Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$1-x+x^2-x^3+..(-1)^nx^n$ I have the following sum:
$1-x+x^2-x^3+..(-1)^nx^n, x\neq -1$
So what I thought was separating it in two cases like this:
Case 1. n is even
$$
1+x^2+x^4+...+x^n-x(1+x^2+...+x^n)
$$
Which I can turn into $\frac{1-x^{n+2}}{1-x^2}-\frac{x(1-x^{n+2})}{1-x^2}=\frac{1-x^{n+2}}{1-x^2}(1-x)$
Case 2. n is odd
$$
1+x^2+x^4+...+x^{n-1}-x(1+x^2+...+x^{n-1})=\frac{1-x^{n+1}}{1-x^2}-x(\frac{1-x^{n+1}}{1-x^2})
$$
My question is: Assuming what I've written is correct, which I'm not entirely sure, how can I combine the two cases for n even and odd into one equation?
| It should be $$\frac{(1-x)(1-(-1)^{n+1}x^{n+1})}{1-x^2} = \frac{1-(-1)^{n+1}x^{n+1}}{1+x}$$
But why make it complicated? This is just a geometric series $\sum_{i=0}^n (-x)^i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the smallest number which when divided by 3, 5 and 7 leaves remainders 2, 4 and 6 respectively Find the smallest number which when divided by 3, 5 and 7 leaves remainders 2, 4 and 6 respectively
I took out the LCM and I thought 105 would be the answer but answer is 104. Can anyone please explain the approch behind solving this question. I will be very grateful
| From Euclid's Algorithm we have $\color{blue}{2 \times 5} - \color{red}{3 \times 3}=1$ showing that $3$ & $5$ are coprime.
To solve
\begin{eqnarray*}
x \equiv \color{red}{2} \pmod{3} \\
x \equiv \color{blue}{4} \pmod{5}
\end{eqnarray*}
$x$ is construct as $x= \color{red}{2} \times \color{blue}{ 2 \times 5} - \color{blue}{4} \times \color{red}{3 \times 3}$ giving $x \equiv 14 \pmod{15}$
Now do the same again with $7$ and $15$ ...$\color{blue}{1 \times 15} - \color{red}{2 \times 7}=1$ showing that $3$ & $5$ are coprime.
To solve
\begin{eqnarray*}
x \equiv \color{blue}{14} \pmod{15} \\
x \equiv \color{red}{6} \pmod{7}
\end{eqnarray*}
$x$ is construct as $x= \color{red}{6} \times \color{blue}{1 \times 15} - \color{blue}{14} \times \color{red}{2 \times 7}$ giving $x \equiv 104 \pmod{105}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the minimum value of $\frac{a_1}{a_2} + \frac{a_2}{a_3} +\cdots + \frac{a_n}{a_1} $ Let $a_1,a_2,\cdots,a_n$ are positive real numbers.
Question: what is the minimum value of
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} +\cdots + \frac{a_n}{a_1} $$
Thought: I have no clue how to proceed. Tried some standard inequalities but in vain.
| By the inequality of arithmetic and geometric means we have that $$\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}}+\dots +\frac{a_{n}}{a_{1}} \geq n \, \left( \frac{a_{1}\cdot a_{2}\cdot \dots \cdot a_{n-1} \cdot a_{n}}{a_{2}\cdot a_{3} \dots \cdot a_{n} \cdot a_{1}}\right)^{1/n} = n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{x+12}{x-3} \gt |x-1| + 1$ Solve $\frac{x+12}{x-3} \gt |x-1| + 1$
I finally got $\frac{(x-6)(x+2)}{x-3} \lt 0$
That means the solutions are $3\lt x \le 6$ and $x\lt -2$ Am I right?
| We can rearrange this equation to give us
\begin{align}\frac{x+12}{x-3}&>|x-1|+1\\
\frac{x+12}{x-3}-1&>|x-1|\\
\frac{x+12}{x-3}-\frac{x-3}{x-3}&>|x-1|\\
\frac{x+12-(x-3)}{x-3}&>|x-1|\\
\frac{15}{x-3}&>|x-1|\end{align}
Clearly $x\neq 3$
We now need to consider three intervals: $(-\infty,1]$, $[1,3)$, $(3,\infty)$
When $x\in(-\infty,1]$ then we have $x-3<0$, and $x-1<0$ so: \begin{align}\frac{15}{x-3}&>-(x-1)\\
15&<(x-3)(1-x)\\
15&<-x^2+4x-3\\
0&<-x^2+4x-18\end{align}
This has no solutions, so we now know that $x>1$, although we can't say anything more specific yet.
Now when $x\in[1,3)$ then we have $x-3<0$ and $x-1>0$ so:
\begin{align}\frac{15}{x-3}&>x-1\\
15&<(x-1)(x-3)\\
15&<x^2-4x+3\\
0&<x^2-4x-12\\
0&<(x+2)(x-6)\end{align}
In the interval $[1,3)$, $(x+2)(x-6)$ is always negative, and thus the inequality is never satisfied. Now we can say that we know $x>3$, although we still cannot say anything more specific.
We turn to the final interval, when $x\in(3,\infty)$ then $x-3>0$ and $x-1>0$ so:
\begin{align}\frac{15}{x-3}&>x-1\\
15&>(x-1)(x-3)\\
15&>x^2-4x+3\\
0&>x^2-4x-12\\
0&>(x+2)(x-6)\end{align}
Remembering we solved this with the assumption that $x>3$, then the solutions to this are $3<x<6$ (without the assumption, the solutions are $-2<x<6$).
Now we need to combine the answers from each of the three intervals. The first two did not have any solutions, so the solution to the initial question is simply $$3<x<6$$
We can plot each side of the inequality and see that this is the case too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles?
How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles?
I already dealt with a similar problem, which is how many ways are there to cut a $1\times 8$ rectangle into $1\times 1$ and $1\times 2$ rectangles. The answer to this problem can be calculated using the number of ways to divide $1\times k$ strip for $3\le k\le8$ sequentially, to arrive at $34$. However, this problem is completely different. How can I solve this problem?
| We are dealing with a $2 \times 16$ grid. The first column of two cells can be filled by a $2 \times 2$, a $2 \times 3$ or a $2 \times 4$ grid. Let us call $f(n)$ the number of ways to fill a $2 \times n$ grid. We then have:
$$f(n) = f(n-2) + f(n-3) + f(n-4), n \geq 4$$
We also know that:
$$f(0) = 1$$
$$f(1) = 0$$
$$f(2) = 1$$
$$f(3) = 1$$
A closed form is not straightforward, but we find:
$$f(4) = 2$$
$$f(5) = 2$$
$$f(6) = 4$$
$$f(7) = 5$$
$$f(8) = 8$$
$$f(9) = 11$$
$$f(10) = 17$$
$$f(11) = 24$$
$$f(12) = 36$$
$$f(13) = 52$$
$$f(14) = 77$$
$$f(15) = 112$$
$$f(16) = 165$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of a function. I am not able to check double derivative. Can someone explain me how $\sin^p x \cos^q x$ attains maximum at $\tan^2 x = \frac pq$.
I am not able to check whether double derivative is positive or negative.
Question
Show that $$\sin^p\theta\cos^q\theta$$ attains a maximum when $$\theta=\tan^{-1}\sqrt{(p/q)}$$
Solution
Ley $y=\sin^p\theta\cos^q\theta$. For a maximum or minimum of $y$, we have $\frac{\text dy}{\text dx}=0$
\begin{align}p\sin^{p-1}\theta\cos^{q+1}\theta-q\sin^{p+1}\theta\cos^{q-1}\theta&=0\\
\sin^{p-1}\theta\cos^{q-1}\theta(p\cos^2\theta-q\sin^2\theta)&=0\end{align}
Therefore \begin{align}\sin\theta&=0\\
&\Downarrow\\
\theta&=0\\
\text{or } \cos\theta&=0\\
&\Downarrow\\
\theta&=\frac \pi 2\\
\text{or }\tan^2\theta&=\frac pq\\
&\Downarrow\\
\theta&=\tan^{-1}\sqrt{p/q}\end{align}
Now $y=0$ at $\theta=0$ and also at $\theta=\frac\pi2$
When $0<\theta<\frac\pi2$, $y$ is positive
Also, $\tan^{-1}\sqrt{p/q}$ is the only value of $\theta$ lying between $0$ and $\frac \pi2$ at which $\frac{\text d}{\text dx}=0$.
Hence $y$ is maximum when $$\theta=\tan^{-1}\sqrt{p/q}$$
This can be seen from the graph of $y$
| Let $f(x) = \sin^{p}(x) \, \cos^{q}(x)$ then
\begin{align}
f'(x) &= f(x) \, (-q \, \tan(x) + p \, \cot(x)) \\
&= -q \, \cot(x) \, f(x) \, \left( \tan^{2}(x) - \frac{p}{q} \right)
\end{align}
Now, the maximum of a function is defined by $f'(x) = 0$ then, seeking values other than $f(x) = 0$,
$$\tan^{2}(x) = \frac{p}{q}$$
or
$$x_{max} = \tan^{-1}\left(\sqrt{\frac{p}{q}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 5
} |
Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of roots we have $$\sum_{cyc}(a-b)^2 = 2\left(\sum_{cyc} a^2-\sum_{cyc}{ab}\right)\\
= 2\left(\sum_{cyc} a\right)^2-6\left(\sum_{cyc} ab\right)\\
= -6p$$
This is coefficient of $x^2$ in the sought cubic.
But now computing coefficient of $x^2$ requires simplifying factors like $(a-b)^2\cdot(b-c)^2$ which becomes very lengthy.
Is there a shorter way around?
