^j}
\\&\phantom{aaaaaaa} = \left(\prod_{j=1}^r (z^{n-(j-1)}-1)\right) (-1)^r
\prod_{j=1}^r \frac{1}{1-z^j}
\\&\phantom{aaaaaaa} = (-1)^r
\prod_{j=1}^r \frac{1}{1-z^j}
\prod_{j=n-(r-1)}^n (z^j-1)
\\&\phantom{aaaaaaa} = \prod_{j=1}^r \frac{1}{z^j-1}
\prod_{j=n-(r-1)}^n (z^j-1).\end{align}$$
We are therefore tasked with showing that
$$\prod_{j=n-(r-1)}^n (z^j-1)
= [w^{n-r}] \frac{1}{1-w} \prod_{j=1}^r \frac{z^j-1}{1-wz^j}.$$
This can be done by induction. We have for $r=1$
$$\begin{align}z^n-1 &= [w^{n-1}] \frac{1}{1-w} \frac{z-1}{1-wz}
\\ &= [w^{n-1}]\left(\frac{z}{1-wz} - \frac{1}{1-w}\right)
= z \times z^{n-1} - 1.\end{align}$$
For the induction step we have for the RHS by the induction hypothesis
multiplying both sides of the equation for $r$ by $z^{n-r}-1$
$$(z^{n-r}-1)
[w^{n-r}] \frac{1}{1-w} \prod_{j=1}^r \frac{z^j-1}{1-wz^j}.$$
The first piece here is
$$\frac{z^{n-r}}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n-r+1}}
\frac{1}{1-w} \prod_{j=1}^r \frac{z^j-1}{1-wz^j} \; dw.$$
Now put $w/z=v$ so that $dw = z\; dv$ to get
$$\begin{align}
&\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{1}{v^{n-r+1}}
\frac{1}{1-vz} \prod_{j=1}^r \frac{z^j-1}{1-vz^{j+1}} \; dv
\\&\phantom{aaaaaaa} = \frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{1}{v^{n-r+1}}
\prod_{j=1}^r \frac{z^j-1}{1-vz^{j}}\frac{1}{1-vz^{r+1}} \; dv.\end{align}$$
Now
$$\frac{1}{1-vz^{r+1}} - \frac{1}{1-v}
= \frac{1}{1-v} \frac{1}{1-vz^{r+1}}
\big(1-v-\left(1-vz^{r+1}\right)\big)
\\ = v \frac{1}{1-v} \frac{z^{r+1}-1}{1-vz^{r+1}}$$
and we obtain
$$\begin{align}
&\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{1}{v^{n-r+1}} \frac{v}{1-v}
\prod_{j=1}^{r+1} \frac{z^j-1}{1-vz^{j}} \; dv
\\&\phantom{aaaaaaa} = \frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{1}{v^{n-r}} \frac{1}{1-v}
\prod_{j=1}^{r+1} \frac{z^j-1}{1-vz^{j}} \; dv.\end{align}$$
This is
$$[v^{n-(r+1)}] \frac{1}{1-v} \prod_{j=1}^{r+1} \frac{z^j-1}{1-vz^j}$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
| Use the classical approximation:
$$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$
With $a = \sqrt{20}$ and $b = \sqrt{20 + \sqrt{20}}$ we have
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{\sqrt{20 + \sqrt{20}}}{2\sqrt{20}} = \sqrt{20} + \frac{\sqrt{400 + 20\sqrt{20}}}{40} $$
Now use the same classical approximation again, this time working with the numerator of the second term. This time with $a=20$ and $b=20\sqrt{20}$ we get
$$\sqrt{400 + 20\sqrt{20}} \approx 20 + \frac{20\sqrt{20}}{40} = 20 + \frac{\sqrt{20}}{2}$$
Combining these, we've got:
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{20 + \frac{\sqrt{20}}{2}}{40} = \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Using similar methods, we get
$$\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Finally, subtracting one from the other we end up with
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} -\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \left( \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80} \right) - \left( \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80} \right)$$
and in this last expression everything cancels out except for
$$\frac{1}{2} - \left(-\frac{1}{2} \right) = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
Help with the integral $\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$ I would like to know what is $$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx.$$ I put $x=\tan(y)$ to get integral of $\displaystyle \int \frac{\sin(y)}{\cos^3(y)}.\sqrt{\cos(2y)}dy$ I don't know whether $\sin(x)=t$ is a good substitution.
| Let $$I =\frac{1}{2} \int \sqrt{\frac{1-x^2}{1+x^2}}\cdot 2xdx$$
Now put $x^2=t\;,$ Then $2xdx = dt$
So $$I = \frac{1}{2}\int\sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int\sqrt{\frac{1-t}{1+t}\times \frac{1-t}{1-t}}dt$$
So $$I = \frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{1}{2}\int\frac{t}{\sqrt{1-t^2}}dt$$
So $$I = \frac{1}{2}\arcsin (t)+\frac{1}{2}\sqrt{1-t^2}+\mathcal{C}$$
So $$I = \frac{1}{2}\arcsin (x^2)+\frac{1}{2}\sqrt{1-x^4}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed form of the summation- $\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}$ I have got the following summation-$$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}.$$
I have to find the closed form or the general form to find the sum of this series. I know upto summation of Telescopic Series and Some special series like those in $AP$ or $GP$.
But I have no idea on how to begin on this problem.
Thanks for any help!!
| Since you requested a more detailed answer than that of @N.S, here you go.
We start with the series you give,
$$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}$$
We now want to simplify this into a series you can deal with; the first step will be to run partial fraction decomposition on the term
$$\frac{r^2}{(r+1)(r+2)}=\frac{r^2+3r+2-3r-2}{(r+1)(r+2)}=\frac{(r+1)(r+2)-3r-2}{(r+1)(r+2)}$$
$$=1-\frac{3r+2}{(r+1)(r+2)} =1+\frac{1}{r+1}-\frac{4}{r+2}$$
Let me know if you need help with this partial fraction decomposition. We can now rewrite the above series as
$$\displaystyle\sum_{r=1}^{n}\left(4^r+\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)=\displaystyle\sum_{r=1}^{n}\left(4^r\right)+\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)\tag{1}$$
The first series is geometric, which you say you know how to evaluate. (It comes out to be $\frac{4}{3}(4^n - 1)$)
To evaluate the second series in $(1)$ we will look at the expanded terms:
$$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)$$
$$= \left(\frac{4}{2}\color{red}{-\frac{4^2}{3}}\right)+\left(\color{red}{\frac{4^2}{3}}\color{green}{-\frac{4^3}{4}}\right)+\left(\color{green}{\frac{4^3}{4}}\color{purple}{-\frac{4^4}{5}}\right)+\cdots+\left(\color{olive}{\frac{4^{n-1}}{n}}\color{fuchsia}{-\frac{4^n}{n+1}}\right)+\left(\color{fuchsia}{\frac{4^n}{n+1}}\color{navy}{-\frac{4^{n+1}}{n+2}}\right)$$
As my above, colored terms hopefully make clear almost all the terms will cancel out with their neighboring term except for the first and last terms (this is called a telescoping series) and so this series comes out to be
$$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right) = \frac{4}{2}-\frac{4^{n+1}}{n+2}=2-\frac{4^{n+1}}{n+2}$$
We can now go plug this back into $(1)$ and rewrite your original series as
$$\frac{4}{3}(4^n - 1)+2-\frac{4^{n+1}}{n+2}= 4^{n+1}\left(\frac{1}{3}-\frac{1}{n+2}\right)+\frac{2}{3}$$
Note that, as $n\to\infty$, this expression becomes closer and closer to being $$4^{n+1}\left(\frac{1}{3}-0\right)+\frac{2}{3}=\frac{4^{n+1}+2}{3}$$
And so the limit towards $+\infty$ clearly diverges to $+\infty$; likewise, the limit towards $-\infty$ diverges towards $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to prove the inequalities between $20^{70^2},30^{60^2},40^{50^2}$ Let $$M=\{ 20^{70^2}, 30^{60^2},40^{50^2}\}$$.
What number is the greatest and which is the smallest?
I thought about beginning by assuming certain inequalities and trying to prove them, for example: $$20^{70^2} < 30^{60^2}$$
However, I do not know exactly what do from here. I took logarithms and expanded the expressions, this took me nowhere. I could get no obvious identity out of them, is there a clever way to do this?
| For a more elementary approach: you could play around with the factors.
$$20^{70^2} = 2^{4900} \cdot 10^{4900} = 32^{980} \cdot 10^{1300}\cdot 10^{3600}, $$
$$30^{60^2} = 3^{3600} \cdot 10^{3600} = 27^{980} \cdot 3^{660} \cdot 10^{3600}.$$
Comparing each factor, it is now obvious that $20^{70^2} > 30^{60^2}$.
Similarly,
$$30^{60^2} = 3^{3600} \cdot 10^{3600} = 3^{268} \cdot 3^{3332} \cdot 10^{1100}\cdot 10^{2500} = 3^{268} \cdot 9^{1666} \cdot 10^{1100}\cdot 10^{2500},$$
$$40^{50^2} = 4^{2500} \cdot 10^{2500} = 2^{5000} \cdot 10^{2500} = 2 \cdot 8^{1666} \cdot 10^{2500},$$
hence $30^{60^2} > 40^{50^2}$.
The trick here is always to find powers of $2$ and $3$ with small difference: $2^5 > 3^3$ and $2^3 < 3^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Assessing the geometric multiplicity of an eigenvalue. Suppose $\lambda_1 = 1$ is an eigenvalue of the hypothetical matrix $\mathbf A$
$$
\mathbf A =
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{11} -1 & a_{12} +1 & a_{13} \\
a_{11} -1 & a_{12} & a_{13} +1 \\
\end{bmatrix}
$$
Let $\mathbf I_3$ represent the identity matrix for $\mathbb R^3$, and let $\mathbf 0_3$ represent the zero vector of $\mathbb R^3$. To find the eigenspace corresponding to $\lambda _1$ it is necessary solve the following linear system of homogeneous equations.
$$
\left(
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{11} -1 & a_{12} +1 & a_{13} \\
a_{11} -1 & a_{12} & a_{13} +1 \\
\end{bmatrix}
- \lambda_1 \mathbf I _3
\right)
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
= \mathbf 0_3
$$
Elementary row operations are used to obtain the following augmented matrix.
$$ \left[
\begin{array}{ccc|c}
a_{11} & a_{12} & a_{13} & 0\\
0&0&0&0 \\
0&0&0&0
\end{array}
\right] \\
$$
Therefore
$$
x_3 = s \quad ; \quad x_2 = t \quad ; \quad a_{11} x_1 = -a_{12}t -a_{13}s
$$
The solution of this system of equations is
$$
s
\begin{bmatrix}
{-a_{13} \over a_{11}} \\
0 \\
1 \\
\end{bmatrix}
+t
\begin{bmatrix}
{-a_{12} \over a_{11}} \\
1 \\
0 \\
\end{bmatrix}
$$
Conclusion: The geometric multiplicity of $\lambda_1 $ is $2$. The eigenspace of $\lambda_1 $ is a two-dimensional subspace of $\mathbb R ^2$ with basis
$$
\left\{
\begin{bmatrix}
{-a_{13} \over a_{11}} \\
0 \\
1 \\
\end{bmatrix}
,
\begin{bmatrix}
{-a_{12} \over a_{11}} \\
1 \\
0 \\
\end{bmatrix}\right\}
$$
Here is the question: Is this analysis correct?
| Not exactly. By row reductions, the correct row echelon form should be
$$
\left[\begin{array}{ccc|c}
a_{11}-1 & a_{12} & a_{13} & 0\\
0&0&0&0 \\
0&0&0&0
\end{array}\right]
$$
instead. Since you don't know the exact values of the $a_{ij}$s, you cannot use as denominators in divisions. All you can say is that the eigenspace of $A$ corresponding to the eigenvalue $\lambda=1$ is the set of all vectors orthogonal to $(a_{11}-1,\ a_{12},\ a_{13})^T$.
Since $(a_{11}-1,\ a_{12},\ a_{13})^T$ may be the zero vector (when $A=I$), the geometric multiplicity of $\lambda=1$ can also be $3$ rather than $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$
So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.)
So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!
| Write
$$
\sqrt{14+8\sqrt{3}}=\sqrt{x\vphantom{b}}+\sqrt{\smash{y}\vphantom{b}}
$$
Then
$$
14+8\sqrt{3}=x+y+2\sqrt{\smash{xy}\vphantom{b}}
$$
so we can look for
$$
xy=48,\qquad x+y=14
$$
that reduces to finding the roots of the equation
$$
z^2-14z+48=0
$$
that are $8$ and $6$. So we can take $x=8$ and $y=6$:
$$
\sqrt{14+8\sqrt{3}}=\sqrt{6}+\sqrt{8}
$$
Alternatively, recall the identity
$$
\sqrt{a+\sqrt{b}}=
\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
In this case $a^2-b=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove: $\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$, where $a^2 + b^2 = c^2$ Could you help me proving this?
$$\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$$
where $c$ is the length of the hypotenuse of a right triangle, and $a$ and $b$ are the lengths of the other sides.
I tried that but not sure if it helps and what to do next:
$$\frac{\log\left(a\right)}{\log\left(c+b\right)}+\frac{\log\left(a\right)}{\log\left(c-b\right)}=\frac{2\log^2\left(a\right)}{\log\left(c+b\right)\log\left(c-b\right)}$$
Thank you in advance!
| The change of base formula for logarithms says that
$$
\log_xy=\frac{1}{\log_yx}
$$
assuming $x$ and $y$ positive and different from $1$. Thus, assuming $a\ne1$, $c-b\ne1$ and $c+b\ne1$, we have
\begin{align}
\log_{c+b}a+\log_{c-b}a
&=
\frac{1}{\log_a(c+b)}+\frac{1}{\log_a(c-b)}\\[6px]
&=
\frac{\log_a(c-b)+\log_a(c+b)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px]
&=
\frac{\log_a\bigl((c-b)(c+b)\bigr)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px]
&=
\frac{\log_a(c^2-b^2)}{\log_a(c+b)\cdot\log_a(c-b)}
\end{align}
Since $c^2-b^2=a^2$, the numerator is $\log_aa^2=2$ and, applying again the change of base,
$$
\log_{c+b}a+\log_{c-b}a=2\log_{c+b}a\cdot\log_{c-b}a
$$
If $a=1$ the equality is obvious. If either $c+b=1$ or $c-b=1$, the equality doesn't make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $x$ is positive, then why does $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$? Given that $x$ is positive, $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$
I've been trying to convert the left side of the equation to the right side:
$$ \frac{1}{\sqrt{x+1} + \sqrt{x}}$$
But then how can I flip this round to be what I have on the right side?
I know that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, so I would think that this would give me
$$(x+1)^{-1/2} + x^{-1/2}$$
Which I thought would then convert to
$$-\sqrt{x+1} + (-\sqrt{x}) $$
so I'm not sure how the first part $\sqrt{x+1}$ got to be positive
| Rationalize the denominator:
\begin{align}
\frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1} - \sqrt{x}} &= \frac{\sqrt{x+1}-\sqrt{x}}{\underbrace{\left(\sqrt{x+1} + \sqrt{x}\right) \cdot \left(\sqrt{x+1} - \sqrt{x}\right)}_{\color{red}{\text{FOIL}}}} \\[0.3cm]
&= \cdots
\end{align}
Some notes:
If you want to show that two things are equal to each other, it's better to manipulate only one of them than it is to start by saying they're equal to each other and then manipulating both sides to verify, etc.
Your idea of converting to $(x+1)^{-1/2} + x^{-1/2}$ won't work because, in general, $\displaystyle \frac{1}{a+b} \ne \frac{1}{a} + \frac{1}{b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $f'(x)$in terms of $f(x)=|\cos(x)|\sqrt{1-\cos(x)}$ I am trying to solve the following exercise :
Let $f$ be the function defined by :
$$\forall x\in]0,\pi[\;\;\;\;\; f(x)=|\cos(x)|\sqrt{1-\cos(x)}$$
calculate $f '(x)$ in terms of $f(x),$ for all $x\in]0,\pi[\setminus \{\frac{\pi}{2}\}.$
My try : we can write $f(x)=\sqrt{\cos^2(x)-\cos(^3(x),}$ then
$$f'(x)=\frac{-2\cos(x)\sin(x)+3\cos^2(x)\sin(x)}{2f(x)} $$
I don't know how I can write $-2\cos(x)\sin(x)+3\cos^2(x)\sin(x)$ in terms of $f(x)$
Any help would be appreciated, Thanks !
| Notice that $\sqrt {1 - \cos x} = \sqrt {2 \sin^2 \frac x 2} = \sqrt 2 | \sin \frac x 2| = \sqrt 2 \sin \frac x 2$ because $\sin \frac x 2 \ge 0$ when $x \in [0, \pi]$.