Thanks!
| $\sum_{cyc}(a-b)^2$ is a symmetric polynomial of degree two, so it must be a polynomial in the coefficients of $x^3+px+r$. The only possibilities to form a polynomial of degree $2$ are $(a+b+c)^2$ and $ab+bc+ca=p.$ Since all terms containing the factor $a+b+c$ must vanish, it can be only a multiple of $p$. From the special case $(a,b,c)=(-1,0,1)$, where $p=-1$ and our three roots are $4,1,1$, we obtain $$\sum_{cyc}(a-b)^2=-6p.$$ Similarly, $\sum_{cyc}(a-b)^2\cdot(b-c)^2$ can only be a multiple of $p^2,$ so from the special case mentioned above, we get
$$\sum_{cyc}(a-b)^2\cdot(b-c)^2=9p^2.$$ Finally, $(a-b)^2\cdot(b-c)^2\cdot(c-a)^2$ is the discriminant of the cubic polynomial, a known expression (en.wikipedia.org/wiki/Discriminant ), so $$(a-b)^2\cdot(b-c)^2\cdot(c-a)^2=-4p^3-27r^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The number $m+n^2$ is divisible by $m+n$. Prove that the number $m+n^3$ is divisible by $m+n$.
$m$ and $n$ are positive integer numbers; $m+n\mid m+n^2$.
Prove that $m+n\mid m+n^3$
So far I have done:
If $m+n\mid m+n^2$, then $m+n\mid (m+n^2)-(m+n)=n^2-n=n(n-1)$.
$m+n^3=n(n^2-n)+(m+n^2)$ and each component is divisible by $m+n$
Proved, thanks for help!
| Hint: If $m + n^2$ is divisible by $m + n$ then so is $(m + n^2) - (m + n) = n^2 - n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y+z)=\frac{20}{x}+\frac{30}{y}$$
$$\implies\frac{20y+30x}{xy}$$
$$\implies\frac{20*\frac{3z}{5}+30*\frac{2z}{5}}{\frac{2z}{5}\frac{3z}{5}}$$
Which gives me
$$\frac{100}{z}$$
Back to the 3rd equation!
| Consider instead $$x(x+y+z)+y(x+y+z)+z(x+y+z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Rotate $xy=1$ by $\frac{\pi}{4}$ in a negative (clockwise) direction. I was studying hyporbolae for the first time and noticed that $y=\frac{1}{x}$ is a rotated hyperbola. I had seen equations like $y=\frac{1}{x}$ before but never noticed they where hyperbolae.
Anyway using geometry and the general form of a hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, I realised that $xy=1$ is $\frac{x^2}{2}-\frac{y^2}{2}=1$ which has been rotated clockwise by $\frac{\pi}{4}$.
I tried to rotate $xy=1$ using the rotation matrix: $ \left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right)$
$$\left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right) \left( \begin{array}{cc} x \\ \frac{1}{x}\end{array} \right)=\left( \begin{array}{cc} x'\\ y' \end{array} \right)$$
$$\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=x', -\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=y'$$
I used two methods. The first worked and the second did not. I'd like to know where I'm going wrong with the second approach.
In the first approach I squared both sides of the above equations and then used simultaneous equations to remove the $x$. Then dividing across by $2$ gives the desired result.
Just out of curiosity I tried another method. I first multiplied across both equations by $\sqrt{2}x$ and rearranged to quadratic form: $$x^2-\sqrt{2}x'x+1=0, x^2-\sqrt{2}y'x-1=0$$
I equated the solutions of each to get: $$\frac{\sqrt{2}x' \pm\sqrt{2x'^2-4}}{2}=\frac{-\sqrt{2}y' \pm\sqrt{2y'^2+4}}{2}$$
The $\pm$s are ugly but I figured I could get rid of them through a series of squaring. I tried a few different ways starting with roots on opposite sides, roots on the same side etc. but I can't quite get it through. Should this work or am I going down a dead-end road ?
Ok I got it: $$\frac{\sqrt{2}x \pm\sqrt{2x^2-4}}{2}=\frac{-\sqrt{2}y \pm\sqrt{2y^2+4}}{2}$$ simplifies to $$\frac{x^2}{2}-\frac{y^2}{2}=1$$
| You can't get a 4th degree equation by an affine transformation of coordinates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| $(x-a)(x-b)(x-c)+a(x-b)(x-c)+bx(x-c)+cx^2=x^3,$
which says that
$$y=\frac{x^3}{(x-a)(x-b)(x-c)}.$$
Thus, $$y(2)=\frac{8}{(2-a)(2-b)(2-c)}=\frac{8}{8-4\cdot3+2\cdot2-(-4)}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.
| Let the general term be $\frac {8r}{4r^4+1}=\frac {2r}{r^4+(\frac {1}{2})^2}=\frac {2r}{(r^2+\frac {1}{2})^2-r^2} =\frac {2r}{(r^2+\frac {1}{2}-r)(r^2+\frac {1}{2}+r)} $ now $2r=r^2+\frac {1}{2}+r-(r^2+\frac {1}{2}-r) $ so iur series is a telescoping one and is equal to $S=\sum _0^{\infty} \frac {1}{r^2+\frac {1}{2}-r}-\frac {1}{r^2+\frac {1}{2}+r}=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Find the domain of $f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ Find the domain of definition of $$f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$$
I found the domain of $\log_{10}(1-\log_7(x^2-5x+13))$ to be $x\in(2,3)$ and the domain of $\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ to be $\sin\frac{9\pi x}{2}=1$ i.e. $x=1,5,9$ etc but the answer given is $\frac{21}{9},\frac{25}{9}$.I dont know where i am wrong.
| $\left|\dfrac3{2+\sin{\frac{9\pi x}2}}\right|\leq1$
$⇔3\leq|2+\sin\frac{9\pi x}2|$
$⇒\sin\frac{9\pi x}2=1$
$⇒\frac{9\pi x}2=2k\pi+\frac{\pi}2$
$⇒x=\dfrac{1+4k}9$
$k=5,6$ satisfy that condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An expression involving the roots of a quadratic polynomial I have the equation $$ 9x^2 - 11x + 1 = 0 $$ whose two roots are $ \alpha $ and $ \beta $ .
I need to evaluate $$ \frac 1 {(9\alpha-11)^2} + \frac{11\beta - 1} 9$$
What I've tried
*
*Expanded the denominator and add them, but nothing simplifies and I get even a complex expression.
*I found that $ \alpha + \beta = \frac{11}{9}
\implies \alpha = \frac{11}{9} - \beta $
How to evaluate the value of the expression in an easier way?
| Since $9\alpha^2-11\alpha+1=0=9\beta^2-11\beta+1$, by Vieta's theorem we have
$$ \frac{1}{(9\alpha-11)^2}+\frac{(11\beta-1)}{9} = \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\left(\frac{11}{9}\right)^2-2\cdot\frac{1}{9}=\color{red}{\frac{103}{81}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
Now what ?
| We need to prove that
$$3(a^3+b^3+c^3)\geq(a^2+b^2+c^2)(a+b+c)$$
and since $(a^2,b^2,c^2)$ and $(a,b,c)$ are the same ordered, our inequality
it's just Chebyshov's inequality:
$$3(a^2\cdot{a}+b^2\cdot{b}+c^2\cdot{c})\geq(a^2+b^2+c^2)(a+b+c).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Set of $x$ such that $\sum \frac{nx^n}{n^2 + x^{2n}}$ converges What is the radius of convergence of $$\sum_{n=1}^\infty \frac{nx^n}{n^2 + x^{2n}}?$$ Here is what I have so far. With the ratio test I want to find values of $x$ for which
$$ \lim_{n\to\infty}x \frac{n+1}{n} \frac{n^2 + x^{2n}}{(n+1)^2 + x^{2(n+1)}} < 1.$$ If $x < 1$, this looks to be true. If $x = 1$, the series is essentially $\sum_{n=1}^\infty \frac{1}{n}$ so diverges. If $x = -1$, $\sum_{n=1}^n \frac{(-1)^n}{n}$ converges by the alternating series test. And for $x > 1$, the series seems to be essentially $$\sum_{n=1}^\infty \frac{1}{x^n}$$ which converges by the ratio test.
| Let's apply the ratio test
$$
\frac{a_{n+1}}{a_{n}} = x\frac{n+1}{n}\frac{n^2 + x^{2n}}{(n+1)^2+x^{2n+2}} =\frac{1}{x}\frac{n+1}{n}\frac{1 + (n/x^n)^2}{1+([n+1]/x^{n+1})^2}
$$
If $|x| < 1$, then clearly the first expression goes to $x$ as $n\rightarrow \infty$, and if $|x|>1$, clearly the second expression goes to $x^{-1}$ as $n\rightarrow\infty$. So if $|x| \ne 1$, the series converges absolutely.
What if $x = \pm 1$? Then the series is $\sum_{n=0}^\infty (\pm 1)^n n/(n^2+1)$. If $x = 1$, this series diverges by the integral test against $f(t) = t/(t^2+1)$. If $x = -1$, this series converges by the alternating series test, though only conditionally.
So the series converges for all $x$ except $x = 1$, though only conditionally for $x = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $ then prove that :
Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that :
$$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}
\geq \frac{ 4}{ a_{1}+ a_{2} + a_{3}+\cdots+ a_{n}}$$
I do not know where to start. please help me .
| Notice that $$\left( \frac{a_1}{a_2^2 + a_3^2 + ... + a_n^2} + \frac{a_2}{a_1^2 + a_3^2 + ... + a_n^2} + \ ... \ \frac{a_n}{a_1^2 + a_2^2+... + a_{n-1}^2}\right)(a_1 + a_2 + ... + a_n) \ge \left( \sqrt{\frac{a_1^2}{a_2^2 + a_3^2 + ... + a_n^2}} + \sqrt{\frac{a_2^2}{a_1^2 + a_3^2 + ... + a_n^2}} + \ ... \ \sqrt{\frac{a_n^2}{a_1^2 + a_2^2... + a_{n-1}^2}}\right)^2$$ by Cauchy-Schwarz inequality. WLOG, take $a_1^2 + a_2^2 + ... + a_n^2 = 1$ (because of homogenity). Thus it remains to prove that $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$Now take $x_m = \frac{a_m}{\sqrt{1-a_m^2}} \ \forall \ 1 \le m \le n$. Then it remains to prove that $$x_1 + x_2 + ... + x_n \ge 2$$ Now since $a_m^2 =\frac{x_m^2}{1+x_m^2}$, we have $$\frac{x_1^2}{1+x_1^2} + \ ... \ + \frac{x_n^2}{1+x_n^2} = 1$$ But $\frac{x_m^2}{1+x_m^2} = x_m \left( \frac{x_m}{1+x_m^2} \right) \le x_m \left( \frac 12\right)$ since $x_m \ge 0$ and it follows that $$1 = x_1 \left( \frac{x_1}{1+x_1^2} \right) + ... + x_n \left( \frac{x_n}{1+x_n^2} \right) \le \frac{x_1}2 + ... + \frac{x_n}2 = \frac 12 \left(x_1 + ... + x_n \right)$$ i.e. $$x_1 + x_2 + ... + x_n \ge 2$$ Done.
Note that $2$ is the minimum attainable value of $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$ (take $a_1 = a_2 =\frac 1{\sqrt 2}$ and all other $a_i $s equal to $0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to deduce the following equation from the other? How, from the equation $(1)$
$(1) a^{2x} + 2a^xb^y + b^{2y} = m - 2a^x - 2b^y$
can we deduce the equation $ (2) a^x + b^y = \sqrt{m + 1} - 1$
I know that the right hand side of $(1)$ is the perfect square $(a^x + b^y)^2 $, but this doesn't help me to deduce $(2)$.
Also I noticed that, by transposition of terms, $(1)$ can be expressed as $(a^x + b^y)(a^x + b^y + 2) = m$.
How to deduce $(2)$ from $(1)$?
| Write $a^x+b^y=s$, then the rearranged form of $(1)$ you have provided is $s(s+2)=m$ or $s^2+2s-m=0$. It remains to solve the quadratic equation for $s$, yielding
$$s=\frac{-2\pm\sqrt{4+4m}}2=-1\pm\sqrt{m+1}$$
Taking the $+$ in the $\pm$ gives $(2)$: $s=a^x+b^y=\sqrt{m+1}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of different natural numbers which are smaller than two hundred million and using only the digits $1$ or $2$ is Number of different natural numbers which are smaller than two hundred million and using only the digits $1$ or $2$ is:
These numbers will be either 1 digit,2 digits,.......9 digits.Number of such $9$ digits numbers are $2^8$,number of such $8$ digits numbers are $2^8$ and so on.So the total numbers are $2^8+2^8+2^7+2^6+.........+2^2+2^1$.
But the answer is $3\times 2^8-2$
| The number of different natural numbers which are smaller than $200,000,000$ and use only the digits $1$ or $2$ is
$$2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1$$
as you stated. However,
\begin{align*}
2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 & = 2^8 + \sum_{k = 1}^{8} 2^k\\
& = 2^8 + 2\sum_{k = 1}^{8} 2^{k - 1}\\
& = 2^8 + 2 \cdot \frac{1 - 2^8}{1 - 2}\\
& = 2^8 + 2(2^8 - 1)\\
& = 3 \cdot 2^8 - 2
\end{align*}
where we have used the geometric series formula
$$\sum_{k = 1}^{n} r^{k - 1} = \frac{1 - r^n}{1 - r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing $3^x+5^x$ is injective How can we show that
$$f(x)=3^x+5^x$$
is injective?
I know that $f$ is strictly increasing in $R$ so $f(x) = k, k \in R$ has only one root.
But I was looking for a proper proof, something I tried was:
Let $x,y \in \mathbb R$. Then for $f$ to be injection, $$3^{x}+5^{x} = 3^{y}+5^{y}$$ must imply that $x=y$.
Now how do we proceed?
| Suppose $3^x+5^x=3^y+5^y.$
Now $$3^x+5^x=3^y+5^y\Rightarrow 3^x-3^y=-(5^x-5^y)$$
Case 1: $x\geq y$, then $5^x-5^y\geq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\leq 0$. But $3^x-3^y\geq 0$. Which lead to $3^x=3^y\Rightarrow x=y$
Case 2: $x\leq y$, then $5^x-5^y\leq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\geq 0$. But $3^x-3^y\leq 0$. Which lead to $3^x=3^y\Rightarrow x=y$
Hence $3^x+5^x=3^y+5^y\Rightarrow x=y\space\space\space \blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why, after squaring, is the smallest root of the quartic the solution of the real problem? Looking at this post, I considered the more general problem of the solution of equation $$x+\sqrt{x+\sqrt{a+x}}=a\tag 1$$ where $a> 0$ .
Using successive squaring processes, this leads to the quartic
$$x^4-2 (2
a+1) x^3+\left(6 a^2+4 a+1\right) x^2-\left(4 a^3+2 a^2+1\right) x+a \left(a^3-1\right)=0\tag2$$ the discriminant of which being
$$\Delta=1024 a^4-1280 a^3-768 a^2-80 a-23$$ which is positive for any $a>\sqrt 3$; then, using the notations used here, since $P=-4(4a+1)$ and $D=-32(4a+1)$, all four roots are real and distinct. Extensive numerical tests showed that all roots are positive.
How could be explained the fact that the only solution of $(1)$ systematically corresponds to the smallest root of $(2)$ ?
How could be explained the fact that, for a few integer values of $a$, one of the solution of $(2)$ is also an integer ? Testing for the range $2 \leq a \leq 1000$, this is the case for the values $a=2,3,5,11,46,68,250,296,566,636$.
| If the real and positive solutions from the 4th-order equations are $x_1 < x_2 < x_3 < x_4$ and we introduce
$$
y_i^{(\alpha,\beta)} = x_i + \alpha \sqrt{x_i + \beta \sqrt{x_i + a}}
$$
where $\alpha,\beta = \pm$ and only take the positive roots. For each $(\alpha,\beta)$ there is a unique $x_i$ that will result in $y_i^{(\alpha,\beta)}=a$.
We can now make two observations:
*
*From the ordered $x_i$ it automatically follows that:
$$
y_1^{(++)} < y_2^{(++)} <y_3^{(++)} <y_4^{(++)}
$$
*For the different possibilities of $x_1$ we have
$$
\max \left( y_1^{(--)}, y_1^{(-+)}, y_1^{(+-)}, y_1^{(++)} \right) = y_1^{(++)}
$$
Hence if the solution of $x + \sqrt{x + \sqrt{x+ a}} =a$ would not correspond to $x_1$ (the smallest) but to $x_j$ one would obtain that $y_1^{(\alpha,\beta)} \leq y_1^{(++)} < a=y_j$, which contradicts the fact that there should be such a combination $(\alpha,\beta)$.
With respect to the solutions you missed the case $a=250$ and $a=296$ for values $a \leq 500$.
We have the following set of equations in integers at the different levels of square roots:
$$
x + a = n^2
$$
$$
x + n = m^2
$$
$$
x + m = a
$$
for some integers $n,m \in \mathbb{Z}$. From the second we obtain $n=m^2-x$ and substitution of this and the third equation in the first gives:
$$
x + (x+m) = (m^2 - x)^2
$$
and hence
$$
x^2 - 2(m^2+1)x + m(m^3 -1) = 0
$$
With two solutions:
$$
x_\pm = (m^2+1) \pm \sqrt{2 m^2 + m + 1}
$$
and the corresponding values for $a$ are:
$$
a_\pm = (m^2+m+1) \pm \sqrt{2 m^2 + m + 1}
$$
In order for this to be integers we only need to find value of $m$ (positive and negative) such that $2 m^2 + m + 1$ is a perfect square.
So the problem simplifies to finding integer solutions of the equation:
$$
2 m^2 + m + 1 = p^2
$$
This can be rewritten as
$$
8 p^2 - (4 m + 1)^2 = 7
$$
which is a type of Pell equation. The succesive solutions of $m_i$ that results in such a perfect square are given by $m_0=0$, $m_1=1$ and the recurrence relation $m_{i+1}=-6 m_{i} - m_{i-1} - 2$. Note that in order to obtain all solutions $(x,a)$ of the problem above, both positive and negative values of $m$ are allowed and that the recurrence relation needs to be followed in both directions.
With the substitution $m_i \rightarrow \frac{1}{4} (u_i - 1)$, a normal recurrence relation is obtained with $u_0=0$, $u_1=5$, and $u_{i+1}=-6 u_{i} - u_{i-1}$. Working this out properly gives:
$$
m_i = \frac{1 - 2 \sqrt{2}}{8} \left( -3 - 2 \sqrt{2} \right)^i + \frac{1 + 2 \sqrt{2}}{8} \left( -3 + 2 \sqrt{2} \right)^i - \frac{1}{4}
$$
and results in
$$
\begin{array}{llllll}
m_0=0 & m_1=1 & m_2=-8 & m_3=45 & m_4=-264 & \dots \\
& m_{-1}=-3 & m_{-2}=16 & m_{-3}=-95 & m_{-4}=552 & \dots
\end{array}
$$
Inserting this sequence in the expressions for the solutions $(x,a)$ gives:
$$
(x_+,a_+)=\dots,(280,296),(14,11),(2,2),(4,5),(76,68),\dots
$$
$$
(x_-,a_-)=\dots,(234,250),(6,3),(0,0),(0,1),(54,46),\dots
$$
which are all integer combinations for which $x + a - ((a-x)^2 - x)^2 == 0$. Only a subset of them will also give $x + \sqrt{x + \sqrt{x+ a}} =a$.