For $x \in [0, \frac \pi 2]$, the function is
$$f(x) = \sqrt 2 \cos x \sin \frac x 2 = \frac {\sqrt 2} 2 [\sin (\frac x 2 + x) + \sin (\frac x 2 - x)] = \frac {\sqrt 2} 2 (\sin \frac {3x} 2 - \sin \frac {-x} 2) = \\
\frac {\sqrt 2} 2 (\sin \frac {3x} 2 + \sin \frac x 2) ,$$
where I have used the formula $\sin a \cos b = \frac 1 2 (\sin (a+b) + \sin (a - b))$.
Similarly, for $x \in [\frac \pi 2, \pi]$, $f(x) = - \sqrt 2 \cos x \sin \frac x 2 = - \frac {\sqrt 2} 2 (\sin \frac {3x} 2 + \sin \frac x 2)$.
This means that, for $x \in [0, \frac \pi 2]$, $f'(x) = \frac {\sqrt 2} 4 (3 \cos \frac {3x} 2 + \cos \frac x 2)$.
It's quite clear that $f'$ has nothing to do with $f$, for two reasons:
*
*first, the functions have changed from $\sin$ to $\cos$; this could be cured by noticing that $\cos x = \sin (\frac \pi 2 \pm x)$, but this means that $f'(x)$ is not expressed in terms of $f(x)$, but in terms of $f(x \pm \frac \pi 2)$
*much worse, there is a $3$ in front of $\cos \frac {3x} 2$ that spoils everything, and that cannot be eliminated by any trick (removing it would in turn alter the coefficient of $\cos \frac x 2$).
I doubt that the problem can be solved as required in its current form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ over positive integers.
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ where $x, y, z \in
\mathbb{N}$
I've found some solutions, like $(0, 0, 1), (1, 1, 3)$ and, more general, $x=k,y=k,z=2k+1$. No idea how to prove or disprove there is no other solutions.
| If we set $x+y=t$, we are just looking for the solutions of
$$ (2t+1)^2+(8x+3) = 4z^2 \tag{1}$$
or:
$$ (8x+3) = (2z+2t+1)(2z-2t-1).\tag{2} $$
If $d$ is a divisor of $8x+3$ that is $\leq \sqrt{8x+3}$, $d$ is odd and
$$ 2z+2t+1=\frac{8x+3}{d},\qquad 2z-2t-1 = d \tag{3} $$
leads to the solution:
$$ z = \frac{1}{4}\left(d+\frac{8x+3}{d}\right),\qquad 2t+1=\frac{1}{2}\left(\frac{8x+3}{d}-d\right).\tag{4} $$
I leave to you to fill the missing details, like, for instance, to find the conditions that ensure the last $z$ and $t$ actually being positive integers with $t\geq x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is this true $\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$? I have a heard time seeing why is this true
$\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$
Could you show me some kind of proof, or an identity that would me to find this?
| $$\begin{align}
\frac{1-y^n}{1-y}
&=\frac{1\color{red}{-y+y}\color{blue}{-y^2+y^2}-y^3+\cdots+y^{n-2}\color{magenta}{-y^{n-1}+y^{n-1}}-y^n}{1-y}\\
&=\frac{(1\color{red}{-y})+y(\color{red}{1}\color{blue}{-y})+y^2(\color{blue}{1}-y)+\cdots+y^{n-2}(1\color{magenta}{-y})+y^{n-1}(\color{magenta}{1}-y)}{1-y}\\
&=\frac{(1-y)(1+y+y^2+\cdots+y^{n-2}+y^{n-1})}{1-y}\\
&=1+y+y^2+\cdots+y^{n-2}+y^{n-1}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $ How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$
I tried the following:
We use the falling factorial power:
$$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$
so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$
Then
$${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} =
\frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$
And
$$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$
$$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$
$$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
$$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
According to the Binomial-coefficients:
$$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$
$$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$
What is wrong ? und How can I continue? :/
|
Hint: The binomial formula with the Cauchy product
\begin{align*}
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}
\end{align*}
does not use falling factorials $x^{\underline{k}}$ resp. $y^{\underline{n-k}}$.
Here is a step by step answer similar to that by @MarkoRiedel. It's convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}
We obtain
\begin{align*}
\sum_{k=0}^{n}&\binom{x-1+n-k}{n-k}\binom{y-1+k}{k}\\
&=\sum_{k=0}^{\infty}\binom{x-1+n-k}{n-k}\binom{-y}{k}(-1)^k\tag{1}\\
&=\sum_{k=0}^\infty [t^{n-k}](1+t)^{x-1+n-k}[z^k](1+z)^{-y}(-1)^k\tag{2}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty(-1)^kt^k(1+t)^{-k}[z^k](1+z)^{-y}\tag{3}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty\left(-\frac{t}{1+t}\right)^k[z^k](1+z)^{-y}\\
&=[t^n](1+t)^{x-1+n}\left(1-\frac{t}{1+t}\right)^{-y}\tag{4}\\
&=[t^n](1+t)^{x+y-1+n}\tag{5}\\
&=\binom{x+y-1+n}{n}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we use the binomial identity $\binom{-p}{q}(-1)^q=\binom{p+q-1}{q}$ and we extend the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.
*In (2) we apply the coefficient of operator twice.
*In (3) we do some rearrangements by using the linearity of the coefficient of operator and we also use the rule
\begin{align*}
[z^{p-q}]A(z)=[z^p]z^{q}A(z)
\end{align*}
*In (4) we apply the substitution rule
\begin{align*}
A(t)=\sum_{k=0}^\infty a_kt^k=\sum_{k=0}^\infty t^k[z^k]A(z)\\
\end{align*}
with $z=-\frac{t}{1+t}$.
*In (5) we do some simplifications.
*In (6) we select the coefficient from $t^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
$\left(\frac{3}{p} \right)=1$ iff $p\equiv 1\pmod{12}$ or $p\equiv -1\pmod{12}$ Let $p\geq 5$ be a prime.
$\left(\frac{3}{p} \right)=1$ iff $p\equiv 1\pmod{12}$ or $p\equiv -1\pmod{12}$.
So $\left(\frac{3}{p} \right)=\left(\frac{p}{3} \right)\cdot (-1)^{(p-1)/2}$ and this is $\left(\frac{p}{3} \right)$ when $p\equiv 1\pmod4$ and it is $-\left(\frac{p}{3} \right)$ if $p\equiv 3\pmod 4$.
*
*Case where $p\equiv 1\pmod 4$: Here $\left(\frac{p}{3} \right)=1$ and therefore $p$ is a square in $\Bbb Z/3\Bbb Z$. This implies $p\equiv 1\pmod 3$, right? I am not sure here. Then the Chinese Remainer Theorem implies that $p\equiv 1\pmod{12}$
*Case where $p\equiv 3\pmod 4$: Here $\left(\frac{p}{3} \right)=-1$, so $p$ is not a square in $\Bbb Z/3\Bbb Z$. Does this imply $p\equiv 2\pmod 3$? We would get $p\equiv 6\mod{12}$ which is not what we want....where did I go wrong?
| In my answer to this question there is an elementary proof that
$$\left(\frac{-3}{p}\right)=1\quad\Longleftrightarrow\quad p\equiv 1\pmod{3}\tag{1}$$
Moreover, it is quite well-known and not difficult to prove that
$$\left(\frac{-1}{p}\right)=1\quad\Longleftrightarrow\quad p\equiv 1\pmod{4}\tag{2}$$
hence the claim follows from the multiplicativity of the Legendre symbol and the Chinese remainder theorem: $\left(\frac{3}{p}\right)=1$ iff both $\left(\frac{-3}{p}\right)$ and $\left(\frac{-1}{p}\right)$ equal $+1$ or both equal $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A$ and $B$ are natural numbers such that $A-B = \dfrac{A}{B}$, what is the value of $A+B$? If $A$ and $B$ are natural numbers such that $A-B = \dfrac{A}{B}$, what is the value of $A+B$?
The answer is $6$. I tried all the basic maths functions but was not able to get it. I would like to know if its a trial and error based question or it can be derived.
| We have:
$A = B + \dfrac{A}{B}\implies AB - B^2 - A = 0$
$\implies A = \dfrac{B^2}{B-1} = B+1 + \dfrac{1}{B-1}$
$\implies B-1 = 1 \implies B = 2$
$\implies A = 4$
$\therefore A+B = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of sum convergence?
$$\sum_{n=1}^{\infty} \frac{2\sqrt{n} + 1}{n^2 + n + 1}$$
This seems like a problem that could be handled by the comparison test.
So we need an $f(n) > \displaystyle \frac{2\sqrt{n} + 1}{n^2 + n + 1}$
So we need $\text{num}f(n) > 2\sqrt{n} + 1$ and $\text{denom}f(n) < n^2 + n + 1$.
So obviously, we can let $\text{denom}f(n) = n^2$ and we can try $\text{num}f(n) = 3n$
So we have $f(n) = \frac{3}{n} > \frac{2\sqrt{n} + 1}{n^2 + n + 1}$
but this fails?
| You're on the right track, but you gave up too much in the numerator. Instead, observe that
$$ \frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$
and $\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\lim_{x\to1}(x^3+5x^2-2)=4$ using the $\epsilon$-$\delta$ definition of a limit I want to prove that the limit of $f(x)=x^3+5x^2-2$ when $x\to 1$ is $4$.
So, I want to show that for any $\epsilon >0$ $\exists \delta_{\epsilon}$ such that for all $x$ that satisfies $|x-1|<\delta$ then $|f(x)-4|< \epsilon$.
Thus, $|x^3+5x^2 - 2 - 4|<|x^3+5x^2|<|x^2(x+5)|=x^2|x+5|<x^2|x-1|<\epsilon.$
And we know that $x^2|x-1|<x^2\delta<\epsilon $. As such, for every $\epsilon >0$ the corresponding $\delta$ is $\min\{\epsilon/x^2, \epsilon\}, x\neq0$.
Is it right?
| I concur with Arthur that your work looks OK minus a few technicalities, but those are rather a big deal in real analysis aren't they? Thus, I would write up your proof as follows:
Given $\epsilon>0$, we need $\delta>0$ such that if $|x-1|<\delta$ then $|(x^3+5x^2-2)-4)|<\epsilon$. Now,
$$
|x^3+5x^2-6|=|(x-1)(x^2+6x+6)|=|x-1||x^2+6x+6|.
$$
If $|x-1|<1$, that is, $0<x<2$, then $x^2+6x+6<(2)^2+6(2)+6=22$ and so
$$
|x^3+5x^2-6|=|x-1||x^2+6x+6|<22|x-1|.
$$
So if we take $\delta=\min\left\{1,\frac{\epsilon}{22}\right\}$, then $|x-1|<\delta$ implies that
$$
|x^3+5x^2-6|=|x-1||x^2+6x+6|<\frac{\epsilon}{22}\cdot22=\epsilon,
$$
as desired. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\lim_{x\to 0}\frac{x+a}{1-3x}=a$, $a>0$, by $\epsilon$-$\delta$ arguments We have
\begin{align}
\left| \frac{x+a}{1-3x} - a \right| &= \left| \frac{x+a-a+3ax}{1-3x} \right| \\[1em]
&= \left| \frac{x(1+3a)}{1-3x}\right|\\[1em]
&= \frac{|x|(1+3a)}{|1-3x|} <\epsilon.
\end{align}
By stating that $|x|<\delta$, we can proceed by
\begin{align}
|x| <\frac{\epsilon|1-3x|}{1+3a}.
\end{align}
By restricting $-1/3<x<1/3$ we have that for such $x$ and for any $\epsilon$, the corresponding $\delta$ is $\min\{1, 2\epsilon/(1+3a) \}$.
Is this right?
| This is one of those cases where starting off with $\delta=1$ is not optimal. If you let $\delta=\frac{1}{6}$, then note that
\begin{align}
|x|<\frac{1}{6}&\Longleftrightarrow -\frac{1}{6}<x<\frac{1}{6}\\[1em]
&\Longleftrightarrow \frac{1}{6}>-x>-\frac{1}{6}\\[1em]
&\Longleftrightarrow \frac{1}{2}>-3x>-\frac{1}{2}\\[1em]
&\Longleftrightarrow \frac{3}{2}>1-3x>\frac{1}{2}\\[1em]
&\Longleftrightarrow \frac{2}{3}<\frac{1}{1-3x}<2\\[1em]
&\Longleftrightarrow \frac{1+3a}{1-3x}<2+3a.
\end{align}
Thus, we should choose $\delta=\min\left\{\frac{1}{6},\frac{\epsilon}{2+3a}\right\}$. To see that this choice of $\delta$ works, consider the following:
Given $\epsilon>0$, let $\delta=\min\left\{\frac{1}{6},\frac{\epsilon}{2+3a}\right\}$. If $|x|<\frac{1}{6}$, then $\left|\frac{1+3a}{1-3x}\right|<2+3a$. Also, $|x|<\frac{\epsilon}{2+3a}$. Hence,
$$\left|\frac{x+a}{1-3x}-a\right|=|x|\left|\frac{1+3a}{1-3x}\right|<\frac{\epsilon}{2+3a}\cdot(2+3a)=\epsilon,
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $a^2+b^2$?
If the range of the function $$f(x)=\frac{x^2+ax+b}{x^2+2x+3}$$ is
$[-5,4]$ then what is the value of $a^2+b^2$ ?
[$a,b$ are natural numbers]
What will the correct approach to this problem?
I tried using the general method as follows:
$({x^2+2x+3})y={x^2+ax+b}$
or,$x^2(y-1)+x(2y-a)+(b-3y)=0$
Then since $x$ is real the discriminant should be greater than equal to 0.
But this method isn't very efficient and quick for this problem.Any shortcuts possible?
|
Note that
$$-5 \leq y \leq 4 \iff (y+5)(y-4) \leq 0 \iff y^2+y-20 \leq 0$$
For $x^2(y-1)+(2y-a)x+(3y-b)=0$ admits real value of $x$,
\begin{align*}
\Delta & \geq 0 \\
(2y-a)^2-4(y-1)(3y-b) & \geq 0 \\
-8y^2+4(3-a+b)y+a^2-4b & \geq 0 \\
8y^2+4(a-b-3)y-a^2+4b & \leq 0 \\
\end{align*}
Comparing coefficients,
$$8y^2+4(a-b-3)y-a^2+4b \equiv 8(y^2+y-20)$$
$$\implies
\left \{
\begin{align*}
a-b-3 &= 2 \\ -a^2+4b &= -160
\end{align*}
\right.$$
$$\implies (a,b)= (14,9) \: \text{ or } \: (-10,-15)$$
Since $a$ and $b$ are natural numbers, reject $\, (a,b)=(-10,-15)$.
$$\therefore \quad a^2+b^2=277$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Distance Between Two Ellipsoids Find the shortest distance between $\begin{align}
&\frac{x_1^2}{a_1^2}+...+\frac{x_n^2}{a_n^2}=1\\
&\frac{x_1^2}{a_1^2}+...+\frac{x_n^2}{a_n^2}=2
\end{align}$.
My thinking is when $n=2$, we have two ellipses where the axes of the second are the same as the first scaled by a factor of $\sqrt{2}$. WLOG, assume $|a_1|<|a_2|$. Then the shortest distance would be $|a_1|(\sqrt{2}-1)$. For the ellipsoid case, I would say the shortest distance is $|a_{\text{min}}|(\sqrt{2}-1)$. I don't know how to prove this formally. I thought about using Lagrange multipliers but had difficulties even setting it up. Any suggestions?
| Let $\frac{x_1^2}{a_1^2}+\dots+\frac{x_n^2}{a_n^2} = 1$ and $\frac{y_1^2}{a_1^2}+\dots+\frac{y_n^2}{a_n^2} = 2$. Then by Cauchy-Schwarz
$$ 1 = \sum\limits_{k=1}^{n}{\frac{y_k^2-x_k^2}{a_k^2}} = \sum\limits_{k=1}^{n}{(y_k-x_k)\frac{y_k+x_k}{a_k^2}}\le\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}. $$
Since $a_k^4\ge(\min{|a_i|})^2a_k^2$ for all $k$, it follows that $\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}\le \frac{1}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}}$, and applying the triangle inequality to the vectors $\vec{x} = \left(\frac{x_1}{a_1},\dots,\frac{x_k}{a_k}\right)$ and $\vec{y} = \left(\frac{y_1}{a_1},\dots,\frac{y_k}{a_k}\right)$ yields
$$ \left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}} = \|\vec{x}+\vec{y}\|\le\|\vec{x}\|+\|\vec{y}\| = 1+\sqrt{2}. $$
Hence, $1\le\frac{1+\sqrt{2}}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}$, i.e. $\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\ge(\sqrt{2}-1)\min{|a_i|}$.
If we assume w.l.o.g. that $\min{|a_i|} = |a_1|$, then $(x_1,x_2,\dots,x_n) = (a_1,0,\dots,0)$ and $(y_1,y_2,\dots,y_n) = (\sqrt{2}a_1,0,\dots,0)$ attains equality in the above inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Determine all functions satisfying $f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$
Let $\Bbb R $ be the set of real numbers .Determine all functions $f:\Bbb R \rightarrow \Bbb R $ satisfying the equation
$$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ for all real numbers $x$ and $y$.
My work
$$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$
Setting $y=0$
$$f(x + f(x)) + f(0) = x + f(x )$$
$$\implies f(x + f(x)) = x + f(x ) -f(0)$$
$$\implies f(t) = t -f(0)$$
Setting $x=0$
$$f(f(y)) + f(0) = 0 + f(y) +yf(0)$$
$$f(f(y)) = f(y) +f(0)(y-1)$$
Setting $y=1$
$$f(x + f(x + 1)) = x + f(x + 1)$$
| Complete solution in the case $f(0) = 2$
Roots of $f$
As you point out, letting $y=0$ yields $$f(x+f(x)) + f(0) = x+f(x)$$
If $f(x)$ is ever $0$, then, we have $f(x+0) + f(0) = x+0$, so $$f(0) = x$$
So $f$ only hits $0$ at most once, and that is at $x = f(0)$; that is, if $f$ ever hits $0$ then $f(f(0)) = 0$ and that is the only time $f$ hits $0$.
Letting $x=y=0$, we see $f(f(0)) + f(0) = f(0)$, so $f(f(0)) = 0$.
Therefore $f$ does indeed hit $0$, so $f$ has exactly one root and it is at $f(0)$.
Moreover, as Batominovski points out, letting $x=0$ and $y=f(0)$ yields $f(0)^2 = 2 f(0)$, so $f(0) = 0$ or $f(0) = 2$.
From this point on, I'll trust Batominovski's proof that if $f(0) = 0$ then the identity is the only possible option; I'll look only at $f(0) = 2$.
Fixed points of $f$
We will do this by interchanging $x$ and $y$.
We'll follow Batominovski's shorthand of writing $P(x,y)$ for the statement $f(x+f(x+y)) + f(xy) = x+f(x+y)+y f(x)$.
We have $P(x,y)$ and $P(y, x)$, so subtracting the two equalities, we obtain the symmetrised expression $$f(x+f(x+y)) - f(y+f(x+y)) = x-y+yf(x)-xf(y)$$
Setting $x=0$ we get $$f(f(y)) + y = f(y+f(y)) + y f(0)$$
Now we will use that $f(0) = 2$.
In this case, $f(2) = 0$ because $f(0)$ is a root of $f$; it is the only root.
If $y$ is a fixed point of $f$, then $2y = f(2y) + 2y$, so $2y$ is a root of $f$, so $y = 1$. (And $1$ is indeed a fixed point.)
So in the case $f(0)=2$, $f$ has exactly one fixed point and it is at $1$.
Hence (by OP's $y=1$ work, which showed that $x+f(x+1)$ is always a fixed point) we have $f(x+1) + x = 1$ for all $x$, so $$f(t) = 2-t$$ for all $t$.
Addendum (ignore, it's a work in progress): Case $f(0) = 0$
(I discovered most of this independently of Batominovski, but I was much too slow with it! I did it as an exercise after Batominovski produced their solution.)
Letting $x=0$, as OP did, we see that $f(f(y)) = f(y)$ for all $y$.
Also the root was unique, so $f(x) = 0$ if and only if $x=0$.
By OP's work, $x+f(x)$ is fixed by $f$, so if $x$ is a fixed point then so is $2x$.
Also letting $y=1$, if $x+1$ is fixed then so is $2x+1$. (That is, if $x$ is fixed then so is $2x-1$.)
Inductively, therefore, all nonpositive integers are fixed, since we know $0$ is fixed already.
Now letting $x=y=-1$, obtain $$f(f(-2) - 1) + f(1) = -1+f(-2)-f(-1)$$
which means $-3+f(1) = -1-2+1$ and so $f(1) = 1$.
Hence again inductively all integers are fixed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Show with comparison test that this series converges: $\sum_{n=17}^{\infty}\frac{7n^{2}+18n+2}{n^{4}+100}$
Show with comparison test that this series converges:
$\sum_{n=17}^{\infty}\frac{7n^{2}+18n+2}{n^{4}+100}$
$$\frac{7n^{2}+18n+2}{n^{4}+100}>\frac{7n^{2}+7n}{n^{4}}=\frac{7}{n^{2}}+\frac{7}{n^{3}}>\frac{7}{n^{2}}$$
$$\Rightarrow \sum_{n=17}^{\infty}\frac{7}{n^{2}}=7\sum_{n=17}^{\infty}\frac{1}{n^{2}}$$
We know that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges, so the complete series will converge, too.
Did I do it right?
| You need to find an upper bound, because having a lower bound that converges does not give you any information at all, since the series could diverge or converge. This is one way of doing things, there are many others:
$$7n^2+18n+2<7(n+4)^2<7(2n)^2=28n^2$$
In addition $n^4+100>n^4$
So $$\frac{7n^2+18n+2}{n^4+100}<\frac {28n^2}{n^4}$$
$$\sum_{n=17}^\infty\frac{7n^2+18n+2}{n^4+100}<\sum_{n=17}^\infty\frac {28n^2}{n^4}=\sum_{n=17}^\infty\frac {28}{n^2}$$
Hence, it does converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| I would like to generalize Behrouz' rather nifty answer:
$$I=\sum_{n=1}^{\infty} \frac{n}{r^n}=\frac{1}{r}+\frac{2}{r^2}+\frac{3}{r^3}+\frac{4}{r^4}$$
The key is now to multiply everything by $r$:
$$rI=1+\frac{2}{r}+\frac{3}{r^2}+\frac{4}{r^3}$$
Now we calculate $rI-I$
$$rI-I=1+(\frac{2}{r}-\frac{1}{r})+(\frac{3}{r^2}-\frac{2}{r^2})+(\frac{4}{r^3}-\frac{3}{r^3})+...$$ simplifies to
$$1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}...$$ And that is our good old geometric sum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 6
} |
What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.
I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:
$$
\sqrt{3 + 2\sqrt{10}i} = x + iy
\quad \longrightarrow \quad
3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.
| $z=3+2i\sqrt{10}$ is a complex number with modulus $7=\sqrt{9+40}$ in the first quadrant ($\text{Re}(x),\text{Re}(y)>0$), hence we have
$$ z=7 e^{i\theta} $$
with
$$\theta = \arctan\frac{2\sqrt{10}}{3}=2\arctan t.$$
Since $\tan(2u)=\frac{2\tan(u)}{1-\tan^2(u)}$, by solving $\frac{2t}{1-t^2}=\frac{2\sqrt{10}}{3}$ with the constraint $t>0$ we get $t=\sqrt{\frac{2}{5}}$, hence the square root of $z$ in the first quadrant is given by:
$$ \sqrt{z} = \sqrt{7} e^{it} = \sqrt{7}\left(\cos\arctan\sqrt{\frac{2}{5}}+i\sin\arctan\sqrt{\frac{2}{5}}\right) $$
or:
$$ \sqrt{z} = \sqrt{7}\left(\sqrt{\frac{5}{7}}+i\sqrt{\frac{2}{7}}\right) = \color{red}{\sqrt{5}+i\sqrt{2}}. $$
Let we check the solution so found:
$$ (\sqrt{5}+i\sqrt{2})^2 = 5-2+2i\sqrt{2\cdot 5} = 3+2i\sqrt{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How can I get from one factored form to another? This came from trying to factor $x^6 + 1$. We know that
$x^4 - x^2 + 1 = (x^4 +2 x^2 + 1) -3x^2 = (x^2+1)^2 - 3x^2 = (x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
I expected that if I factored the $x^4 - x^2 + 1$ completely into $4$ factors and then regrouped them $2$ by $2$, I could arrive at $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
$$\left(x+ \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x+ \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right)$$
But that didn't happen. No matter how I group them into quadratics, I always end up with a complex coefficient. How can I get $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$ from the $4$ factors?
| $$\left(\frac{\sqrt{3}\pm i}{2}\right)^2 = \frac{3-1\pm 2\sqrt{3}i}{4} = \frac{1\pm\sqrt{3}i}{2} \implies \underbrace{\sqrt{\frac{1\pm \sqrt{3}i}{2}} = \frac{\sqrt{3}\pm i}{2}}_{\text{assume taking principal branch of } \sqrt{\;}}
\\
{\Huge\Downarrow}
\\
\left(x \pm \sqrt{\frac{1 + \sqrt{3}i}{2}}\right)
\left(x \pm \sqrt{\frac{1 - \sqrt{3}i}{2}}\right)\\
\| \\
x^2 \pm \left(\frac{\sqrt{3} + i}{2} + \frac{\sqrt{3} - i}{2}\right) +
\left(\frac{\sqrt{3} + i}{2}\right)\left(\frac{\sqrt{3} - i}{2}\right)\\
\| \\
x^2 \pm \sqrt{3}x + 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt{2}+\sqrt{3}$ is irrational. Problem: Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
The book where I encountered this problem had the following hint:
We make a polynomial with integer coefficients called $f(x)$ that $f(\sqrt{2}+\sqrt{3})=0$. (Why?)
Accepting this I solved the problem like this:
If $x=\sqrt{2}+\sqrt{3}$ then $x^2=5+2\sqrt{6}$ and so $(x^2-5)^2=24$ thus:
$$x^4-10x^2-1=0$$
But I want to know the reason that we should do this.
| More generally,
suppose
$r
=\sqrt{a}+\sqrt{b}
$
is rational,
where
$a$ and $b$
are positive integers.
Then
$r(\sqrt{a}-\sqrt{b})
=a-b
$
so
$\sqrt{a}-\sqrt{b}
=\dfrac{a-b}{r}
$
is also rational.
Adding and subtracting these,
$\sqrt{a}$ and $\sqrt{b}$
are rational.
Therefore,
if either or both
of
$\sqrt{a}$ and $\sqrt{b}$
are irrational,
then
$\sqrt{a}+\sqrt{b}$
is irrational
(and similarly for
$\sqrt{a}-\sqrt{b}$).
This might form a basis
for a proof that
$\sum_{k=1}^n \sqrt{a_k}
$
is irrational
under suitable assumptions on the
$a_k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?
| When you have this:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
It it incorrect to have this next:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
Just like when you have (for example) $3 \cdot 2x = x$, it is incorrect to then divide both sides by $x$ and end up with $6 = 1$ (you lose the solution $x = 0$ when you do this).
The correct next step is to subtract $\dfrac{2\tan x}{1-\tan^2 x}$ from both sides and continue as follows:
\begin{align}
-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} &= \frac{2\tan{x}}{1-\tan^2{x}}\\[0.3cm]
-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} - \frac{2\tan{x}}{1-\tan^2{x}} &= 0\\[0.3cm]
\frac{2\tan{x}}{1-\tan^2{x}} \left(-\tan^2 x - 1\right) &= 0
\end{align}
So then we have the following two equations:
$$\frac{2\tan x}{1-\tan^2 x} = 0 \qquad \text{or} \qquad -\tan^2 x - 1 = 0$$
The first equation is trivial. The second one is perhaps even more so because the second one has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Show that, for all $n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$ I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:
Show that, for all $n > 1$ $$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$$
I tried to follow the below steps in order to prove the RHS inequality:
Proving that $f < g$ on $I$ from $a$ to $b$:
Step $1$. Prove that $f' < g'$ on $\operatorname{Int}(I)$.
Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$
Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x > 1$. One has
$$f'(x) = -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$
We note that, for every $x > 1$, $f'(x) > g'(x)$. Moreover, $f(1^+) = \log(2) < 1 = g(1^+).$
My problem is that I got the wrong inequality sign in Step $1$.
Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.
| It's given in the question that $n>1$. I'm assuming $n\in\mathbb Z$.
$x_n=\left(1+\frac{1}{n}\right)^n$ is strictly increasing. Proof: $x_n>0,$ $\forall n$.
$$\frac{x_{n+1}}{x_n}=\frac{n+2}{n+1}\left(1-\frac{1}{(n+1)^2}\right)^n>$$
By Bernoulli's inequality:
$$>\frac{n+2}{n+1}\left(1-\frac{n}{(n+1)^2}\right)=$$
$$=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1$$
By Binomial theorem: $$\left(1+\frac{1}{n}\right)^n=1+\frac{1}{n}+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^2+$$
$$+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^3+\cdots+\left(\frac{1}{n}\right)^n<$$
$$<1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}<$$
$$<1+1+\frac{1}{2}+\frac{1}{2\cdot 3}+$$
$$+\frac{1}{3\cdot 4}+\cdots+\frac{1}{(n-1)n}=$$
Use the formula/identity $\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}$, $\forall k$, $k\neq 0$, $k\neq 1$.
$$=1+1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+$$
$$+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n-1}-\frac{1}{n}=$$
$$=3-\frac{1}{n}<3$$
Therefore $x_n$ converges. $e$ is sometimes defined as:
$$\lim_{n\to +\infty} x_n=e$$
$y_n=\left(1+\frac{1}{n}\right)^{n+1}$ is strictly decreasing. Proof: $y_n>0,$ $\forall n$.
$$\frac{y_{n-1}}{y_n}=\frac{n}{n+1}\left(1+\frac{1}{n^2-1}\right)^n>$$
By Bernoulli's inequality:
$$>\frac{n}{n+1}\left(1+\frac{n}{n^2-1}\right)=$$
$$=\frac{n^3+n^2-n}{n^3+n^2-n-1}>1$$
because $n^3+n^2-n-1>0$ because $n^3>n$, $n^2>n$, $n>1$.
$$\lim_{n\to +\infty}y_n=\lim_{n\to +\infty}\left(x_n\left(1+\frac{1}{n}\right)\right)=$$
$$=e(1+0)=e$$
$$x_n<e<y_n$$
$$\ln x_n< 1<\ln y_n$$
$$n\ln\left(1+\frac{1}{n}\right)<1<(n+1)\ln\left(1+\frac{1}{n}\right)$$
$$\frac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\frac{1}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Solving an Absolute Value Inequality The Question:
$|x^2+x-1|-|3x+1|<x^2-2$
I don't really know how to work with absolute values, besides basic ones.
Here is my attempt at simplifying, can someone tell me where I messed up and give some hints on how to continue?
$|x^2+x-1|-|3x+1|<x^2-2$
$|x^2+x-1|<x^2-2+|3x+1|$
$x^2+x-1<x^2-2+3x+1$ or $x^2+x-1>-x^2+2-3x-1$
$2x>1$ or $x^2+2x-1>0$
Answer:
$x<-1 $ or $x>-1+\sqrt{2}$
| We have four cases which depend on the sign of $(x^2+x-1)$ and $(3x+1)$ (the arguments of the absolute values).
$1)$ If $x\geq -1/3$ then you have to solve
$$|x^2+x-1|-(3x+1)<x^2-2$$
or $|x^2+x-1|<x^2+3x-1$.