Note that also the $(x_i,a_i)$ could be written in a similar form, but more terms, as done for $m_i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
3 variable multiplication with 1 constraint lagrange multiplier Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^4y^6z^2$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$
So, $f(x,y,z) = x^4y^6z^2 $
and $g(x,y,z) = x^2 + y^2 + z^2 - 1$
then i've done the partial derivatives
$$\frac{\partial f}{\partial x}(x,y,z)=\lambda\frac{\partial g}{\partial x}$$
which gives
$$4x^3y^6z^2 = 2xλ$$
$$6x^4y^5z^2 = 2yλ$$
$$2x^4y^6z = 2zλ$$
which i subsequently go on to find that
$3x^2 = 2y^2 = 6z^2 $
This is where i've hit a dead end. Where do i go from here? or am i doing it all wrong?
Thanks.
| The minimal value is $0$ for $x=0$.
The maximal value we can find by AM-GM:
$$x^4y^6z^2=108\left(\frac{x^2}{2}\right)^2\left(\frac{y^2}{3}\right)^3z^2\leq108\left(\frac{2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}+z^2}{6}\right)^6=\frac{1}{432}.$$
The equality occurs for $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$,
which says that $\frac{1}{432}$ is a maximal value.
Done!
If you wish to use the Lagrange multipliers method you need to add the following words.
Let $F(x,y,z,\lambda)=x^4y^6z^2+\lambda(x^2+y^2+z^2-1)$ and $A=\left\{(x,y,z,\lambda)|x^2+y^2+z^2=1\right\}$.
Sinse $F$ is a continuous function and $A$ is a compact, we see that $F$ gets on $A$ the maximal value and gets on $A$ the minimal value, which happens for solutions of your system.
The rest is to solve the system $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$ for $xyz\neq0$ and to solve your system for $xyz=0$.
Now, you can choose, that you want.
I think that to solve our problem by AM-GM is much better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove the inequality $\sqrt{\frac{x}{y^2+z^2}}+\sqrt{\frac{y}{z^2+x^2}}+\sqrt{\frac{z}{x^2+y^2}}\ge \frac{2\sqrt{6}}3$ with $x+y+z=3,x,y,z\ge 0$
Let $x\geq0$, $y\geq0$ and $z\ge 0$ such that $x+y+z=3$. Show that
$$\sqrt{\dfrac{x}{y^2+z^2}}+\sqrt{\dfrac{y}{z^2+x^2}}+\sqrt{\dfrac{z}{x^2+y^2}}\ge \dfrac{2\sqrt{6}}{3}.$$
I tried C-S,Holder but without success.
$$\left(\sum_{cyc}\sqrt{\dfrac{x}{y^2+z^2}}\right)^2(\sum x(y^2+z^2))\ge (x+y+z)^3$$
| Holder helps!
$$\left(\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}\right)^2\sum_{cyc}x^2(y^2+z^2)\geq(x+y+z)^3.$$
Thus, it remains to prove that
$$(x+y+z)^3\geq\frac{8}{3}\sum_{cyc}x^2(y^2+z^2)$$ or
$$(x+y+z)^4\geq16\sum_{cyc}x^2y^2,$$ which is true by AM-GM:
$$(x+y+z)^4=\left(\sum_{cyc}(x^2+2xy)\right)^2\geq\left(2\sqrt{\sum_{cyc}x^2\cdot2\sum_{cyc}xy}\right)^2=$$
$$=8\sum_{cyc}(x^3y+x^3z+x^2yz)\geq8\sum_{cyc}(2x^2y^2)=16\sum_{cyc}x^2y^2.$$
Done!
Another way.
By AM-GM and C-S we obtain:
$$\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}=\sum_{cyc}\frac{\sqrt6x}{2\sqrt{1.5x(y^2+z^2)}}\geq$$
$$\geq\sum_{cyc}\frac{\sqrt6x}{1.5x+y^2+z^2}=2\sqrt6\sum_{cyc}\frac{x^2}{3x^2+2xy^2+2xz^2}\geq$$
$$\geq\frac{2\sqrt6(x+y+z)^2}{\sum\limits_{cyc}(3x^2+2x^2y+2x^2z)}.$$
Thus, it remains to prove that
$$3(x+y+z)^2\geq\sum\limits_{cyc}(3x^2+2x^2y+2x^2z)$$ or
$$3(xy+xz+yz)\geq\sum_{cyc}(x^2y+x^2z)$$ or
$$(x+y+z)(xy+xz+yz)\geq\sum_{cyc}(x^2y+x^2z)$$ or
$$3xyz\geq0.$$
Done againe!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the best way to solve modular arithmetic equations such as $9x \equiv 33 \pmod{43}$? What is the best way to solve equations like the following:
$9x \equiv 33 \pmod{43}$
The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.
Is there a more efficient way ?
Help would be greatly appreciated!
| $43$ is prime. And even if it weren't, $\gcd(9,43)= 1$.
That means we know that $9^{-1}_{43}$ exists. That is there is a number so that $9\cdot 9^{-1}_{43} \equiv 1 \pmod {43}$. And we could calculate (or guess) that $9^{-1}_{43}\equiv 24$ (because $9\cdot 24\equiv 1 \pmod {43}$) and we can solve $9x \equiv 33\pmod {43} \implies 24\cdot 9 x\equiv x \equiv 24\cdot 33\pmod {43}$.
But we don't have to. If $\gcd(9,43) =1$ and if $9^{-1}$ exists, then in this case we are allows to know that division is acceptable, and we are, in this case, allowed to divide both sides by $3$.
$9x \equiv 33 \pmod {43}$
$3x\equiv 11 \pmod {43}$ (we wouldn't be allowed to do this if $\gcd(3,43)\ne 1$)
Now $3$ is such a small number that we know that one these: $11, 11 + 43, 11+2*43$ is divisible by $3$.
As it turns out $3x \equiv 11 \equiv 11 + 43 \equiv 54\pmod {43}$.
And we can divide both sides by $3$ to get
$x \equiv 18\pmod {43}$.[1]
..........
Alternatively we can do baby steps.
$9x \equiv 33 \pmod {43} \implies$ there is a $k$ so that
$9x = 33 + 43k$. Divide by $3$ and we get
$3x = 11 + 43\frac k3$. That will be an integer so long as we choose a $k$ that is divisible by $3$. Relable $m= \frac k3$
$3x = 11 + 43m$
$3x =11 + 43 + 43(m-1)$
$3x = 54 +43(m-1)$
$x = 18 + 43\frac {m-1}3$ which is an integer so long as we select $m-1$ so that $m-1$ is divisible by $3$. (For example the easiest example is $m-1=0$, $m=1$, $k = 3$ an $9*18 = 33 + 3\cdot 43$. But we don't need to find the actually $k$)
......
[1] Notice our earlier answer $24\cdot 33\equiv 24\cdot 32 + 24\equiv 48\cdot 16 + 24 \equiv 5\cdot 16 + 24\equiv 80 + 24 \equiv 86 + 18 \equiv 18\pmod {43}$. So our answers are consistent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Contour integration getting wrong result.
By means of contour integration, evaluate
$$
\int_0^\infty \frac{(\log_e(u))^2}{u^2+1} \, du \, .
$$
When we apply Cauchy Residue theorem, we get $+i$ as a valid pole.
The residue for this pole is $\frac{(i(\pi)^2)}{8}$.
The final answer is $2(\pi)i *$ (sum of residues), which gives answer as $\frac{(-(\pi)^3)}{4}$.
Changing limits to $0$ to infinity, we get $\frac{(-(\pi)^3)}{8}$.
But the answer from other the regular integration is $\frac{(\pi^3)}{8}$.
Why am I getting an extra minus ($-$)?
| As far as I can see, you've accurately calculated the integral
$$\int_{-\infty}^\infty{\frac{\log^2{u}}{u^2+1}du} = -\frac{\pi^3}{4}$$
I suspect the problem comes because you're assuming that
$$\int_{-\infty}^\infty{\frac{\log^2{u}}{u^2+1} du} = 2 \int_{0}^\infty{\frac{\log^2{u}}{u^2+1}du} $$
This is not true since, for positive $u$, $$\log(-u) = \log(u)+i\pi$$.
Using this, we can get the answer you need by realising that
$$\int_{-\infty}^\infty{\frac{\log^2{u}}{u^2+1}du} = \int_{0}^\infty{\frac{\log^2{u}}{u^2+1}du} + \int_{0}^\infty{\frac{(\log{u}+i\pi)^2}{u^2+1}du}$$
Expanding, this gives us
$$2 \int_{0}^\infty{\frac{\log^2{u}}{u^2+1}du} - \pi^2 \int_0^\infty \frac{1}{u^2 + 1}du + 2 \pi i \int_0^\infty \frac{\log u}{u^2 + 1}du = -\frac{\pi^3}{4}$$
Clearly, the RHS is real, and so even without evaluating it you can be sure that the imaginary part of the LHS should integrate out to be zero (you could verify this if you want). Furthermore, the integral
$$\int_0^\infty \frac{1}{u^2 + 1}du = \frac{\pi}{2}$$
Thus, we have that
$$2 \int_{0}^\infty{\frac{\log^2{u}}{u^2+1}du} = \frac{\pi^3}{2} -\frac{\pi^3}{4} = \frac{\pi^3}{4}$$
Or in other words,
$$\int_{0}^\infty{\frac{\log^2{u}}{u^2+1}} = \frac{\pi^3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability that at least $2$ people will not receive any ace.
I've a deck with 52 french cards ($13$ values for each of $4$ suits) and $4$ players. Randomly dealing out all cards, what's the probability that at least
$2$ people will not receive any ace?