Two subcases:
$1.1)$ For $-1/3\leq x< (\sqrt{5}-1)/2$, it becomes
$$-(x^2+x-1)<x^2+3x-1$$
or $0<2x^2+4x-2$ which holds in this interval for $\sqrt{2}-1<x$.
$1.2)$ For $x\geq (\sqrt{5}-1)/2$, we have
$$(x^2+x-1)<x^2+3x-1$$
or $0<2x$, $x>0$ which always satisfied in this interval.
$2)$ If $x< -1/3$ then you have to solve
$$|x^2+x-1|+(3x+1)<x^2-2$$
or $|x^2+x-1|<x^2-3x-3$.
Two subcases:
$2.1)$ Now for $-(\sqrt{5}+1)/2<x<-1/3 $, it becomes
$$-(x^2+x-1)<x^2-3x-3.$$
or $0<2x^2-2x-4$ which holds in this interval for $x<-1$.
$2.2)$ For $x\leq -(\sqrt{5}+1)/2$, we have
$$(x^2+x-1)<x^2-3x-3.$$
or $0<-2x-2$, $x<-1$ which always satisfied in this interval.
Finally, putting all together, we have that the solution is:
$$\mbox{$x>\sqrt{2}-1$ (1.1 and 1.2) or $x<-1$ (2.1 and 2.2).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$ $$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$
So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for $f^{(15)}(0)$ we get the result of $2$?
| It is more precise to write $$\begin{align*} f(x) &= \sum_{n=0}^\infty 2\left(-\frac{x}{2}\right)^n + 6x(-x^2)^n \\
&= \sum_{n=0}^\infty 2 (-1/2)^n x^n + \sum_{m=0}^\infty 6(-1)^m x^{2m+1} \\
&= \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k.
\end{align*}$$ Thus, $f^{(15)}(0)$ is $15!$ times the coefficient of $x^{15}$ in the series expansion. Note this corresponds to the choices $n = 15$ and $m = 7$; i.e., $$f^{(15)}(0) = 15! \left( 2 (-1/2)^{15} + 6 (-1)^{7} \right).$$ This is because the coefficient of $x^{15}$ is found at different values of the indices of the series expansion of each term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How do I determine the values of $k$ for which $3x^2 + kx+12 = 0$ has no real solutions, $1$ real solution, and $2$ real solutions? I know that a point has to be $(0,12)$ just because of the $C$ term always being the y-intercept.
Please just tell me how to approach the problem/give tips, so that you are not doing my homework for me.
| By the distributive property the equation is equivalently:
$$0=3\left(x^2+\frac{k}{3}x\right)+12$$
Subtracting $12$ from both sides of the equation.
$$-12=3\left(x^2+\frac{k}{3}x\right)$$
Dividing by $3$ on both sides of the equation.
$$-4=x^2+\frac{k}{3}x$$
Using the fact that $\left(x+\frac{y}{2}\right)^2=x^2+2\left(x\right)\left(\frac{y}{2}\right)+\frac{y^2}{4}$:
$$-4=\left(x+\frac{k}{6}\right)^2-\frac{k^2}{36}$$
$$\frac{k^2}{36}-4=\left(x+\frac{k}{6}\right)^2$$
Note now that if $u^2=c$ then $c=0$ gives one solution $0$, $c>0$ gives $2$ solutions $\pm \sqrt{c}$, and $c<0$ gives no real solutions.
With a little more thinking we can see that If we want $1$ real root we need:
$$\frac{k^2}{36}-4=0$$
If we want $2$ real solutions we need:
$$\frac{k^2}{36}-4>0$$
And finally no real solutions:
$$\frac{k^2}{36}-4<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A confusing question related to three variables with fractions.. Recently, I had a mock-test of a Mathematics Olympiad. There was a question which not only I but my friends too were not able to solve. The question goes like this:
If,
$$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c} $$
Then what is the value of
$$ \frac{1}{a^5} + \frac{1}{b^5} + \frac{1}{c^5} $$
To, solve this question, I used a variety of ways like:
1) transposing variables in the first equation and,
2) putting a whole power of five to both the sides of the equation one. But, I was unable to find the solution.
The options were -- (a) 1 , (b) 0 , (c) $ \frac{1}{a^5 + b^5 + c^5} $ , (d) None of them.
So, I require any possible help. And, a complete answer would be most welcome. Thanks in advance.
| Options $1$ and $2$ are wrong, take $a=1,b=-1,c=-1$.
Now notice given non-zero real number $a,b,c$ we have:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\overbrace{\iff}^{\text{multiply by a+b+c}} 3+2(ab+bc+ac)=1\iff ab+bc+ac=-1$.
Analogously we have $\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^5+b^5+c^5}\iff a^5b^5+b^5c^5+a^5c^5=-1$.
Taking $a=2,b=2,c=\frac{-5}{4}$ satisfies $ab+bc+ac=-1$ but not $a^5b^5+b^5c^5+a^5c^5=-1$.
So the answer is $d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Ugly Differentiation If $$y=\frac{ax^2}{(x-a)(x-b)(x-c)} + \frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$$
Then prove that $$\frac{dy}{dx}=\frac{y}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$$
What is the smartest way to solve this?
| HINT:
$$\dfrac c{x-c}+1=\dfrac x{x-c}$$
$$\dfrac{bx}{(x-b)(x-c)}+\dfrac x{x-c}=x\cdot\dfrac{b+(x-b)}{(x-b)(x-c)}=\dfrac{x^2}{(x-b)(x-c)}$$
$$\frac{ax^2}{(x-a)(x-b)(x-c)}+\dfrac{x^2}{(x-b)(x-c)}=x^2\cdot\frac{a+(x-a)}{(x-a)(x-b)(x-c)}=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
Now take logarithm in both sides and differentiate
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve a system of quadratic equations in two variables for integral solutions. Let's say I have two equations like:
$x^2 - y^2 = 6$ or
$x^2 + y^2 = 6$
What is the best way to solve these sort of equations for finding only positive integral solutions?
| As a starting point, I suggest plotting the solutions to each equation to see where the intersections (if any) between them are. For the pair of equations given in the question, we see that there appear to be two points where the circle $x^2 + y^2 = 6$ (red) intersects with the hyperbola $x^2 - y^2 = 6$ (blue), and the intersection is at approximately $x = \pm 2.4, y=0$.
You can find the solutions analytically by adding and subtracting the two equations:
$$(x^2 + y^2) + (x^2 - y^2) = 6 + 6 \implies x^2 = 6 \implies x = \pm \sqrt{6} \approx 2.449$$
and
$$(x^2 + y^2) - (x^2 - y^2) = 6 - 6 \implies y^2 = 0 \implies y = 0.$$
Therefore, the solutions to the system of equations are $(\pm \sqrt{6}, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $a$ and $b$ are prime to each other and $n$ is prime then$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor. Question is:
If $a$ and $b$ are prime to each other and $n$ is prime then prove that $\frac{a^n+b^n}{a+b}$ and
$a+b$ have no common factor unless $a+b$ is a multiple of $n$.
This is what I thought so far:
$a^n$ is also prime to $b^n$ because of the theorem (If $a$ is prime to $b$ then $a^n$ is also prime to $b^n$), Also $a^n$ and $b^n$ is a divisor of $(a+b)$ therefore by theorem (If $a$ is prime to $b$, and each of these numbers is a divisors of $N$, then $ab$ is a divisor of $N$) $a^nb^n$ is a divisor of $(a+b)$.
Thereafter no idea, How should I proceed further if I'm thinking in a right way and If not, How should I prove this?
Source: Higher Algebra by Barnard and Child
| Counterexample: $a=1, b=2, n=9$.
$a$ and $b$ are coprime, and $a+b$ is not a multiple of $n$, but $(a^n + b^n)/(a+b) = 171$ and $a+b$ have a common factor of $3$. I would guess that the question should read "unless $a+b$ has a common factor with $n$", or else $n$ should be prime, or "has no common factor with" should be replaced by "is not divisible by" as in DougM's answer.
Edit: Assuming $n$ is prime, then the statement holds, by a routine calculation rather similar to DougM's answer:
We rule out the possibility that $n$ is even, since if $a+b \mid a^n+b^n$ then $a+b$ divides both $2a^n$ and $2b^n$, hence it divides $(2a^n,2b^n) = 2$, forcing $a=b=1$ which is trivial to verify.
If $n$ is odd, then $(a^n+b^n)/(a+b)$ may be written as $\sum_{i=1}^n a^{n-i} (-b)^{i-1}$. Evaluating this mod $a+b$, we see that every term is congruent to $a^{n-1}$ so the sum simplifies to $na^{n-1}$. We know that $a+b$ and $a$ are coprime so $(a+b, (a^n+b^n)/(a+b)) = (a+b, na^{n-1}) = (a+b,n)$. The claim follows immediately from the fact that if $n$ is prime then $(a+b,n) = 1$ iff $n$ does not divide $a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}$. In an attempt to answer this thread, I discovered an identity involving binomial coefficients. However, I am not able to find a proof. All tricks are welcome.
Let $n$ and $k$ be nonnegative integers with $k\leq n$. Prove that $$\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}\,.$$
| It's convenient to use the coefficient of operator $[z^j]$ to denote the coefficient of $z^j$ in a series. This way we can write e.g.
\begin{align*}
[z^j](1+z)^n=\binom{n}{j}
\end{align*}
We recall the Euler transformation formula of a series
\begin{align*}
A(z)=\sum_{n= 0}^\infty a_nz^n\qquad\qquad
\frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^n\binom{n}{j}a_j\right)z^n
\end{align*}
This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev.
This formula is useful, but it can be considerably generalized
\begin{align*}
\text{from}&\qquad&\sum_{j=0}^n\binom{n}{j}a_j&=[z^n]\frac{1}{1-z}A\left(\frac{z}{1-z}\right)\\
\text{to}&\qquad&\sum_{j=0}^n\binom{n+pj}{m+qj}a_j&=[z^n]\frac{z^m}{(1-z)^{m+1}}A\left(\frac{z^{q-p}}{(1-z)^q}\right)\qquad q>p\tag{1}
\end{align*}
A proof of (1) can be found in Theorem 2.1 in Riordan arrays and combinatorial sums by R. Sprugnoli.
We can now show:
The following is valid
\begin{align*}
\sum_{j=k}^n(-1)^{j-k}\binom{j}{k}\binom{2n-j}{j}2^{2(n-j)}=\binom{2n+1}{2k+1}\tag{2}
\end{align*}
$$ $$
We start by rearranging the left hand side of (2)
\begin{align*}
\sum_{j=k}^n&(-1)^{j-k}\binom{j}{k}\binom{2n-j}{j}2^{2(n-j)}\\
&=(-1)^k4^n\sum_{j=k}^n\binom{2n-j}{j}\binom{j}{k}\left(-\frac{1}{4}\right)^j\tag{3}
\end{align*}
In order to use (1) we derive a generating function $A(z)$ for
\begin{align*}
a_j=\binom{j}{k}\left(-\frac{1}{4}\right)^j
\end{align*}
We obtain
\begin{align*}
A(z)&=\sum_{j= 0}^\infty a_jz^j\\
&=\sum_{j=k}^\infty\binom{j}{k}\left(-\frac{1}{4}\right)^jz^j\\
&=\sum_{j=0}^\infty\binom{j+k}{j}\left(-\frac{1}{4}\right)^{j+k}z^{j+k}\tag{4}\\
&=\left(-\frac{z}{4}\right)^k\sum_{j=0}^\infty\binom{-k-1}{j}\left(\frac{z}{4}\right)^{j}\tag{5}\\
&=\frac{\left(-\frac{z}{4}\right)^k}{\left(1+\frac{z}{4}\right)^{k+1}}\tag{6}
\end{align*}
Comment:
*
*In (4) we shift the index so that $j$ starts from $0$.
*In (5) we apply the binomial identity $\binom{-u}{v}=\binom{u+v-1}{v}(-1)^v$.
*In (6) we apply the binomial series expansion.
Since we have derived $A(z)$ we can proceed with the main calculation.
We obtain from (3) and (6)
\begin{align*}
(-1)^k4^n&\sum_{j=k}^n\binom{2n-j}{j}\binom{j}{k}\left(-\frac{1}{4}\right)^{j}\\
&=(-1)^k4^n[z^{2n}]\frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)\tag{7}\\
&=(-1)^k4^n[z^{2n}]\frac{1}{1-z}\left(-\frac{z^2}{4(1-z)}\right)^k\frac{1}{\left(1+\frac{z^2}{4(1-z)}\right)^{k+1}}\tag{8}\\
&=4^{n-k}[z^{2n-2k}]\frac{1}{\left(1-z+\frac{z^2}{4}\right)^{k+1}}\tag{9}\\
&=4^{n-k}[z^{2n-2k}]\frac{1}{\left(1-\frac{z}{2}\right)^{2(k+1)}}\\
&=4^{n-k}[z^{2n-2k}]\sum_{j=0}^\infty\binom{-2k-2}{j}\left(-\frac{z}{2}\right)^j\tag{10}\\
&=4^{n-k}[z^{2n-2k}]\sum_{j=0}^\infty\binom{2k+j-1}{j}\left(\frac{z}{2}\right)^j\tag{11}\\
&=4^{n-k}\binom{2n+1}{2n-2k}\frac{1}{4^{n-k}}\tag{12}\\
&=\binom{2n+1}{2k+1}
\end{align*}
and the claim follows.
Comment:
*
*In (7) we use the representation (1) with the parameter settings
$
(n,m,p,q)\rightarrow (2n,0,-1,1)
$.
*In (8) we use the representation (6) of $A(z)$.
*In (9) we do some simplifications and use the rule $[z^{u+v}]A(z)=[z^u]z^{-v}A(z)$.
*In (10) we apply the binomial series expansion.
*In (11) we use the binomial identity $\binom{-u}{v}=\binom{u+v-1}{v}(-1)^v$.
*In (12) we select the coefficient of $z^{2n-2k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
If $t = \frac{(2-x)(3-x)}{3x^2 -7x + 6 }$, then find $\frac{-x^4 + 2x^3 - 3x^2 + 2x}{3x^2 -7x + 6 }$? Suppose that $t =\frac{(2-x)(3-x)}{3x^2 -7x + 6 }$. Can you, please, give me any hints to express the following in terms of $t$:
$$\frac{-x^4 + 2x^3 - 3x^2 + 2x}{3x^2 -7x + 6 }?$$
| 1-Firstly, expand numerator of the given equation like
$$t=\frac{x^2-5x+6}{3x^2-7x+6}$$
2-Don't do anything with the denominator because denominator is same in the required answer.
3-Multiply $t$ by $-x^2$, to get $-x^4 $ term.
4-Then, make the same terms which you required by adding and subtracting terms.
5-At the end, you have$$-x^2t+\frac{-3x^3+3x^2+2x}{3x^2-7x+6}=\frac{-x^4+2x^3-3x^2+2x}{3x^2-7x+6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$.
Prove the given trigonometric identity:
$$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$
My Approach:
\begin{align*}
\text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\
&=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\\
&=\cos^32\beta+3\cos^2\beta\cdot\sin^2\beta\cdot\cos2\beta
\end{align*}
Please help me to continue further.
| $$(\cos^2\beta)^3 -(\sin^2\beta)^3=(\cos^2\beta-\sin^2\beta)(\cos^4\beta +\sin^4\beta +\cos^2\beta\sin^2\beta)$$
$$=\cos2\beta(1-\sin^2\beta\cos^2\beta)$$
$$=\cos2\beta\left(1-\frac{\sin^2 2\beta}{4}\right)$$
$$=\cos2\beta\left(1+\frac{\cos 4\beta -1}{8}\right)$$
Now use $2\cos A.\cos B=\cos(A+B) +\cos(A-B)$
$$=\frac18\left(7\cos2\beta +\cos 2\beta\cos 4\beta\right)$$
$$=\frac18\left(7\cos 2\beta +\frac{\cos 6\beta + \cos 2\beta}{2}\right)$$
$$=\frac{1}{16}\left(15\cos 2\beta + \cos 6 \beta\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to find the center of a circle and its radius in a 3D space given 3 points I'm working on writing a code in Octave (C++) for a helical spring. I need to figure out the center line of this spring in an effort to find any trends between different spring platforms as the geometry changes throughout the manufacturing processes.
I have a file with the center line of the wire, which is made up of 2000 or so points. Since I learned that you need 3 points to make a circle, I'm figuring i will be able to approach it that way (but haven't figured out how) so....
Given 3 points in 3 Dimensional space, how could I find the center and the radius of a circle in a code friendly manor.
Thanks
| Let $A = (A_1, A_2, A_3)^T, \,\, B = (B_1, B_2, B_3)^T, \,\, C = (C_1, C_2, C_3)^T$ be the three points and $X = (x,y,z)^T$ be the center we are looking for. The point $X$ lies on the two planes passing through the midpoints of the segments $AB$ and $AC$. In fact $X$ also lies on the plane passing through the midpoint of segment $BC$, the three planes intersecting at the common line passing trough the center of the circle through $A, B, C$ and orthogonal to the plane defined by the three points. Also, $X$ lies on the plane through $A, B, C$. Thus, $X$is the unique intersection point of three planes, which constitute a system of three linear equations with three unknown variables $X = (x, y, z)^T$. Let us write the equations of the three planes:
\begin{align}
\overrightarrow{AB} \cdot \left(X - \frac{1}{2}(B+A)\right) &= 0\\
\overrightarrow{AC} \cdot \left(X - \frac{1}{2}(C+A)\right) &= 0\\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big) \cdot \left(X - A\right) &= 0
\end{align}
which turn into
\begin{align}
(B - A)^T \left(X - \frac{1}{2}(B+A)\right) &= 0\\
(C - A)^T \left(X - \frac{1}{2}(C+A)\right) &= 0\\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big) \cdot \left(X - A\right) &= 0
\end{align}
and as a linear system it looks like this:
\begin{align}
2(B - A)^T X &= (B - A)^T (B+A) \\
2 (C - A)^T X &= (C - A)^T (C+A) \\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)^T X &= \Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)^T A
\end{align}
So if we put the vectors multiplying $X$ in as rows in the $3 \times 3$ matrix
\begin{align}
M=
\left[ {\begin{array}{c}
2 (B - A)^T \\
2 (C - A)^T \\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)^T
\end{array} } \right] = \left[ {\begin{array}{c c c}
2 B_1 - 2 A_1 & 2 B_2 - 2 A_2 & 2 B_3 - 2 A_3\\
2 C_1 - 2 A_1 & 2 C_2 - 2 A_2 & 2 C_3 - 2 A_3 \\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_1 & \Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_2 & \Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_3
\end{array} } \right]
\end{align}
and the numbers in the right hand side of the equations into a $3 \times 1$ row vector
\begin{align}
K&=
\left[ {\begin{array}{c}
(B - A)^T (B+A) \\
(C - A)^T (C+A) \\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)^T A
\end{array} } \right] = \left[ {\begin{array}{c}
B^TB - A^TA\\
C^TC - A^TA \\
\det \Big|\overrightarrow{AB} \,\, \overrightarrow{AC} \,\, A \Big|
\end{array} } \right] \\ &= \left[ {\begin{array}{c}
B_1^2 + B_2^2 + B_3^2 - A_1^2 - A_2^2 - A_3^2\\
C_1^2 + C_2^2 + C_3^2 - A_1^2 - A_2^2 - A_3^2\\
\Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_1 A_1 + \Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_2 A_2 + \Big(\overrightarrow{AB} \times \overrightarrow{AC}\Big)_3
A_3 \end{array} } \right]
\end{align}
Then the center of the circle is $$X = M^{-1}K$$ and the radius is $$R = \sqrt{(X - A)^T(X-A)} = \sqrt{(x-A_1)^2 + (y-A_2)^2 + (z-A_3)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove that $a^3+b^3+c^3=1+3abc$, where $a$, $b$ and $c$ are power series. Let $a=1+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\ldots$, $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\ldots$ and $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\ldots$ I want to show that $a^3+b^3+c^3=1+3abc$.
I tried to identify which functions $a$, $b$ and $c$ represent. For instance, $a=\sinh x-x+1$, but I don't know who $b$ and $c$ are. I also thought about using the equality $a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-ac-bc)(a+b+c)$, but obtained nothing.
I would like you to give me a hint to solve this question.
| Note that $a = \sinh x - x + 1$ and $$b' = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + \cdots$$ Now we can see that $$e^{w x} = 1 + w x + \frac{w^{2}x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots = b' + wb + w^{2}c$$ and $$e^{w^{2}x} = b' + w^{2}b + wc$$ and $$e^{x} = b' + b + c$$ so that $$b = \frac{w^{2}e^{wx} + we^{w^{2}x} + e^{x}}{3}$$ and $$c = \frac{we^{wx} + w^{2}e^{w^{2}x} + e^{x}}{3}$$
Note that $w$ above denotes a cube root of unity with $w \neq 1$. It is easily seen that if $d = b'$ then we have $$b^{3} + c^{3} + d^{3} - 3bcd= (d + c + b)(d + cw + bw^{2})(d + cw^{2} + bw) = e^{x + wx + w^{2}x} = 1$$ Thus $b^{3} + c^{3} + d^{3} = 1 + 3bcd$. I think the expression for $a$ should be equal to $d = 1 + x^{3}/3! + x^{6}/6! + \cdots$ instead of $1 + x^{3}/3! + x^{5}/5! + \cdots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$ I have this equation that I have to solve:
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$
Maybe it involves means? The square roots are leaning towards it... Could someone help me?
| the inequality is given by
$$\sqrt{(3-x)}>1/2+\sqrt{x+1}$$ and we have $$-1\le x\le 3$$
after squaring we get
$$\frac{7}{4}-2x>\sqrt{x+1}$$
Now it must be $$\frac{7}{8}>x\geq -1$$
squaring again we have
$$x^2-2x+\frac{33}{64}>0$$
Can you proceed?
solving the last inequality we obtain $$-1\le x<\frac{1}{8}(8-\sqrt{31})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Problem with Implicit Differenation Problem:
Find $\frac{d^{2}y}{dx^2}$ in terms of $x$ and $y$.
\begin{equation*}
y^2 = x^3
\end{equation*}
Answer:
\begin{eqnarray*}
2y \frac{dy}{dx} &=& 3x^2 \\
\frac{dy}{dx} &=& \frac{3x^2}{2y} \\
\frac{d^2y}{dx^2} &=& \frac{2y(6x) - 3x^2(2)}{4y^2} \\
\frac{d^2y}{dx^2} &=& \frac{6xy - 3x^2}{2y^2} \\
\frac{d^2y}{dx^2} &=& \frac{6xy - 3x^2}{2x^3} \\
\end{eqnarray*}
However, the back of the book gets:
\begin{equation*}
\frac{d^{2}y}{dx^2} = \frac{3x}{4y}
\end{equation*}
I am hoping that somebody here can point out my mistake.
| The issue started in the third step: the numerator should be $2y(6x) - 3x^2(2\, \color{red}{dy/dx})$, not $2y(6x) - 3x^2(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real-Analysis Methods to Evaluate $\int_0^\infty \frac{x^a}{1+x^2}\,dx$, $|a|<1$.
In THIS ANSWER, I used straightforward contour integration to evaluate the integral $$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^a}{1+x^2}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)}$$for $|a|<1$.
An alternative approach is to enforce the substitution $x\to e^x$ to obtain
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\
&=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\
&=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\
&=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\
&=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \tag 1\\\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2
\end{align}$$
Other possible ways forward include writing the integral of interest as
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{0}^1 \frac{x^{a}+x^{-a}}{1+x^2}\,dx
\end{align}$$
and proceeding similarly, using $\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$.
Without appealing to complex analysis, what are other approaches one can use to evaluate this very standard integral?
EDIT:
Note that we can show that $(1)$ is the partial fraction representation of $(2)$ using Fourier series analysis. I've included this development for completeness in the appendix of the solution I posted on THIS PAGE.
| I'll assume $\lvert a\rvert < 1$. Letting $x = \tan \theta$, we have
$$\int_0^\infty \frac{x^a}{1 + x^2}\, dx = \int_0^{\pi/2}\tan^a\theta\, d\theta = \int_0^{\pi/2} \sin^a\theta \cos^{-a}\theta\, d\theta$$
The last integral is half the beta integral $B((a + 1)/2, (1 - a)/2)$, Thus
$$\int_0^{\pi/2}\sin^a\theta\, \cos^{-a}\theta\, d\theta = \frac{1}{2}\frac{\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{a+1}{2} + \frac{1-a}{2}\right)} = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)$$
By Euler reflection,
$$\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \pi \csc\left[\pi\left(\frac{1+a}{2}\right)\right] = \pi \sec\left(\frac{\pi a}{2}\right)$$
and the result follows.
Edit: For a proof of Euler reflection without contour integration, start with the integral function $f(x) = \int_0^\infty u^{x-1}(1 + u)^{-1}\, du$, and show that $f$ solves the differential equation $y''y - (y')^2 = y^4$, $y(1/2) = \pi$, $y'(1/2) = 0$. The solution is $\pi \csc \pi x$. On the other hand, $f(x)$ is the beta integral $B(1+x,1-x)$, which is equal to $\Gamma(x)\Gamma(1-x)$. I believe this method is due to Dedekind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 5,
"answer_id": 2
} |
RMO inequality problem If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?
| Using Cauchy-Scwarz to
\begin{align*}
\frac{a_1}{\sqrt{x_1}}, \frac{a_2}{\sqrt{x_2}}, \ldots, \frac{a_n}{\sqrt{x_n}} \\
\sqrt{x_1}, \sqrt{x_2}, \ldots, \sqrt{x_n}
\end{align*}
we get
\begin{align*}
\frac{a_1^2}{x_1}+\frac{a_2^2}{x_2}+\cdots + \frac{a_n^2}{x_n} \geq \frac{(a_1+a_2+\cdots + a_n)^2}{x_1+x_2+\cdots+x_n}
\end{align*}
Hence we have
\begin{align*}
\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{d^2}{e-1}+\frac{c^2}{a-1}+\frac{e^2}{d-1} \geq \frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}
\end{align*}
Putting $x=a+b+c+d+e$, we need to prove
\begin{align*}
\frac{x^2}{x-5} \geq 20
\end{align*}Since $x > 5$, this can be written as $x^2 -20x + 100 \geq 0$. This follows readily since
\begin{align*} x^2-20x+100 = (x-10)^2\end{align*}
The proof also shows that the numerators can be any permutation of $a^2,b^2,c^2,d^2,e^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
The inequality $p<\frac{a+m_b}{b+m_a}
Let in triangle $ABC$ $BC=a$ and $AC=b$, $m_a, m_b -$ the medians drawn from the vertex $A$ and $B$. Find the greatest real number $p$ and the lowest real $q$, such that inequality
$$p<\frac{a+m_b}{b+m_a}<q$$
hold.
My attempt:
I used formula:
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$
| Hint for $p=\frac{1}{4}$:
We know
\begin{align}
p<\frac{a+m_b}{b+m_a}
\end{align}
holds for all triangles. Fix the corners $A,C$ of the triangle and let $B$ to converge to $C$. Since we always have
\begin{align*}
b + \frac{a}{2} > m_a > b - \frac{a}{2} \\
\frac{b}{2} + a > m_b > \frac{b}{2} - a
\end{align*}
when $a \rightarrow 0$ we are going to have
\begin{align*}
m_a &\rightarrow b \\
m_b &\rightarrow \frac{b}{2} \\
a &\rightarrow 0
\end{align*}
So we should have
\begin{align}
p &\leq \frac{b/2}{b+b} \\
p &\leq \frac{1}{4}
\end{align}
If $p > \frac{1}{4}$ you can find some $a,b$ such that the inequality doesn't satisfy.
Edit: Verifying $p=\frac{1}{4}$ (with some cheap inequalities)
Case-1: $b\geq m_a$
We know $m_b > \frac{b}{2} - a$, therefore
\begin{align*}
4m_b + 4a &> 2b - 4a + 4a \\
&> 2b > b + m_a
\end{align*}
Case-2: $b < m_a$
Call the intersection of medians from $A,B$ as $G$. Define the foot of the median from $B$ as $E$. The triangle inequality in $AGE$ gives us $\frac{m_b}{3} > \frac{2m_a}{3} - \frac{b}{2}$. Using this we have
\begin{align*}
4m_b + 4a &> 8m_a - 6b +4a \\
&> 8m_a - 6b > m_a + b
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1$ for positive integers $x,y$? I have to prove that
$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1\;?$$
For any positive integers $x, y$.
Can anyone help me?
| Let $x=py+q$, with $0\le q<y$.
Then if $q=0$,
$$\left\lceil\frac{x}y\right\rceil=p,\left\lfloor\frac{x-1}y\right\rfloor+1=p+\left\lfloor\frac{y-1}y\right\rfloor+1=p.$$
and if $q>0$,
$$\left\lceil\frac{x}y\right\rceil=p+1,\left\lfloor\frac{x-1}y\right\rfloor+1=p+\left\lfloor\frac{q-1}y\right\rfloor+1=p+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Can someone give me a hint on how to solve this problem? Problem with indefinite integral, can someone give a hint to do right substitute?
$$ \int \left(\frac{x^{\frac{1}{3}}-1}{x}\right)^{\frac{5}{3}}dx$$
| By setting $x=y^3$ we have $dx=3y^2dy$ and
$$ \int\left(\frac{x^{1/3}-1}{x}\right)^{5/3}\,dx = 3\int \frac{(y-1)^{5/3}}{y^3}\,dy $$
By setting $y=1+z^3$ the last integral turns into:
$$ 9 \int \frac{z^7}{(1+z^3)^3}\,dz $$
and by partial fraction decomposition (yes, that is tedious but algorithmic) the last integral equals
$$\small \frac{1}{6} \left[\frac{9 z^2}{\left(1+z^3\right)^2}-\frac{24 z^2}{1+z^3}+10 \sqrt{3}\arctan\left(\frac{2z-1}{\sqrt{3}}\right)-10\log(z+1)+5\log\left(1-z+z^2\right)\right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| Your first way is wrong, and the mistake is that you in fact argued that
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1\implies \lim_{x\to0}\sin^2x=x^2$$
and this is, of course, wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Maximizing hyper-cylinder volume I would like to calculate the volume of a $n$-dimensional hyper-cylinder which is inside a unit $n$-dimensional hyper-sphere. Any ideas on how to approach this problem?
| Volume of a n-hypersphere:
$V = \frac {\pi^{\frac n2}}{\Gamma(\frac n2)} R^n$
Volume of a n-hypercyilnder: volume of the n-1 hpersphere - height.
equation of a hyperspehre:
$x_1^1 + x_2^2 + \cdots x_n^2 = R^2$
Hyper-cylinder:
$x_2^2 + x_3^2 + \cdots x_{n}^2 = r^2$
subtracting one from the other
$x_1^2=R^2-r^2$
with $x_1$ as our height.
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} r^{n-1} \sqrt {R^2 - r^2}$
To maximize V.
$\frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})}$ is a constant. And we can treat $R$ as constant and find $r$ in terms of $R.$
$\frac {dv}{dt} = $$(n-1) r^{n-2} \sqrt {R^2 -r^2} - r^{n}(\sqrt {R^2 -r^2})^{-1} = 0\\
(n-1) (R^2 -r^2) = r^2\\
r = \frac {R}{\sqrt n}$
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} \frac {R^n}{n^{\frac{n}2}} \sqrt {n-1}$
Since this is a unit hypersphere $R = 1$
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} \frac { (n-1)^\frac 12}{n^{\frac{n}2}} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality $\frac{ab}c+\frac{ac}b+\frac{bc}a\ge a^2+b^2+c^2-abc+1$
Let $a,b,c \in [10;100]$. Prove the inequality
$$\frac{ab}c+\frac{ac}b+\frac{bc}a\ge a^2+b^2+c^2-abc+1$$
My work so far:
$$\frac{ab}c+\frac{ac}b+\frac{bc}a\ge a^2+b^2+c^2-abc+1 \Leftrightarrow$$
$$\Leftrightarrow (ab-c)(ac-b)(bc-a)\ge0$$
| $ab-c\geq0$, $ac-b\geq0$, $bc-a\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(b+c)^2\geq a^2+4\cdot AD^2$
Let $ABC$ be a triangle and $AD$ be the altitude through $A$. Prove that $$(b+c)^2\geq a^2+4\cdot AD^2$$
(where $a=BC$, $b=CA$, $c=AB$).