My try:
$$p=\frac{\frac{4!}{2!}\binom{48}{13,13,12,10}+\binom{4}{2}\binom{48}{13,13,11,11}+\frac{4!}{3!}\binom{48}{13,13,13,9}}{\binom{52}{13,13,13,13}}$$
Where:
$\binom{48}{13,13,13,9}$ is the case $A$ has all $4$ aces,
$\binom{48}{13,13,12,10}$ the case $A$ has $1$ ace $B$ has $3$ aces,
$\binom{48}{13,13,11,11}$ the $A$ has $2$ aces and same for $B$,
$\frac{4!}{2!}$ arrangements of $4$ people to be $A$ and $B$,
$\binom{4}{2}$ combination of $4$ people to be $A$ and $B$,
$\frac{4!}{3!}$ arrangements of $4$ people to be $A$
Am I right? If yes, is there a more elegant solution than mine?
| There are $\binom42$ ways to choose two of the people who have 26
available "slots" between them.
Consider only the distribution of aces, where the rest of the cards go doesn't matter.
The probability that all the aces fall in one or other of the 26
slots of the chosen two is
$\binom42\cdot \frac{26}{52}\cdot\frac{25}{51}\cdot\frac{24}
{50}\cdot\frac{23}{49}$
There are $3$ groups of two containing A, say, viz. AB, AC and AD, so cases where A has all the aces have been counted three times, and similarly for the others, so to correct for this, subtract
$2\cdot4 \cdot \frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}$
The final answer comes out as $\frac{76}{245}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse laplace transform of complicated rational function
Find the inverse laplace $$\mathcal{L}^{-1}\left\{ \frac{2s + 1}{2s^2 + s + 2} \right\}$$
I don't see a simple way to take the inverse laplace transform, convolution theorem?
| This is a case where you "massage" the expression to fit it into a more standard form
First get rid of the leading coefficient in the denominator.
$$ \frac{2s+1}{2s^2+s+2}=\frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}$$
Then complete the square on the denominator.
$$ \frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}=\frac{s+\frac{1}{2}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}} $$
Then re-write the numerator to obtain the result
$$ \frac{2s+1}{2s^2+s+2}= \frac{\left(s+\frac{1}{4}\right)+\frac{1}{4}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}} $$
Set this up for solving in terms of $\sin$ and $\cos$ and $e^{-t/4}$
$$ \mathcal{L}^{-1}\left(\frac{\left(s+\frac{1}{4}\right)}{\left(s+\frac{1}{4}\right)^2+\left(\frac{\sqrt{15}}{4}\right)^2}\right)+\frac{1}{4}\cdot\frac{4}{\sqrt{15}}\mathcal{L}^{-1}\left(\frac{\frac{\sqrt{15}}{4}}{\left(s+\frac{1}{4}\right)^2+\left(\frac{\sqrt{15}}{4}\right)^2}\right) $$
to obtain the result
$$f(t)=\left[\cos\left(\frac{\sqrt{15}}{4}t\right)+\frac{\sqrt{15}}{15}\sin\left(\frac{\sqrt{15}}{4}t\right)\right]e^{-t/4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by contradiction:
Assume $c$ is divisible by $3$ and $a$ or $b$ is not divisible by $3$.
Since $c$ is divisible by $3$ we can write $c$ as $ \ c = 3m \implies c^{2} = 9m^{2} \implies 9 | c^{2}$.
Since $a$ and $b$ are not divisible by $3$, $\ a = 3k+1$ and $ \ b = 3n+1$ for some integers $\ k,n.$
Then,
$ a^{2} + b^{2} = (3k+1)^{2} + (3n+1)^{2} = 9k^{2} + 6k +9n^{2} + 6n + 2$.
I am stuck here. I can't find a contradiction. How can I show that $ a^{2} + b^{2} $ is not divisible by $9$.
| Assume that the remainders of the divisions $a:3$ and $b:3$ are $r$ and $s$, respectively. Write $a=3p+r$, $b=3q+s$
Then
$$a^2+b^2=9(p^2+q^2)+6(pr+qs)+r^2+s^2$$
Since $r$ and $s$ are $0$, $1$ or $2$, $r^2+s^2$ is $0$, $1$, $4$, $2$, $5$ or $8$. Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Divide triangle into two equivalent parts by line perpendicular to the base. Originally the problem wants me to divide into 3 equivalent(equal areas) parts but I thought that I should do 2 first. Tried 2 and could not after many attempts. DF below is movable so it is not really the correct solution.
| The Math
\begin{align}
\text {Area of }\triangle \text{ ABH} &= \frac{mbh}{2} \\[1.5ex]
\text {Area of }\triangle \text{ AXY} &= \frac{mx^2}{2} \\[1.5ex]
\text{For } \triangle \text { AXY to be half the area of } \triangle { ABH:} \\[1.5ex]
\frac{mx^2}{2} &= \frac{1}{2} \times \frac{mbh}{2} \\[1.5ex]
\text{giving } x \text{ (independent of gradient } m \text{)}: \\[1.5ex]
x^2 &= \frac{bh}{2}\\
x &= \sqrt{\frac{bh}{2}}
\end{align}
The Construction
1. Draw a line $\perp$ to AC through A, and a line $\parallel$ to AC through B. Mark D, the point of intersection.
2. Mark E, the midpoint of AC. Then with centre A and radius AE, draw a circle. Mark F, the point of intersection of the circle and the perpendicular.
\begin{align}
\text{DF} &= \text{AD} + \text{AF} = h \text{ (height)} + \frac{b \text{ (base)}}{2}
\end{align}
*Mark G, the midpoint of DF. Then with centre G and radius DG, draw a circle cutting AC in H.
AH = $\sqrt{\frac{bh}{2}}$, which is the required length (see Note 1 below).
*Draw the line $\perp$ to AC through H, cutting AB in I.
*$\triangle$ AHI is half the area of $\triangle$ ABC
Notes:
*
*In step 3 above,
\begin{align}
\text {AD } & = h \\
\text {AF } & = \frac{b}{2} \\[1.5ex]
\text {DG } & = \frac{h + \frac{b}{2}}{2} = \frac{h}{2} + \frac{b}{4} \\[1ex]
\text {So, AG} &= h - (\frac{h}{2} + \frac{b}{4}) = \frac{h}{2} - \frac{b}{4} \\[1ex]
\text {And, GH} & = DG = \frac{h + \frac{b}{2}}{2} = \frac{h}{2} + \frac{b}{4} \\[1ex]
\text{Now, let AH } &= x \\[1ex]
\text{Then by Pythagoras' Theorem (} \triangle \text{AGH):} \\
x^2 & = (\frac{h}{2} + \frac{b}{4})^2 - (\frac{h}{2} + \frac{b}{4})^2 \\[1ex]
& = \frac{bh}{2} \\
\text{And, } x &= \sqrt{\frac{bh}{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| The condition gives $$(a+b)ab=a^2+b^2$$ or
$$(b-1)a^2+b^2a-b^2=0.$$
If $b=1$ we obtain $a=1$ and $a^2+b^2=2$.
If $b\neq1$ we need $b^4+4b^2(b-1)=m^2$, where $m\in\mathbb N$ or
$$b^2+4b-4=n^2,$$
where $n\in\mathbb N$ or
$$(b+2)^2=8+n^2$$ or
$$(n-b-2)(n+b+2)=-8$$ and since $n>0$, we obtain two cases only:
*
*$n-b-2=4$ and $n+b+2=-2$, which gives $n=1$, $b=-5$, $a=\frac{5}{2}$ or $a=\frac{5}{3},$ which is impossible;
*$n-b-2=-2$ and $n+b+2=4$, which gives $n=1$, $b=1,$ which is impossible here.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Probability of truth in a chain of statements
Three individuals $A$,$B$ and $C$ tell the truth with probability $1/3$.
(A) $C$ makes a statement and $A$ claims that it is true. What is the probability that the statement is true.
(B) $C$ makes a statement and $A$ tells you that $B$ claims the statement is true. What is the probability that the statement is true.
Let $T_A$ be the event that $A$ tells the truth and similarly $T_B \ \& \ T_C$
My answer in the first case $\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}$.
I am confused with part $B$ of the question.
My reasoning: Probability of $C$ telling the truth is $1/3$. Now we don't know what $B$ said so we assume two cases. Therefore we would have the probability as $P(T_A \ T_B \ T_C \cup T_A \ T_B^C \ T_C^C) = \frac{1}{3}\cdot \frac{1}{3} \cdot \frac{1}{3}+\frac{1}{3}\cdot \frac{2}{3} \cdot \frac{2}{3}$.
Is this correct? This problem appears as an exercise to Bayes theorem so I am confused as to how the theorem is applicable.
Thanks in advance for any assistance.
| Careful. We need to distinguish between "$X$ tells the truth" and "$X$ claims that $Y$ told the truth". The former always happens with probability $1/3$, while the latter depends on whether or not $Y$ is true or false. Let's revisit part (a) and redefine our events. Define:
*
*$X_C$: $C$ makes a true statement.
*$X_A$: $A$ claims that $C$ makes a true statement.
We're asked to find the probability that $C$ made a true statement, given that $A$ claims that $C$ made a true statement:
\begin{align*}
\Pr[X_C \mid X_A]
&= \frac{\Pr[X_CX_A]}{\Pr[X_A]} \\
&= \frac{\Pr[X_CX_A]}{\Pr[X_CX_A] + \Pr[\overline{X_C}X_A]} \\
&= \frac{\frac{1 \cdot 1}{9}}{\frac{1 \cdot 1}{9} + \frac{2 \cdot 2}{9}} \\
&= \frac{1}{1 + 4} \\
&= \frac{1}{5}
\end{align*}
Can you see how to generalize this for part (b)?
EDIT: Here's part (b). Define:
*
*$X_C$: $C$ makes a true statement.
*$X_B$: $B$ claims that $C$ makes a true statement.
*$X_A$: $A$ claims that $B$ makes a true statement.