I used Apollonius theorem and Pythagoras theorem every where. I guess that we can do it using these two theorems but I can't process.
| Suppose $DB=x$ and $DC=y$. Then $a\leq x+y$ (since $a=x+y$ or $a=|x-y|$), $b=\sqrt{x^2+AD^2},c=\sqrt{y^2+AD^2}$. It suffices to show $\sqrt{(x^2+AD^2)(y^2+AD^2)}\geq xy+AD^2$, which follows from Cauchy-Schwarz inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$ Q: Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$.
My working:
$(x+x^2+x^3+...)^3$
$=x^3(1+x+x^2+...)^3$
$= x^3 \sum_{r=0}^\infty \begin{pmatrix} {r+3-1}\\{r}\end{pmatrix}x^r$
$= \sum_{r=0}^\infty \begin{pmatrix} {r+2}\\{r}\end{pmatrix}x^{3+r}$
Hence the number of ways is $\begin{pmatrix} {2n-1}\\{2n-3}\end{pmatrix}$=$\begin{pmatrix} {2n-1}\\{2}\end{pmatrix}$.
However, the actual answer is $\begin{pmatrix} {2n+2}\\{2}\end{pmatrix}$$-3$$\begin{pmatrix} {n+1}\\{2}\end{pmatrix}$. Did i go wrong somewhere in my proof? Or is it some conceptual understanding gone wrong? Thanks!
| Since we are selecting $2n$ balls from a total of $3n$ balls,
we can count how many ways we can choose the $n$ balls we are leaving out.
This is the number of solutions in nonnegative integers to $a+b+c=n$,
which is given by $\color{blue}{\dbinom{n+2}{2}}$.
Alternate solution:
We want to find the number of solutions to $x_1+x_2+x_3=2n$ where $0\le x_i\le n$ for each $i$.
If S is the set of all solutions in nonnegative integers and if $A_i$ is the set of solutions with $x_i>n$,
we have $\big|\overline{A_1}\cap\overline{A_2}\cap\overline{A_3}\big|=\big|S\big|-\big|A_1\big|-\big|A_2\big|-\big|A_3\big|=\color{blue}{\dbinom{2n+2}{2}-3\dbinom{n+1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find bounds in derivation of Stirling's Formula I'm reading a very nice note by Terry Tao on Stirling's Formula (https://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/).
At one point in the derivation he states:
$$ \text{“ } n \log\left(1 + \frac{x}{\sqrt{n}}\right) - \sqrt{n}x = -\int_0^x \frac{(x-y) \, dy}{(1 + y/ \sqrt{n})^2}$$
This gives a uniform upper bound
$$ n \log\left(1 + \frac{x}{\sqrt{n}}\right) - \sqrt{n} x \leq -cx^2$$
for some $c>0$, when $|x| \leq \sqrt{n}$ and
$$n \log\left(1 + \frac{x}{\sqrt{n}}\right) - \sqrt{n} x \leq -c|x|\sqrt{n} $$
for $|x|> \sqrt{n}. \text{''}$.
I've been playing around with the first equation a bit, and I don't understand where the upper bounds come from.
How to tell at a glance (as Tao seems to) that these upper bounds hold?
| Assume $|x|\leq \sqrt{n}$. First, consider the case $x>0$. Observe
\begin{align}
-\int^x_0 \frac{(x-y)\ dy}{(1+y/\sqrt{n})^2} \leq -\int^x_0 \frac{(x-y)}{4}\ dy = -\frac{1}{8}x^2
\end{align}
because
\begin{align}
\left(1+\frac{y}{\sqrt{n}}\right) \leq \left(1+\frac{x}{\sqrt{n}}\right) \leq 2.
\end{align}
In the case when $x<0$. Observe
\begin{align}
-\int^x_0 \frac{(x-y) dy}{(1+y/\sqrt{n})^2}= -\int^0_{-|x|} \frac{(|x|+y) dy}{(1+y/\sqrt{n})^2} \leq - \int^0_{-|x|} (|x|+y) dy = -\frac{x^2}{2}
\end{align}
since
\begin{align}
\left(1+\frac{y}{\sqrt{n}}\right) \leq 1.
\end{align}
Now, when $|x|> \sqrt{n}$. Consider the case $x>0$ and observe
\begin{align}
f(x):=n\log\left(1+\frac{x}{\sqrt{n}}\right)-\frac{3}{4}\sqrt{n} x \leq 0
\end{align}
because $f(\sqrt{n}) = n(\log 2- 3/4) \leq 0$ and $f'(x) \leq 0$ for all $x>\sqrt{n}$. Thus, it follows
\begin{align}
n\log\left( 1+ \frac{x}{\sqrt{n}}\right)-\sqrt{n} x< -\frac{1}{4}\sqrt{n}x.
\end{align}
Lastly, when $x<-\sqrt{n}$ we need to be cautious what Tao actually means because $\log(1-|x|/\sqrt{n})$ is not defined unless we look at the function $\log||x|/\sqrt{n}-1|$ instead. But then the left-hand side will be positive. So...Tao probably made a typo.
Edit: He probably looked at the Taylor expansion to see that
\begin{align}
n\log\left(1+\frac{x}{\sqrt{n}}\right)= \sqrt{n} x - \frac{1}{2}x^2 + \frac{x^3}{3\sqrt{n}}-\ldots = \sqrt{n} x+ \mathcal{O}(x^2).
\end{align}
So when $x$ is small (in his case when $x/\sqrt{n}\leq 1$), it should follow that
\begin{align}
n\log\left(1+\frac{x}{\sqrt{n}}\right)-\sqrt{n} x= \mathcal{O}(x^2).
\end{align}
However, Taylor expansion approximation doesn't really work when you are far away from where you want to expand. But when $x$ is big (in your case $x/\sqrt{n}>>1$), it should be clear that
\begin{align}
n\log\left(1+ \frac{x}{\sqrt{n}}\right)-\sqrt{n} x
\end{align}
is dominated by $-\sqrt{n}x$ because $n\log(1+\frac{x}{\sqrt{n}})$ doesn't grow fast enough to overtake $-\sqrt{n}x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the 3.3.3 equation $x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$ have a complete integer or rational solution? Consider the equation
$$
X_1^3+X_2^3+X_3^3 = Y_1^3+Y_2^3+Y_3^3. \tag{$\star$}
$$
Does this have a complete solution (a.k.a. parameterization) in integers? If not, does it have a complete solution in rational numbers?
I know the [integral and/or rational] solutions to other related equations — such as the 3.2.2 equation
$$
X_1^3+X_2^3 = Y_1^3+Y_2^3
$$
and the 3.1.3 equation
$$
X_1^3 = Y_1^3+Y_2^3+Y_3^3.
$$
(see, for example, Choudhry’s paper) — but I have yet to find any paper giving the complete rational or integral solution to ($\star$).
|
Update:
Yes, there is a complete rational solution. Since we are dealing with a homogeneous equation, it is best to consider rational points on,
$$x_1^2+x_2^2 =1\tag1$$
$$x_1^3+x_2^3+x_3^3 =1\tag2$$
$$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1\tag3$$
to reduce the number of variables. The complete rational solution to $(1)$ and $(2)$ are well-known. It turns out the complete rational solution for $(3)$ is,
$$\color{blue}{\big(a + (a + b)\gamma\big)^3 + (a \gamma - b)^3 + \big(c + (c + d)\gamma\big)^3 + (c \gamma - d)^3 + \gamma^3 = 1}\tag4$$
where,
$$\gamma=\frac{-a^3 + b^3 - c^3 + d^3 + 1}{a^3 + (a+b)^3+c^3+(c+d)^3+1}\tag5$$
and can be found by generalizing the approach used by Choudhry in the paper cited by the OP.
Proof:
Equating the five terms of $(3)$ and $(4)$, one can easily solve for the five unknowns $a,b,c,d,\gamma$ as,
$$a = \frac{x_1+x_2x_5}{\beta},\quad b = \frac{-x_2+(x_1-x_2)x_5}{\beta}$$
$$c = \frac{x_3+x_4x_5}{\beta},\quad d = \frac{-x_4+(x_3-x_4)x_5}{\beta}$$
$$\gamma = x_5, \quad\quad\beta = x_5^2+x_5+1$$
Substituting these into $(5)$, the equality holds only if the $x_i$ satisfies $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1$, thus proving the assertion.
Example:
The smallest solution to $(3)$ is,
$$\big(\tfrac{1}{4})^3+\big(\tfrac{1}{4})^3+\big(\tfrac{2}{4})^3+\big(\tfrac{3}{4})^3+\big(\tfrac{3}{4})^3=1\tag6$$
Equating terms with $(4)$ one finds $a,b,c,d,\gamma$ as,
$$a=\tfrac{7}{37},\;b=\tfrac{-4}{37},\;c=\tfrac{17}{37},\;d=\tfrac{-15}{37},\;\gamma=\tfrac{3}{4}$$
If there is a zero term,
$$0+\big(\tfrac{1}{7})^3+\big(\tfrac{1}{7})^3+\big(\tfrac{5}{7})^3+\big(\tfrac{6}{7})^3=1\tag7$$
then,
$$a=\tfrac{6}{127},\;b=\tfrac{-13}{127},\;c=\tfrac{37}{127},\;d=\tfrac{-59}{127},\;\gamma=\tfrac{6}{7}$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| Hint: For $a$
$$L.H.S=\sin^4 A+ \cos^4 A$$
$$=(\sin^2 A+ \cos^2 A)^2-2\sin^2A.\cos^2A$$
$$=1-\frac {1}{4}(2\sin A.\cos A)^2$$
$$=1-\frac {1}{4} \sin^2 2A$$
$$=\frac {4-\sin^2 2A}{4}$$
$$=\frac {4-1+\cos 4A}{4}$$
$$=\frac {3+\cos 4A}{4}=R.H.S$$
Proved.
Just consider that $\theta=A$.
Follow similar steps above for $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
What transformation maps $y=\frac{1}{4}x-2$ to $y=-3x+6$? What transformation maps $y=\frac{1}{4}x-2$ to $y=-3x+6$?
I have tried many things, rotating around the origin, reflecting about common lines, and nothing seems to work.
Any help is appreciated. Thanks!
| Have a look at the geometrical situation:
(Large version)
We have
$f(x) =\frac{1}{4}x-2$ (green line) and $g(x) =-3x+6$ (red line).
We see that a rotation around the intersection $A$ of both lines would do most of the job.
Easiest, rotations around the origin are formulated, so we
*
*first translate the scene such that $A$ turns into the origin $O$ via $T$, then
*rotate the line of $f$ into the line of $g$ via $R$ around the origin, and finally
*retranslate the origin $O$ to $A$ again via $T^{-1}$.
These individual transformations compose into the transformation
$$
M = T^{-1} R T
$$
In Detail:
As we want to use translations, in other words: affine transformations, and still use the convenient matrix formulation, we use homogeneous coordinates $u = (x, y, 1)^\top$.
Then the transformation matrices are:
$$
T =
\begin{pmatrix}
1 & 0 & -a_x \\
0 & 1 & -a_y \\
0 & 0 & 1
\end{pmatrix}
$$
where $A = (a_x, a_y)$. Test:
$$
T u =
\begin{pmatrix}
1 & 0 & -a_x \\
0 & 1 & -a_y \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
1
\end{pmatrix}
=
\begin{pmatrix}
x-a_x \\
y-a_y \\
1
\end{pmatrix}
$$
so $T(u_A) = (0,0,1)^\top$, as desired. The rotation around the origin about an angle $\varphi$ is
$$
R =
\begin{pmatrix}
\cos(\varphi) & -\sin(\varphi) & 0 \\
\sin(\varphi) & \cos(\varphi) & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
The inverse transformation of $T$ is
$$
T^{-1} =
\begin{pmatrix}
1 & 0 & a_x \\
0 & 1 & a_y \\
0 & 0 & 1
\end{pmatrix}
$$
Calculation of the parameters:
We can describe the two lines by
\begin{align}
(1/4) x - y &= 2 \\
3x + y &= 6
\end{align}
which we can solve as
$$
\left[
\begin{array}{rr|r}
1/4 & -1 & 2 \\
3 & 1 & 6
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
3 & 1 & 6
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
0 & 13 & -18
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
0 & 1 & -18/13
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & 0 & 32/13 \\
0 & 1 & -18/13
\end{array}
\right]
$$
thus $A = (32/13,-18/13) = (2.4615,-1.3846)$.
We can determine the angle between the lines from
$$
d_f \cdot d_g = \cos(\angle(d_f, d_g)) = \cos(-\varphi)
$$
and the unit direction vectors of the lines, which are
$$
d_f
= \frac{(1, 1/4)^\top}{\lVert (1, 1/4) \rVert}
= \frac{(4,1)^\top}{\sqrt{17}}
$$
and
$$
d_g
= \frac{(1,-3)^\top}{\lVert (1,-3) \rVert}
= \frac{(1,-3)^\top}{\sqrt{10}}
$$
so $\varphi = -\arccos(1/\sqrt{170})=-85.6^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of an integral (SEEMOUS 2014 problem 4) Please show me how to solve this problem. I tried to make the substitution $x/n=y$ but I don't know how to go further.
(a) Prove that $$\lim_{n\to\infty}n\int_0^n\frac{\arctan\frac{x}{n}}{x(x^2+1)}\,dx=\frac{\pi}{2}$$
(b) Find the limit $$\lim_{n\to\infty}n\left(n\int_0^n\frac{\arctan\frac{x}{n}}{x(x^2+1)}\,dx-\frac{\pi}{2}\right)$$
| About the first integral, we have
$$ \int_{n}^{+\infty}\frac{dx}{1+x^2}=O\left(\frac{1}{n}\right) $$
and for any $x\in\mathbb{R}^+$
$$ \lim_{n\to +\infty}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}=1 $$
hence by the dominated convergence theorem
$$ \lim_{n\to +\infty}\int_{0}^{n}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}\cdot\frac{dx}{1+x^2}=\int_{0}^{+\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}.$$
In a similar way, since in a right neighbourhood of the origin we have
$$ \frac{\arctan x}{x} = 1-\frac{x^2}{3}+\frac{x^4}{5}-\ldots $$
the second limit equals
$$ -1+\sum_{k\geq 1}\frac{(-1)^k}{(2k-1)(2k+1)}=\color{red}{-\frac{\pi+2}{4}}.$$
Explanation:
$$\begin{eqnarray*}\lim_{n\to +\infty}n\left(\int_{0}^{n}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}\cdot\frac{dx}{1+x^2}-\frac{\pi}{2}\right)&=&\lim_{n\to +\infty}n\left(\int_{0}^{n}\left(\frac{\arctan\frac{x}{n}}{\frac{x}{n}}-1\right)\cdot\frac{dx}{1+x^2}-\arctan\frac{1}{n}\right)\\&=&-1+\lim_{n\to +\infty}n^2\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\cdot\frac{dx}{1+n^2 x^2}\\&=&-1+\lim_{n\to +\infty}\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\cdot\frac{dx}{x^2+\frac{1}{n^2}}\\(DCT)\quad &=& -1+\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\frac{dx}{x^2}\\(Taylor)\quad&=&-1+\int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^n x^{2n-2}}{2n+1}\,dx\\&=&-1+\frac{1}{2}\sum_{k\geq 1}(-1)^k\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\\&=&-1+\frac{1}{2}\left[-\arctan(1)-\left(\arctan(1)-1\right)\right]\\&=&-\arctan(1)-\frac{1}{2}.\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of square roots........... $$
\mbox{If}\quad S =
1 + \,\sqrt{\,\frac{1}{2}\,}\, + \,\sqrt{\,\frac{1}{3}\,}\, +
\,\sqrt{\,\frac{1}{4}\,}\, +
\,\sqrt{\,\frac{1}{5}\,}\, + \cdots + \,\sqrt{\,\frac{1}{100}\,}\,\,,
$$
then what is the value of $\left\lfloor\,S\,\right\rfloor$ ?.