We're asked to find the probability that $C$ made a true statement, given that $A$ claims that $B$ made a true statement:
\begin{align*}
\Pr[X_C \mid X_A]
&= \frac{\Pr[X_CX_A]}{\Pr[X_A]} \\
&= \frac{\Pr[X_CX_BX_A] + \Pr[X_C\overline{X_B}X_A]}{\Pr[X_CX_BX_A] + \Pr[X_C\overline{X_B}X_A] + \Pr[\overline{X_C}X_BX_A] + \Pr[\overline{X_C} \, \overline{X_B}X_A]} \\
&= \frac{\frac{1 \cdot 1 \cdot 1}{27} + \frac{1 \cdot 2 \cdot 2}{27}}{\frac{1 \cdot 1 \cdot 1}{27} + \frac{1 \cdot 2 \cdot 2}{27} + \frac{2 \cdot 2 \cdot 1}{27} + \frac{2 \cdot 1 \cdot 2}{27}} \\
&= \frac{1 + 4}{1 + 4 + 4 + 4} \\
&= \frac{5}{13}
\end{align*}
This diagram might help:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$
$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$
Apply L'Hôpital's rule:
$\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$
Apply L'Hôpital's rule again:
$\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$ $a=-2$
Is my approach right?
| We need two limits below (which are easily obtained and the second one necessitates the use of Taylor series or L'Hospital's Rule) $$\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2},\,\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}$$ The limit in question can be written as $$\lim_{x\to 0}(a+1)\cdot\frac{x-\sin x} {x^{3}}-a\frac{1-\cos x} {x^{2}}-\frac{(b-a-1)\sin x} {x^{3}}=1$$ This means that $$\lim_{x\to 0}\frac{(b-a-1)\sin x}{x^{3}}=1+\frac{a}{2}-\frac{a+1}{6}\tag{1}$$ or (upon multiplication by $x^{2}$) $$\lim_{x\to 0}\frac{(b-a-1)\sin x} {x} =0$$ ie $b-a-1=0$. Using this in $(1)$ we get $$1+\frac{a}{2}-\frac{a+1}{6}=0$$ ie $a=-5/2,b=a+1=-3/2,a+b=-4$.
Your approach has a fundamental issue when you replace $(b\sin x) /x^{3}$ by $b/x^{2}$. This is not allowed by any theorem on limits and is an invalid step in the evaluation of a limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve $y''+y=\cos x$?
Solve $y''+y=\cos x$.
After first solving the homogeneous equation we know that the solution to it is $y=c_1\cos x+c_2\sin x$.
We can guess that the private solution to non-homogeneous equation will be of form: $y_p=x(A_1\cos x+A_2\sin x)$.
Then:
$$
y_p'=A_1\cos x-A_1x\sin x+A_2\sin x+A_2x\cos x\\
y_p''=-2A_1\sin x-A_1x\cos x+2A_2\cos x-A_2x\sin x
$$
If we plug these into the original equation we get:
$$
\cos x(A_1+A_2x-A_1x+2A_2)+\sin x(A_2-A_1-2A_2-A_2x)=\cos x \quad\ast
$$
We can try to solve the system:
$$
\begin{cases}
x(A_2-A_1)+A_1+2A_2=1\\
x(-A_1-A_2)+A_2-2A_1=0
\end{cases}
$$
But there're 3 unknowns in the system so I don't see how to find out the values of $A_1$ and $A_2$.
The solution says that we get $2(A_2\cos x-A_1\sin x)=\cos x$ from which is follows that $A_1=0$ and $A_2=0.5$. But how do we get to this conclusion? Is there some trick I missed?
I checked my calculations in Wolfram Alpha and they match.
| So the corresponding auxiliary equation to $y''+y=\cos x$ is $m^2+1=0$, so $$y_c=c_1 \cos x + c_2 \sin x,$$
so things are fine so far. Now since our RHS is $\cos x$, like you said, we assume that the particular solution is of the form $A \sin x+B \cos x$. But since $A\sin x$ is already accounted for in $y_c$, we take $Ax \sin x+Bx \cos x$. Thus,
\begin{cases} y_p=Ax \sin x+Bx \cos x \\ {y_p}'=Ax \cos x+B \cos x-Bx \sin x +A \sin x \\ {y_p}''= -Bx \cos x+2A \cos x-Ax\sin x-2B \sin x \end{cases}
$\implies \left( -Bx \cos x+2A \cos x-Ax\sin x-2B \sin x \right) + \left( Ax \sin x+Bx \cos x \right) = \cos x$
$\implies 2A \cos x - 2B \sin x = \cos x \implies \begin{cases} A=\frac{1}{2} \\ B =0 \end{cases}$
Therefore our solution $y=y_c + y_p$ is
$$y=c_1 \cos x + c_2 \sin x+\frac{1}{2}x \sin x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$
Each of these "not yet a number" terms can be expanded to give
$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$
$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$
$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$
Adding each of these terms together and factoring out the two we find that they equal
$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$
So our total is $9$.
This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method.
| It is an olympiad question. Otherwise picking suitable numbers will do:
$$a=1,b=2,c=-3$$
$$S=\left(5-2+\frac13 \right)\left(\frac15 -\frac12+3\right)=\frac{10}{3}\cdot \frac{27}{10}=9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$.
Then recall that in evaluating a discontinuous integrand, the integral is defined by the relations $$\int_a ^b f(x) = lim_{x->b^-} \int_a ^x f(x) dx $$
if $x = b$ is the discontinuous point or $$\int_a ^b f(x) = lim_{x->a^+} \int_x ^b f(x) dx $$
if $x = a$ is the discontinuous point.
With that in mind:
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$
$$\int _0 ^3\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} + \int _3 ^4\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} $$
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} + \int _3 ^4\frac{dx}{(x-1)^2 -(2)^2} $$
Remembering that $\int \frac{du}{u^2 +a^2} = \frac{1}{a} \arctan \left( \frac{u}{a}\right) + C, $
we get:
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} +\int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$
equals
$$lim_{x->3^+} \int_0 ^x \frac{dx}{(x-1)^2 -(2)^2} + lim_{x->3^-} \int_x ^4 \frac{dx}{(x-1)^2 -(2)^2} $$
equals
$$lim_{x->3^+} \left( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_3 ^x + lim_{x->3^-} \left(\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_x ^4$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{3-2}{2}\right) \right ) + \left ( \frac{1}{2} \arctan \left( \frac{4-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)- 0.2318 \right ) + \left ( \frac{\pi}{8}-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
which makes $$\int _0 ^4\frac{dx}{x^2 - 2x - 3} = 0.1609$$
But in my book, it said there is no value of $\int _0 ^4\frac{dx}{x^2 - 2x - 3}.$
How do you prove that the integral $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ has no value?
| In fact, you should use the expression for integral of the function $\dfrac{1}{x^2-a^2}$:
$$
\int_3^4 \frac{dx}{(x-1)^2-2^2} = \int_2^3 \frac{dx}{x^2-2^2} = \frac{1}{4}\lim_{x\to 2^-}\left (\log\frac{1}{5} -\log\frac{x-2}{x+2}\right ).
$$
The above limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A Curious binomial identity While playing around with random binomial coefficients , I observed that the following identity seems to hold for all positive integers $n$:
$$ \sum_{k=0}^{2n} (-1)^k \binom{4n}{2k}\cdot\binom{2n}{k}^{-1}=\frac{1}{1-2n}.$$
However, I am unable to furnish a proof for it ( though this result is just a conjecture ). Any ideas/suggestions/solutions are welcome.
| When going through Jack's nice answer I did some intermediate steps to better see what's going on. Here is a somewhat more elaborated version, which might also be convenient for other readers.
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{(-1)^k\binom{4n}{2k}\binom{2n}{k}^{-1}}\\
&=\sum_{k=0}^{2n}\binom{4n}{2k}(2n+1)\int_0^1(1-x)^kx^{2n-k}\,dx\tag{1}\\
&=(2n+1)\int_0^1x^{2n}\sum_{k=0}^{2n}\binom{4n}{2k}\left(-\frac{1-x}{x}\right)^k\,dx\tag{2}\\
&=(2n+1)\int_0^1x^{2n}\cdot\frac{1}{2}
\left(\left(1+i\sqrt{\frac{1-x}{x}}\right)^{4n}+\left(1-i\sqrt{\frac{1-x}{x}}\right)^{4n}\right)\,dx\tag{3}\\
&=\frac{2n+1}{2}\int_0^1\left(\sqrt{x}+i\sqrt{1-x}\right)^{4n}+\left(\sqrt{x}-i\sqrt{1-x}\right)^{4n}\,dx\tag{4}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}
\left[(\cos \theta+i\sin \theta)^{4n}+(\cos\theta-i\sin \theta)^{4n}\right]\cos \theta\sin \theta\,d\theta\tag{5}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}
\left[e^{4ni\theta}+e^{-4ni\theta}\right]\cos \theta\sin \theta\,d\theta\tag{6}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}\cos(4n\theta)\sin(2\theta)\,d\theta\tag{7}\\
&=\frac{2n+1}{2}\int_{0}^{\frac{\pi}{2}}\left[\sin( (4n+2)\theta)-\sin ((4n-2)\theta))\right]\,d\theta\tag{8}\\
&=\frac{2n+1}{2}\left[-\frac{1}{4n+2}\cos((4n+2)\theta)
+\frac{1}{4n-2}\cos((4n-2)\theta\right]_0^{\frac{\pi}{2}}\\
&=\frac{2n+1}{2}\left(\frac{1}{2n+1}-\frac{1}{2n-1}\right)\\
&\color{blue}{=\frac{1}{1-2n}}
\end{align*}
Comment:
*
*In (1) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}
*In (2) we do some rearrangements in order to apply the binomial theorem.
*In (3) we consider the even function derived from $(1+z)^{2n}$
\begin{align*}
\sum_{k=0}^{n}\binom{2n}{2k}z^{2k}=\frac{1}{2}\left((1+z)^{2n}+(1-z)^{2n}\right)
\end{align*}
Replacing $n$ with $2n$ and $z$ with $i\sqrt{\frac{1-x}{x}}$ and the application of de Moivre's theorem in (6) becomes plausible.
*In (4) we do some simplifications.
*In (5) we substitute $x=\cos ^2\theta, dx=-2\cos\theta\sin\theta\,d\theta$.
*In (6) we apply De Moivre's theorem and in (7) and (8) trigonometric sum formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Conversion of Parametric Equation to a Rectangular Equation I was looking for the rectangular equation of the given parametric equation $$ x = \tan \theta$$
$$ y = \tan 2\theta$$
My work:
$$x = \tan \theta$$
$$\theta = \arctan x$$
Then, substituting it to $ y = \tan 2\theta$, it becomes:
$$y = \tan 2\theta$$
$$y = \tan (2(\arctan x))$$
At this point, I'm stuck. How do you get the Rectangular equation of the above Parametric equation?
| From trigonometry we know that $\displaystyle \tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}.$
If $\alpha$ happens to be the same as $\beta$ then this says $\displaystyle\tan(\alpha+\alpha) = \frac{\tan\alpha + \tan\alpha}{1-\tan\alpha\tan\alpha} = \frac{2\tan\alpha}{1 - \tan^2\alpha}.$
In other words $\displaystyle\tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha}.$
That is the double-angle formula for the tangent function.
So $\displaystyle y = \tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2x}{1-x^2}.$
So you're looking for the graph of $y = \dfrac{2x}{1-x^2}.$
(Note that $\tan\theta$ goes all the way from $-\infty$ to $+\infty,$ and therefore so does $x.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the Galois group of $x^6+x^5-8x^4-5x^3+19x^2+4x-11=0$? A while ago I happened on this equation: $$f(x)=x^6+x^5-8x^4-5x^3+19x^2+4x-11=0.$$
$f$ does not have linear factors, as can be seen by drawing its graph. I am not certain about whether it has quadratic or cubic factors over $\mathbb{Q}$, but over $\mathbb{Q}(\sqrt{5}$) it factors into $$(x^3+\frac{1}{2}(1-\sqrt{5})x^2+\frac{1}{2}(-\sqrt{5}-7)x+\frac{1}{2}(1+3\sqrt{5}))*$$ $$(x^3+\frac{1}{2}(1+\sqrt{5})x^2+\frac{1}{2}(\sqrt{5}-7)x+\frac{1}{2}(1-3\sqrt{5})). $$
This seems to say that one can first extend $\mathbb{Q}$ to $\mathbb{Q}(\sqrt{5})$, then extend somehow to include the roots of the two cubics. I do note that the polynomial $f(x^2-3)$ is divisible by $f(x)$ with no remainder. This means that if $r$ is a root of $f(x)$; i.e. a root of either of these two cubics, then $r^2-3$ is also a root. In fact I constructed the polynomial by evaluating $g(g(g(x)))-x$, where $g(x)=x^2-3$, and throwing out the quadratic factor.
I evaluated numerically the roots of the equation and came up with
(1) -2.22918
(2) 1.96925
(3) 0.877962
(4) -2.10553
(5) 1.43326
(6) -0.94576
$g$ permutes these roots; in fact, $g$ as a permutation is (123)(456), using the labels above. So $g$ is an element of the Galois group. There is also an order-2 mapping of the roots, namely the mapping $h$ that takes 1 to 1 and $\sqrt{5}$ to $-\sqrt{5}$; the non-zero element of the Galois group of $\mathbb{Q}(\sqrt{5})$ over $\mathbb{Q}$. I have tried generating subgroups of $S_6$ by using $g$ and some triple transposition (for $h$) for several possibilities of the triple transposition, and in each case I came up with $\mathbb{Z}_6$ as the Galois group. But if that's the case, then a cubic extension of $\mathbb{Q}(\sqrt{5})$ to an extension field $E$ would split the equation; $E$ would be the splitting field. But that then says the roots of one of the cubics above can be written as combinations of the roots of the others using only numbers from $\mathbb{Q}(\sqrt{5})$ as coefficients, but I don't see how this is done. It still seems to me the Galois group could be a group of order 18. How does one show that this is not the case?
| Irreducible over the rationals. Galois group as Robert said; I added two degree six polynomials where the roots are also real of modest absolute value, but come out very pretty, method originally due to Gauss.
================================================================
? f = x^6 + x^5 - 8 * x^4 - 5 * x^3 + 19 * x^2 + 4 * x - 11
%13 = x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11
? factor(f)
%14 =
[x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11 1]
? polgalois(f)
%2 = [18, -1, 1, "F_18(6) = [3^2]2 = 3 wr 2"]
? poldisc(f)
%3 = 2415125
? factor(poldisc(f))
%4 =
[5 3]
[139 2]
===================================================================
? f = x^6 + x^5 - 5 * x^4 - 4 * x^3 + 6 * x^2 + 3 * x - 1
%5 = x^6 + x^5 - 5*x^4 - 4*x^3 + 6*x^2 + 3*x - 1
?
? polgalois(f)
%6 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%7 = 371293
? factor(poldisc(f))
%8 =
[13 5]
===========================================================
?
?
? f = x^6 + x^5 - 15 * x^4 - 28 * x^3 + 15 * x^2 + 38 * x - 1
%9 = x^6 + x^5 - 15*x^4 - 28*x^3 + 15*x^2 + 38*x - 1
? polgalois(f)
%10 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%11 = 8390618797
? factor(poldisc(f))
%12 =
[11 2]
[37 5]
============================================================
One of them has roots $2 \cos \left( \frac{2\pi}{13} \right),2 \cos \left( \frac{4\pi}{13} \right), $ and so on. The last one has a root $2 \cos \left( \frac{2\pi}{37} \right) +2 \cos \left( \frac{20\pi}{37} \right) +2 \cos \left( \frac{22\pi}{37} \right), $ the other five are also sums of three cosine terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Closed-form for Floor Sum 1 Does a closed form exist for the following sum?
$$\sum_{k=0}^n \lfloor \sqrt{k} + \sqrt{k + n} \rfloor$$
If not, why is this sum so radically different than the sums below?
Closed forms do exist for the following sums*:
$$\sum_{k=0}^n \lfloor \sqrt{k + n} \rfloor$$
$$\sum_{k=0}^n \lfloor \sqrt{k} \rfloor$$
There is this floor functional identity:
$$\lfloor \sqrt{k} + \sqrt{k + 1} \rfloor = \lfloor\sqrt{4k+2}\rfloor$$
Don't know if this will help.
Thanks
*Existing closed forms
$$\sum_{k=0}^n \lfloor \sqrt{k} \rfloor=2\left(\sum_{k=0}^{\lfloor \sqrt{n} \rfloor-1}k^2\right)+\left(\sum_{k=0}^{\lfloor \sqrt{n} \rfloor-1}k\right)+\lfloor\sqrt{n}\rfloor\left(n-\lfloor\sqrt{n}\rfloor^2+1\right)$$
$$\left(\sum_{k=0}^n k^2\right)=\frac{2n^3+3n^2+n}{6}$$
$$\left(\sum_{k=0}^n k\right)=\frac{n^2+n}{2}$$
$$\sum_{k=1}^n \lfloor \sqrt{k+C} \rfloor=\sum_{k=C+1}^{C+n} \lfloor \sqrt{k} \rfloor=\sum_{k=0}^{C+n} \lfloor \sqrt{k} \rfloor-\sum_{k=0}^{C} \lfloor \sqrt{k} \rfloor$$
| The "difference" is actually that
$$
\eqalign{
& \left\lfloor {x + y} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 - \left\{ x \right\} \le \left\{ y \right\}} \right] \cr}
$$
where
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
$$
and where $[P]$ denotes the Iverson bracket
So
$$
\eqalign{
& \left\lfloor {\sqrt k + \sqrt {k + n} } \right\rfloor = \cr
& = \left\lfloor {\sqrt k } \right\rfloor + \left\lfloor {\sqrt {k + n} } \right\rfloor + \left[ {1 - \left\{ {\sqrt k } \right\} \le \left\{ {\sqrt {k + n} } \right\}} \right] \cr}
$$
and since you know the first two terms, the difficulty is
to establish when the condition in the Iverson bracket is met.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Formulation of the Wallis product for $\pi/2$ using the $\Gamma$ function I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.
This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots $$
Now
$$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot \dots\ (2n)(2n)}{1\cdot1\cdot3\cdot3\cdot5\cdot5\cdot\dots\ (2n-1)(2n-1)(2n+1)}\cdots $$
$$=\frac{1\cdot2\cdot1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot2\cdot3\cdot2\cdot \hspace{12pt} \dots\hspace{12pt}n\cdot2\hspace{12pt}\cdot\hspace{12pt} n\cdot2\hspace{24pt}}{\frac{1}{2}\cdot2\cdot\frac{1}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\dots\ (n-\frac{1}{2})\cdot2\cdot(n-\frac{1}{2})\cdot2\cdot(2n+1)}\cdots $$
$$=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot \hspace{12pt} \dots\hspace{12pt}n\hspace{12pt}\cdot\hspace{12pt} n\hspace{24pt}}{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot(n-\frac{1}{2})\cdot(2n+1)}\cdots $$
$$=\big(\frac{1\cdot2\cdot3\cdot\ \dots\ n\hspace{24pt}}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot\sqrt{2n+1}}\cdots\big)^2 $$
$$ = \lim_{n\rightarrow\infty}\;\;\Bigg(\frac{\Gamma{(n+1)}}{\frac{\Gamma{(n+\frac{1}{2})}}{\sqrt{\pi}}}\Bigg)^2\frac{1}{(2n+1)}$$
$$=\lim_{n\rightarrow\infty}\;\; \frac{\pi}{2n+1}\Big(\frac{\Gamma{(n+1)}}{\Gamma{(n+\frac{1}{2})}}\Big)^2$$
$$= \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2.$$
Question 1) Is this a correct fully equivalent reformulation of rhs of WP?