Here $\left\lfloor\,S\,\right\rfloor$ is the greatest integer function which is less than or equal to $S$.
| The Euler-Maclaurin Summation Formula yields
$$\begin{align}
\sum_{k=1}^{100}k^{-1/2}&=1 +\int_1^{100} x^{-1/2}\,dx+\frac12\left(\frac{1}{\sqrt{100}}-1\right)+\frac1{24} \left(1-\frac{1}{(100)^{3/2}}\right)+R_2 \\\\
&=18.591625+R_2
\end{align}$$
where
$$\begin{align}
\left|R_2\right|&\le \frac{2\zeta(2)}{(2\pi)^2}\int_1^{100} \left|\frac{d^2 x^{-1/2}}{dx^2}\right|\,dx\\\\
&= \frac{2\zeta(2)}{(2\pi)^2}\,\frac12 \left(1-\frac{1}{(100)^{3/2}}\right)\\\\
&=0.041625
\end{align}$$
Therefore, the integer part of the sum of interest is indeed $18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$
Prove that the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1, y=0$ or $\frac{y^2}{a^2-b^2}+\frac{z^2}{a^2}=-1, z=0$.
EDIT:
Here $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is the base of the cone. Let the vertex of the cone be $(x_1,y_1,z_1)$. Let the generator be $$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.$$
Then $z=0$ implies any point on ellipse be $(x_1-lz_1/n, y_1-mz_1/n,0)$. It lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then the point satisfies the ellipse we get $$\frac{(nx_1-lz_1)^2}{a^2}+\frac{(ny_1-mz_1)^2}{a^2}=n^2$$
Eliminating $l,m,n$ we get the equation of the cone as $$\frac{1}{a^2}(zx_1-xz_1)^2+\frac{1}{b^2}(zy_1-yz_1)^2=(z-z_1)^2.$$ How to get the locus of vertex $(x_1,y_1,z_1)$?
Edit 2
How to get the locus of the vertices in the given two forms using purely mathematical way?
I not able to solve the problem. Please help.
| Considering that
*
*the ellipse on the $x,y$ plane is symmetrical with repect to both axis, and also it is symmetrical with respect to the $z,x$ and $z,y$ planes.
*a circular cone whose axis is incident to the $x,y$ plane, will intercept on it a quadric that is symmetrical to the plane containing the cone axis and parallel to the $z$ axis.
we conclude that the axis of the cone shall lie either on the $z,x$ or in the $z,y$ plane, and so shall thus be for the vertex .
case: cone axis in the $z,x$ plane
Let the vertex be $V=(x_v,0,z_v)$, and the unit vector defining the axis $\mathbb {a} =(cos \alpha, 0, sin\alpha)$.
A point $P$ on the circular cone will obey to
$$
\frac{{\mathop {VP}\limits^ \to }}
{{\left| {\mathop {VP}\limits^ \to } \right|}} \cdot \mathbf{a} = const. = \cos \beta
$$
and in particular a point on the $x,y$ plane will obey to:
$$
\left( {\left( {x - x_{\,v} } \right)\cos \alpha + \left( {0 - z_{\,v} } \right)\sin \alpha } \right)^{\,2} = \left( {\left( {x - x_{\,v} } \right)^{\,2} + \left( {0 - z_{\,v} } \right)^{\,2} + y^{\,2} } \right)\cos ^{\,2} \beta
$$
that is (duly excluding the cases $\beta=\pi /2$ and $\beta=\alpha$ which can be dealt apart)
$$
\begin{gathered}
\left( {z_{\,v} ^{\,2} + y^{\,2} } \right)\cos ^{\,2} \beta = \hfill \\
= \left( {\cos ^{\,2} \alpha - \cos ^{\,2} \beta } \right)\left( {x - x_{\,v} + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \alpha - \cos \beta } \right)}}} \right)\left( {x - x_{\,v} + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \alpha + \cos \beta } \right)}}} \right) \hfill \\
\end{gathered}
$$
$$
\left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)\left( {x - x_{\,v} - z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta - \cos \alpha } \right)}}} \right)\left( {x - x_{\,v} + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta + \cos \alpha } \right)}}} \right) + y^{\,2} = - z_{\,v} ^{\,2} \tag{1}
$$
First step for this to represent the targeted ellipse is to get rid of the term in $x^1$, which means
that is shall be:
$$
- x_{\,v} - z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta - \cos \alpha } \right)}} = - \left( { - x_{\,v} + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta + \cos \alpha } \right)}}} \right)
$$
i.e.
$$
- z_{\,v} \left( {\frac{{\sin \alpha \;\cos \alpha }}
{{\cos ^{\,2} \beta - \cos ^{\,2} \alpha }}} \right) = x_{\,v} \tag{2}
$$
So that eq. (1) can be rewritten as:
$$
\left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)\left( {x + x_{\,v} \frac{{\cos \beta }}
{{\cos \alpha }}} \right)\left( {x - x_{\,v} \frac{{\cos \beta }}
{{\cos \alpha }}} \right) + y^{\,2} = - z_{\,v} ^{\,2}
$$
leading to:
$$
\begin{gathered}
\left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)x^{\,2} + y^{\,2} = x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} - z_{\,v} ^{\,2} = \hfill \\
= x_{\,v} ^{\,2} \left( {\frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} - \frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)^{\,2} }}
{{\sin ^{\,2} \alpha \;\cos ^{\,2} \alpha }}} \right) = \hfill \\
= x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{1 - \cos ^{\,2} \beta }}
{{1 - \cos ^{\,2} \alpha }} \hfill \\
\end{gathered} \tag{3}
$$
Comparing this to the target we obtain
$$
\left\{ \begin{gathered}
x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = b^{\,2} \hfill \\
\left( {\frac{{\cos ^{\,2} \beta - \cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right) = \frac{{b^{\,2} }}
{{a^{\,2} }} \hfill \\
\end{gathered} \right.
$$
With some algebraic manipulation
$$
\left\{ \begin{gathered}
x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = b^{\,2} \hfill \\
x_{\,v} ^{\,2} \frac{{\cos ^{\,2} \beta }}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = a^{\,2} \hfill \\
x_{\,v} ^{\,2} \frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = a^{\,2} - b^{\,2} \hfill \\
\frac{{\left( {\cos ^{\,2} \beta - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} = \frac{{b^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}} \hfill \\
\left( {\frac{{\cos ^{\,2} \beta - \cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right) = \frac{{b^{\,2} }}
{{a^{\,2} }} \hfill \\
\frac{{\cos ^{\,2} \beta }}
{{\cos ^{\,2} \alpha }} = \frac{{a^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}} \hfill \\
\end{gathered} \right.
$$
we finally arrive to rewrite the eq. (2) as:
$$
z_{\,v} ^{\,2} = \frac{{b^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}}\left( {1 - \frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }}} \right)x_{\,v} ^{\,2} = \frac{{b^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}}x_{\,v} ^{\,2} - b^{\,2}
$$
$$
\frac{{x_{\,v} ^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}} - \frac{{z_{\,v} ^{\,2} }}
{{b^{\,2} }} = 1
$$
case: cone axis in the $z,y$ plane
The overall scheme is the same, just we have to exchange $x$ with $y$ and $a$ with $b$, giving
$$
\frac{{y_{\,v} ^{\,2} }}
{{\left( {b^{\,2} - a^{\,2} } \right)}} - \frac{{z_{\,v} ^{\,2} }}
{{a^{\,2} }} = 1
$$
or if you prefer:
$$
\frac{{y_{\,v} ^{\,2} }}
{{\left( {a^{\,2} - b^{\,2} } \right)}} + \frac{{z_{\,v} ^{\,2} }}
{{a^{\,2} }} = - 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
complex number equation I have this equation:
$$z^2=-i$$
All I can figure out from my knowledge is that
$$z^2=r^2\cos(2x)+i\sin(\cos(2x))$$ in this case is:
$$\cos(3\pi/4)+i\sin(3\pi/4))$$
and that will be $$e^{3\pi/4}$$
because $-i =0-i$ so that $\cos2x=0$ and $\sin2x=-1$.
Am I on the right track? I feel like I'm missing something.
The answer should be $\pm(-1+i)/\sqrt{2}$.
where does the $\sqrt{2}$ come from?
| Different ways to represent complex numbers:
$z = x + iy$
$z = \rho (\cos\theta + i \sin \theta)$
$z = \rho e^i\theta$
$z^2 = (x^2-y^2) + i(2xy)$
$z^2 = \rho^2 (\cos2\theta + i \sin 2\theta)$
$z^2 = \rho^2 e^{2i\theta}$
This is important background, and if you are not comfortable bouncing between these, then keep reviewing until it clicks.
lets use this one:
$z^2 = \rho^2 (\cos2\theta + i \sin 2\theta) = -i\\
z^2 = (\cos \frac {3\pi}{2} + i \sin \frac {3\pi}{2}) = -i\\
2\theta = \frac {3\pi}{2}\\
\theta = \frac {3\pi}{4}\\
z = (\cos \frac {3\pi}{4} + i \sin \frac {3\pi}{4})\\
z = -\frac {\sqrt 2}{2} + i \frac {\sqrt 2}{2})\\
z = \frac{\sqrt 2}{2} (-1 + i)$
$z^2 = -i$ has two solutions. You can choose to think if it as $\pm$ as was so handy with real numbers. But with complex numbers, I think of it as rotating clockwise and counter clockwise.
So I think of the second solution as:
$z = (\cos \frac {7\pi}{4} + i \sin \frac {7\pi}{4})$ that is half-way between $\frac {3\pi}{2}$ and $2\pi$
$z = \frac{\sqrt 2}{2} (1 - i)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2 x}}+2^{\frac{1}{\sin^2 x}}+3^{\frac{1}{\sin^2 x}}+\cdots + n^{\frac{1}{\sin^2 x}}\right]^{\sin^2x}$
$$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]^{\sin^2x}$$
Limit is of form $(\infty)^{0} $
$$\lim_{x\to 0}e^{\sin^2x\log{ {\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]}}}$$
I don't know how to proceed further.
| Hint:
$$n^y\leq 1^y+2^y+3^y+\cdots + n^y\leq n^y+n^y+\cdots+n^y=n^{y+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Calculating inverse matrices directly I'm trying to find the inverse matrix of
$$ \begin{bmatrix}
3 & 4 \\
6 & 2
\end{bmatrix}$$
I set up the following equation:
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\,
\begin{bmatrix}
3 & 4 \\
6 & 2
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
So
\begin{cases}
3a+6b = 1 \\[4px]
4a+2b = 0 \\[4px]
3c+6d = 1 \\[4px]
4c+2d = 0
\end{cases}
I solve $a,b,c,d$ and get
\begin{cases}
a = \dfrac{1}{-9} \\[4px]
b = \dfrac{4}{18} \\[4px]
c = \dfrac{1}{-9} \\[4px]
d = \dfrac{4}{18}
\end{cases}
But I get $c,d$ wrong. They are supposed to be
\begin{cases}
c = \dfrac{6}{18} \\[4px]
d = \dfrac{-3}{18}
\end{cases}
Why does my way not work?
| You just switched $1$ with $0$ in the last two equations.
However, for $2\times2$ matrices there's a very simple method. Provided $ad-bc\ne0$, we have
$$
\begin{bmatrix} a & b \\ c & d\end{bmatrix}^{-1}=
\frac{1}{ad-bc}
\begin{bmatrix} d & -b \\ -c & a\end{bmatrix}
$$
Switch the diagonal elements and change the sign to the nondiagonal one, then divide by the determinant.
This can be verified by doing the multiplication.
In your case, $ad-bc=6-24=-18$, so
$$
\begin{bmatrix} 3 & 4 \\ 6 & 2 \end{bmatrix}^{-1}=
-\frac{1}{18}
\begin{bmatrix} 2 & -4 \\ -6 & 3\end{bmatrix}=
\begin{bmatrix}-1/9 & 2/9 \\ 1/3 & -1/6 \end{bmatrix}
$$
Important note. This cannot be extended to bigger matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Left Inverses of a Matrix I am confused about left inverses after examining two particular problems. From what I understand in order to find the left inverse of an $m\times n$ matrix I must first transpose it then augment it with the identity matrix that would result after multiplying $A$ by an unknown $X$. After this I should perform RREF as far as I can and whatever I'm left with (barring any inconsistent rows) will be the columns of $X$ (the inverse of $A$). At that point it should be true that $XA = I$, however I do not get that for the first problem. I checked to make sure the first matrix does not have a right inverse so I don't know why $AX$ produces the identity matrix when it was the left inverse that was being solved for.
$A = \begin{bmatrix}1 & 2 \\ 2 & 5 \\ 3 & 7\end{bmatrix}$
$\begin{bmatrix}1 & 2 & 3 & | & 1 & 0\\ 2 & 5 & 7 & | & 0 & 1\\\end{bmatrix} => RREF => \begin{bmatrix}1 & 0 & 1 & | & 5 & | & -2\\0 & 1 & 1 & | & -2 & | & 1\end{bmatrix}$
Therefore,
$X = \begin{bmatrix}5-r & -2-t \\ -2-r & 1-t \\ r & t\end{bmatrix}$ (See correction)
However $XA$ does not produce the identity matrix while $AX$ does. Is this because I transposed $A$ in order to augment it and find $X$? As far as I know, $X$ should be the left inverse, so $XA$ should work. What really confuses me is that for this second matrix, $B$, I attempt the exact same thing and find that $XB = I$ as expected.
$B = \begin{bmatrix}2 & -1 \\ 4 & -1 \\ 2 & 2\end{bmatrix}$
$\begin{bmatrix}2 & 4 & 2 & | & 1 & 0\\ -1 & -1 & 2 & | & 0 & 1\end{bmatrix} => RREF => \begin{bmatrix}2 & 4 & 2 &| & 1 & 0 \\ 0 & 1 & 3 & | & 1/2 & 1\end{bmatrix}$
Therefore,
$X = \begin{bmatrix}-1/2+5s & 1/2-3s & s \\ -2+5t & 1-3t & t\end{bmatrix}$
Could someone please explain what is happening? I strongly suspect it has to do with taking the transpose, but why then does it work for one and not the other? The problems are both "cooked" so I know there are definitely left inverses for both.
**** Correction / Solved ****
$X = \begin{bmatrix}5-r & -2-r & r\\ -2-t & 1-t & t\\\end{bmatrix}$
| The matrix and Moore-Penrose pseudoinverse are
$$
\mathbf{A} =
%
\left(
\begin{array}{cc}
1 & 2 \\
2 & 5 \\
3 & 7 \\
\end{array}
\right), \qquad
%
\mathbf{A}^{\dagger} =
\frac{1}{3}
\left(
\begin{array}{ccc}
12 & -9 & 3 \\
-5 & 4 & -1 \\
\end{array}
\right)
%
$$
The reduced row echelon form is
$$
\mathbf{E}_{\mathbf{A}} =
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
0 & 0 \\
\end{array}
\right)
$$
The matrix $\mathbf{A}$ has rank $\rho = 2$. Because the number of columns $n=2$ matches the rank, $\mathbf{A}$ has full column rank $n=\rho$. Therefore the nullspace $\mathcal{N}\left( \mathbf{A} \right)$ is trivial. Therefore the pseudoinverse is a left inverse:
$$
\mathbf{A}^{\dagger} = \mathbf{A}^{L}.
$$
That is,
$$
\begin{align}
\mathbf{A}^{L} \mathbf{A} &= \mathbf{I}_{2} \\
%
\frac{1}{3}
\left(
\begin{array}{rrr}
12 & -9 & 3 \\
-5 & 4 & -1 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
1 & 2 \\
2 & 5 \\
3 & 7 \\
\end{array}
\right)
%
&=
%
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
%
\end{align}
$$
The matrix $\mathbf{A}$ has a row rank deficiency so the nullspace is nontrivial.
$$
\mathcal{N}\left( \mathbf{A}^{*} \right) = \text{span }
\left\{\,
\left(
\begin{array}{r}
-1 \\
-1 \\
1
\end{array}
\right)
\, \right\}
$$
The matrix product
$$
\mathbf{A} \mathbf{A}^{\dagger} =
\left(
\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & 1 \\
1 & 1 & 2 \\
\end{array}
\right)
$$
is not an identity matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a + b + c + a^2 + b^2 + c^2 \leq4$ given that $a,b,c\geq-1$ and $a^3 + b^3 + c^3=1$ Let $a,b,c\geq-1$ be real numbers with $a^3 + b^3 + c^3=1$.
Prove that $$a + b + c + a^2 + b^2 + c^2 \leq4.$$
| Hint:
This inequality is "separable" and it is enough to note that for $x> -1$,
$$f(x) = x^3-x^2-x+1 = (1+x)(1-x)^2 \geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The Laplacian of a Dyad in Cartesian using Indicial Notation
*
*The problem statement, all variables and given/known data
Given the dyad formed by two arbitrary position vector fields, u and v, use indicial notation in Cartesian coordinates to prove:
$$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$
*Relevant equations
Per my professor's notes, the Laplacian of a dyad (also a tensor) is given as:
$$
\nabla^2 {\mathbf {S}} = \nabla \cdot {S_{ij,k} \mathbf{e_{i}e_{j}e_{k}}} = S_{ij,kk} \mathbf{e_{i}e_{j}}
$$
*The attempt at a solution
$$
\nabla^2 {\mathbf {uv}} = (u_{i}v_{j})_{,kk} = u_{i,kk}v_{j} + u_{i}v_{j,kk} \\
u_{i,kk}v_{j} + u_{i}v_{j,kk} = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v}
$$
I don't know where the following terms come from:
$$
2\nabla {\vec u} \cdot {(\nabla \vec v)}^T
$$
Does anyone have any suggestions? I feel that I am missing a step or something.
| We need to evaluate $\nabla^2 \{ \mathbf{a} \mathbf{b}\}$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors
Let's write $\nabla^2$ as $\nabla_x^2 + \nabla_y^2 + \nabla_z^2$
First, let's evaluate $\nabla_x^2 \{\mathbf{a} \mathbf{b} \}$
$$
\begin{align}
\nabla_x^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_x \nabla_x \{\mathbf{a} \mathbf{b}\} \\\\
&= \nabla_x \{ \left( \nabla_x \mathbf{a} \right) \mathbf{b} + \mathbf{a} \nabla_x \mathbf{b} \} \\\\
&= \left(\nabla_x^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_x \mathbf{a} \right) \left( \nabla_x \mathbf{b} \right) + \mathbf{a} \nabla_x^2 \mathbf{b}
\end{align} \tag{1}$$
Similarly we evaluate the y and z components of the laplacian as under
$$
\begin{align}
\nabla_y^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_y \nabla_y \{\mathbf{a} \mathbf{b}\} \\\\
&= \nabla_y \{ \left( \nabla_y \mathbf{a} \right) \mathbf{b} + \mathbf{a} \nabla_y \mathbf{b} \} \\\\
&= \left(\nabla_y^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_y \mathbf{a} \right) \left( \nabla_y \mathbf{b} \right) + \mathbf{a} \nabla_y^2 \mathbf{b}
\end{align} \tag{2}$$
$$
\begin{align}
\nabla_z^2 \{\mathbf{a} \mathbf{b}\} &= \nabla_z \nabla_z \{\mathbf{a} \mathbf{b}\} \\\\
&= \nabla_z \{ \left( \nabla_z \mathbf{a} \right) \mathbf{b} + \mathbf{a} \nabla_z \mathbf{b} \} \\\\
&= \left(\nabla_z^2 \mathbf{a} \right) \mathbf{b} + 2 \left( \nabla_z \mathbf{a} \right) \left( \nabla_z \mathbf{b} \right) + \mathbf{a} \nabla_z^2 \mathbf{b}
\end{align} \tag{3} $$
Now, combining the above equations and noting that
$$
\begin{align}
\left( \nabla_x \mathbf{a} \right) \left( \nabla_x \mathbf{b} \right) + \left( \nabla_y \mathbf{a} \right) \left( \nabla_y \mathbf{b} \right) + \left( \nabla_z \mathbf{a} \right) \left( \nabla_z \mathbf{b} \right) = \left(\vec{\nabla} \mathbf{a} \right) \cdot \left( \vec{\nabla} \mathbf{b} \right)
\end{align} \tag{4}$$
, we get
$$
\begin{align}
\nabla^2 \{ \mathbf{a} \mathbf{b} \} &= \left(\nabla^2 \mathbf{a} \right) \mathbf{b} + 2 \left(\vec{\nabla} \mathbf{a} \right) \cdot \left( \vec{\nabla} \mathbf{b} \right) + \mathbf{a} \nabla^2 \mathbf{b}
\end{align} \tag{5}$$
Remarks :-
*
*Note that $\nabla_x, \nabla_y$ and $\nabla_z$ are scalars and hence we could write $\nabla_x \{ \mathbf{a} \mathbf{b} \} = \left( \nabla_x \mathbf{a} \right) \mathbf{b} + \mathbf{a} \{ \nabla_x \mathbf{b}\}$ and similarly for the y and z components.
*Since $\mathbf{a} \mathbf{b}$ is a dyadic product, we made sure that we don't change this order throughout the calculation.
*My final result differs from that by Mark Viola because in evaluating the term $\hat x_i\hat x_j v_j\nabla ^2(u_i)$ eq(4) of his answer, he has changed the order of the dyadic product $\hat x_i\hat x_j$, which is not allowed. According to me the said term should be evaluated as
$$
\hat x_i\hat x_j v_j\nabla ^2(u_i) = \nabla ^2(u_i) v_j \hat x_i\hat x_j = \nabla ^2(u_i) \hat x_i v_j \hat{x_j} = \left( \nabla^2 \vec{u} \right) \vec{v} \tag{6}
$$
Note that since $v_j$ and $\nabla^2 \left( u_i \right)$ are scalars and hence their orders could be swapped but the order of the dyadic product $\hat{x_i} \hat{x_j} $ couldn't be and it is because of this that my final result differs from Mark and the one in the question itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve this problem for a unit circle Given $541$ points in the interior of a circle of unit radius, show that there must be a subset
of $10$ points whose diameter (the maximum distance between any pair of points) is less
than $\sqrt{2}/4$.
| We assume the circle is centered on the origin $(0,0)$
You can draw a $8$ by $8$ grid of squares each with side length $\frac{1}{4}$. The top-left square will have the 4 coordinates $(-1,1),(-1,\frac{3}{4}),(-\frac{3}{4},1),(-\frac{3}{4},\frac{3}{4})$ and the bottom-right square will have the 4 coordinates $(1,-1),(1,-\frac{3}{4}),(\frac{3}{4},-1),(\frac{3}{4},-\frac{3}{4})$ etc.
We know that this grid will completely cover the entire circle since the diameter of the circle is equal to the side length of the entire grid (i.e. $2$). We can also rule out any points being in the 4 corner squares of the grid. This is because the distance from the closest point in each of these squares to the origin is $\sqrt{\frac{3}{4}^2 + \frac{3}{4}^2} = \sqrt{\frac{18}{16}} > 1$. Thus no point inside the circle can be in these 4 corner squares.
From here, we can use Pigeonhole Principle. Since there are only $64 - 4 = 60$ squares for the $541$ points to be in, there is at least $1$ square with at least $10$ points within the square (along the sides is fine). The maximum distance between any $2$ points in such a square is clearly the diagonal length, which is $\sqrt{\frac{1}{4}^2 + \frac{1}{4}^2} = \sqrt{\frac{2}{16}} = \frac{\sqrt 2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $abc$ where $a^2+b^2+c^2=144$ and $ab+bc+ca=144$ Total surface area of a cuboid is 288 sq.cm. and length of a diagonal of it is 12 cm. Find its volume.
This is the question. We know that diagonal =$a^2+b^2+c^2$ and surface area =$2(ab+bc+ca)$ and volume =$abc$ . So finally we have to find the value of $abc$ where $$a^2+b^2+c^2=144$$ and $$ ab+bc+ca =144 .$$ Somebody please help me.
| Hint:
$$a^2+b^2+c^2=ab+bc+cc \Leftrightarrow a=b=c \Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What should I do further. Please suggest.
|
What should I do further. Please suggest.
Multiply the both sides by $(1-\tan^2A)(1-3\tan^2A)$ and factorize the left-hand side.
The following way might be easier.
Since
$$-\tan A-\tan(2A)=\tan(3A)=\tan(A+2A)=\frac{\tan A+\tan(2A)}{1-\tan A\tan(2A)}$$
we get
$$(\tan A+\tan(2A))\left(1+\frac{1}{1-\tan A\tan(2A)}\right)=0$$
$$\tan A+\tan(2A)=0\quad\text{or}\quad \tan A\tan(2A)=2$$
$$t+\frac{2t}{1-t^2}=0\quad\text{or}\quad \frac{2t^2}{1-t^2}=2$$
where $t=\tan A$, which should be easy to deal with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Formal Power Series as Linear Operators Let $t^k$ act as the $k$-th derivative operator on the set of polynomials. So
$$t^k(x^n)=t^k x^n=(n)_kx^{n-k}$$
where $(n)_k=n(n-1)(n-2)...(n-k+1)$ is the falling factorial. Then with a formal power series, $f(t)=\sum_{k\ge 0}a_k\frac{t^k}{k!}$, the linear operator $f(t)$ acts as such that
$$f(t)(x^n)=f(t)x^n=\sum_{k=0}^n\binom{n}{k}a_k x^{n-k}$$
Therefore, depending on the coefficients of the power series, we can get some interesting binomial identites. For example, if $f(t)=e^{yt}$, since the coefficients $a_n=y^n$, we get
$$e^{yt}x^n=\sum_{k=0}^n\binom{n}{k}y^k x^{n-k}=(x+y)^n$$
by linearity,
$$(e^{yt}-1)x^n=(x+y)^n-x^n=\sum_{k=1}^{n}\binom{n}{k}y^k x^{n-k}$$
and perhaps not as obvious
$$\left(\frac{e^{yt}-1}{t}\right)x^n=\int_{x}^{x+y}u^ndu$$
Now suppose that $f(t)=e^{yt}-1-yt$. Then
$$(e^{yt}-1-yt)x^n=(x+y)^n-x^n-ynx^{n-1}=\sum_{k=2}^{n}\binom{n}{k}y^k x^{n-k}$$
Obviously there is a nice formed forward difference equation in the previous case that is not happening here. But there is a relationship with subtracted terms of the binomial expansion. What i would really like help understanding is whether or not a possible analogous integral representation exists for the following operator:
$$\left(\frac{e^{yt}-1-yt}{t^2}\right)x^n=\left(\sum_{k=0}^\infty\frac{y^{k+2}}{(k+2)(k+1)}\frac{t^k}{k!}\right)x^n=\sum_{k=0}^n\binom{n}{k}\frac{y^{k+2}}{(k+1)(k+2)}x^{n-k}$$
$$=\sum_{k=0}^n\binom{n+2}{k+2}\frac{y^{k+2}}{(n+1)(n+2)}x^{n-k}=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$
It is not as simple. Clearly $\frac{d^2}{dx^2}\frac{x^{n+2}}{((n+2)(n+1)}$. If I integrated below I think the math is correct
$$\int_x^{x+y}{\frac{u^{n+1}}{n+1}}du=\frac{1}{(n+1)(n+2)}\sum_{k=1}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$
Which is really close, but the lower bound on the summation is $1$, not $2$. Does any one have any insight in how i can fix this, if possible?
| Note: OPs calculations are quite ok and it shows the operators are closely related, but different. I don't think there is a necessity to fix anything.
I skimmed through the classic The Umbral Calculus by Steven Roman, but there was no indication that something more is going on regarding OPs question. Another source I've checked without success was The Calculus of Finite Differences by C. Jordan.
It might be helpful to list a few higher powers of the operators under consideration.
In the following I use OPs notation which is precisely the same used by Steven Roman.
Translation operator: $e^{yt}$
Since this operator satisfies:
\begin{align*}
e^{yt}x^n&=\sum_{k=0}^\infty \frac{y^k}{k!}t^kx^n
=\sum_{k=0}^n \frac{y^k}{k!}(n)_kx^{n-k}=\sum_{k=0}^n\binom{n}{k}y^kx^{n-k}\\
&=(x+y)^n
\end{align*}
we obtain
\begin{align*}
\left(e^{yt}\right)^2 x^n=e^{yt}(x+y)^n=(x+2y)^n
\end{align*}
and in general for $j\geq 1$
\begin{align*}
e^{jyt}x^n=(x+jy)^n
\end{align*}
Since $x^n, n\geq 0$ form a basis of the vector space of all polynomials $p$ in a single variable $x$ and the translation operator is linear, we obtain
\begin{align*}
e^{jyt}p(x)=p(x+jy)\tag{1}
\end{align*}
hence the name translation operator.
Forward difference operator: $e^{yt}-1$
Here we obtain for polynomials $p$ using (1)
\begin{align*}
\left(e^{yt}-1\right)p(x) = p(x+y)-p(x)
\end{align*}
The next one is
Operator: $\frac{\exp(yt)-1}{t}$
We obtain
\begin{align*}
\left(\frac{e^{yt}-1}{t}\right)x^n&=\sum_{k=1}^\infty \frac{y^k}{k!}t^{k-1}x^n\\
&=\sum_{k=1}^{n+1}\frac{y^k}{k!}(n)_{k-1}x^{n-(k-1)}\\
&=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}y^kx^{n+1-k}\\
&=\frac{1}{n+1}\left((x+y)^{n+1}-x^{n+1}\right)\tag{2}\\
&=\frac{1}{n+1}\int_x^{x+y}u^n\,du
\end{align*}
Similarly we obtain from (2) by linearity
\begin{align*}
\left(\frac{e^{yt}-1}{t}\right)^2x^n
&=\left(\frac{e^{yt}-1}{t}\right)\frac{1}{n+1}\left((x+y)^{n+1}-x^{n+1}\right)\\
&=\frac{1}{n+1}\left[\frac{1}{n+2}\left((x+2y)^{n+2}-(x+y)^{n+2}\right)\right.\\
&\qquad\qquad\quad\left.-\frac{1}{n+2}\left((x+y)^{n+2}-x^{n+2}\right)\right]\\
&=\frac{1}{(n+1)(n+2)}\left((x+2y)^{n+2}-2(x+y)^{n+1}+x^{n+2}\right)\\
&=\frac{1}{(n+2)_2}\left(\int_{x+y}^{x+2y}u\, du-\int_x^{x+y}u\,du\right)\tag{3}
\end{align*}
Here at (2) and (3) we can see quite nicely how the operator $\frac{\exp(yt)-1}{t}$ is connected with the integral operator. It can be extended to higher powers without too much effort and the relationship with the integral operator looks plausible.
Operator: $\frac{\exp(yt)-1-t}{t^2}$
Now we take a look at the operator which is on the focus of OP and its generalisation.
\begin{align*}
\left(\frac{e^{yt}-1-t}{t^{2}}\right)x^n
&=\sum_{k=2}^\infty\frac{y^k}{k!}t^{k-2}x^n\\
&=\sum_{k=2}^{n}\frac{y^{k}}{k!}(n)_{k-2}x^{n-(k-2)}\\
&=\frac{1}{(n+2)_2}\sum_{k=2}^n\binom{n+2}{k}y^kx^{n+2-k}\\
&=\frac{1}{(n+2)_2}\left((x+y)^{n+2}-x^{n+2}-nyx^{n+1}\right)
\end{align*}
Comparing the final expression with (3) we do not see a plausible representation via integrals since the term $nyx^{n+1}$ don't provide anything nicely of the form
\begin{align*}
\text{integrated expression (end point) - integrated expression (starting point)}
\end{align*}
This impression becomes more strongly when looking at the general case. We obtain for $j\geq 1$
\begin{align*}
\left(\frac{e^{yt}-1-\frac{t^2}{2}-\cdots-\frac{t^{j-1}}{(j-1)!}}{t^j}\right)x^n
&=\sum_{k=j}^\infty\frac{y^k}{k!}t^{k-j}x^n\\
&=\frac{1}{(n+j)_j}\sum_{k=j}^\infty\binom{n+j}{k}y^kx^{n+j-k}\\
&=\frac{1}{(n+j)_j}\left((x+y)^{n+j}-\sum_{k=0}^{j-1}\binom{n+j}{k}y^kx^{n+j-k}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.