When using WP we would get:
$$\frac{\pi}{2} = \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2,$$
$$\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2 = 1,$$ which implies
$$\frac{\Gamma(n)}{\Gamma(n-\frac{1}{2})} \overset{n\rightarrow \infty}{\longrightarrow} \sqrt{n}. \tag{1}$$
Alternatively if (1) is known otherwise (e.g. from the Sterling approximation (Question 2) Is it possible to derive (1) from the Sterling approximation or from any other elementary considerations?)) one could derive the WP easily from that.
Question 3) Is this conclusion correct?
| By Stirling,
$$\frac{\Gamma(n)}{\Gamma(n-\frac12)}\approx\frac{\sqrt{2\pi n}\left(\dfrac ne\right)^n}{\sqrt{2\pi (n-\frac12)}\left(\dfrac {n-\frac12}e\right)^{n-\frac12}}.$$
After simplifications,
$$\frac1{\left(1-\dfrac1{2n}\right)^n}\left(\frac {n-\frac12}e\right)^{1/2}\to e^{1/2}\sqrt {\frac ne}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve in the xy-plane. Now if I could get y as a function of x, I could easily parametrize the curve but I am not able to do that.
Is there an easier way to solve the problem?
| $2x^2+2xy+2y^2=1$ can be rewritten as $(2x+y)^2+3y^2=2$ and this can be parameterised
\begin{eqnarray*}
\frac{2x+y}{\sqrt{2}} &=& \cos \theta \\
\sqrt{\frac{3}{2}} y &=& \sin \theta . \\
\end{eqnarray*}
now substitute these into $x+y+z=0$ and we have
\begin{eqnarray*}
x&=&\frac{1}{\sqrt{2}} \cos \theta- \frac{1}{\sqrt{6} } \sin \theta \\
y &=&\sqrt{ \frac{2}{3}} \sin \theta \\
z&=& - \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{6}} \sin \theta.
\end{eqnarray*}
Alternatively
\begin{eqnarray*}
\left(\begin{array}{c}
x \\ y \\ z
\end{array}\right)=\left(\begin{array}{c}
\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}\right) \cos \theta+\left(\begin{array}{c}
-\frac{1}{\sqrt{6}} \\ \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{6}}
\end{array}\right) \sin \theta
\end{eqnarray*}
now observe that these are orthogonal unit vectors and so this is a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
| Note that
$$\cos u + \cos v = 2\cos((u + v)/2)\cos((u − v)/2)$$
and
$$\cos u - \cos v = -2\sin((u + v)/2)\sin((u − v)/2).$$
Hence
\begin{align*}
1 + \cos 2C - \cos 2A - \cos 2B&=1 + \cos (2\pi-2(A+B)) - 2\cos(A+B)\cos(A-B)\\
&=2\cos^2 (A+B) - 2\cos(A+B)\cos(A-B)\\
&=2\cos(A+B)(\cos(A+B)-\cos(A-B))\\
&=2(-\cos C)(-2\sin A\sin B)=4\sin A\sin B\cos C.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$. We have
$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$
$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$
$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$
and we have to prove that
$$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$
squaring (1) & (2), and adding them we have
$$x^2 + y^2 - z^2 = -2xy \cos(\theta - \phi)$$
multiplying (1) & (3), we have
$$x^2 + y^2 - z^2 = -xy (\cos\theta \sec\phi + \cos\phi \sec\theta)$$
from here I can't move forward
help me
| We denote $a,b,c$ the angles $\theta, \phi, \psi$. The condition is
$$\begin{bmatrix} \cos a &\cos b &\cos c\\\sin a &\sin b &\sin c\\\sec a &\sec b &\sec c\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ The determinant of the matrix is equal to
$$\frac{\sin(b-a)}{\cos c}+\frac{\sin(a-c)}{\cos b}+\frac{\sin(c-b)}{
\cos a}$$
This determinant is in general non-zero as it is easy to verify.
Therefore the solution of the linear system above has a unique solution which is the obvious one $x=y=z=0$ in which case the equality to prove is trivially true.
The conclusion is that it remains to add some condition on the angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this:
Problem
Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote the highest ranking achieved by a woman. (For instance, $X = 1$ if the top-ranked person is female.) Find $P(X = i),i = 1, 2, 3, . . ., 8, 9, 10$.
Solution given was:
*
*$P(X=1)= \frac{5}{10}= \frac{1}{2}$ because there are 5 women and total of 10 to choose from
*$P(X=2)=\frac{5}{10}\times \frac{5}{9}=\frac{5}{18}$ because for rank1 thereare 5 men and total of 10 to choose from, for rank 2 (we want awoman) we still have 5 women but only a total of 9 to choosefrom.
*$P(X=3)=\frac{5}{10}\times \frac{4}{9}\times \frac{5}{8}=\frac{5}{36}$
*$P(X=4)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}=\frac{10}{168}$
*$P(X=5)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{252}$
*$P(X=6)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{5}{5}=\frac{1}{252}$
My solution was
*
*$P(X=1)=\frac{5\times 9!}{10!}=\frac{1}{2}$ because there are five women to occupy 1st rank and then there remained 9 which can permute in $9!$ ways. There are total $10!$ ways to permute $10$ people
*$P(X=2)=\frac{5\times \binom{5}{4}\times 8!}{10!}=\frac{5}{18}$ because there are five women to occupy 2st rank. The 1st rank will be of a man. So we have to select $4$ out of $5$ men which will be ranked after $2$nd rank. These four men and remaining 4 women can be permuted in $8!$ ways.
*$P(X=3)=\frac{5\times \binom{5}{3}\times 7!}{10!}=\frac{5}{72}$
*$P(X=4)=\frac{5\times \binom{5}{2}\times 6!}{10!}=\frac{5}{504}$
*$P(X=5)=\frac{5\times \binom{5}{1}\times 5!}{10!}=\frac{5}{6048}$
*$P(X=6)=\frac{5\times \binom{5}{0}\times 4!}{10!}=\frac{1}{30240}$
Doubts
*
*Where my logic went wrong?
*When I compared the two approach, I realized that the books solution is permuting ranks higher than the highest ranked girl, while my solution is permuting ranks lower than the highest ranked girl. So I was guessing what makes book solution not permute lower ranks and my solution not permuting higher ranks. Shouldn't we permute on both sides of highest ranked girl?
| I would like to suggest another way of getting the results. For me this way is "combinatorically more intuitive":
*
*If $X = i$, then there are $10-i$ ranks left for the remaining $4$ women. So. there are $\binom{10-i}{4}$ ways of choosing 4 further ranks.
*As each permutation of the $5$ men and $5$ women gives another way of ranking we need to multiply by $5! \cdot 5!$
*All together (it gives exactly the values from the solution presented): $$P(X = i) = \frac{\binom{10-i}{4}\cdot5! \cdot 5!}{10!} = \frac{\binom{10-i}{4}}{252} \mbox{ for } i = 1,...,6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponentiating Matrices - Am I doing it correctly? I recently learned how to exponentiate matrices by extending the power series of $e^x$ to matrices:
$$e^A=A^0+A^1+\frac{A^2}{2!}+\frac{A^3}{3!}+...$$
I thought that this was pretty cool, so I decided to try it with the matrix
$$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
I then realized that I would have to find a general formula for $A^n$. I ended up getting
$$A^n=\begin{pmatrix}2^{n-1}&2^{n-1}\\2^{n-1}&2^{n-1}\end{pmatrix}$$
for $n\ge 1$, and for $n=0$, I have that $A^0=I$. So my final answer would be in the form
$$e^A=\begin{pmatrix}\alpha&\alpha-1\\\alpha-1&\alpha\end{pmatrix}$$
where
$$\alpha=1+2^0+\frac{2^1}{2!}+\frac{2^2}{3!}+...$$
$$\alpha=\frac{1}{2}\bigg(1+2^0+2^1+\frac{2^2}{2!}+\frac{2^3}{3!}+...\bigg)$$
$$\alpha=\frac{1}{2}(e^2+1)$$
and so my answer would be
$$e^A=\begin{pmatrix}\frac{e^2+1}{2}&\frac{e^2-1}{2}\\\frac{e^2-1}{2}&\frac{e^2+1}{2}\end{pmatrix}$$
Is this correct?
| Your method is correct, as you've been told. But here's another way of doing this that may interest you. Let$$M=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$Then$$M^{-1}=\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\text{ and }M^{-1}.A.M=\begin{pmatrix}2&0\\0&0\end{pmatrix}.$$Therefore$$M^{-1}.e^A.M=e^{\begin{pmatrix}2&0\\0&0\end{pmatrix}}=\begin{pmatrix}e^2&0\\0&1\end{pmatrix}.$$So,$$e^A=M.\begin{pmatrix}e^2&0\\0&1\end{pmatrix}.M^{-1}=\begin{pmatrix}\frac{e^2+1}{2}&\frac{e^2-1}{2}\\\frac{e^2-1}{2}&\frac{e^2+1}{2}\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...}$$
How can I get a finite value?
| $$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$a^2=7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}$$
$$a^4=98\sqrt{7\sqrt{2\sqrt{...}}}$$
so
$$a^4=98a$$
and, assuming $a$ is nonzero,
$$a=\sqrt[3]{98}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
$$b^2=2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}$$
$$b^4=28\sqrt{2\sqrt{7\sqrt{...}}}$$
$$b^4=28b$$
and, assuming $b$ is nonzero,
$$b=\sqrt[3]{28}$$
so
$$ab=\sqrt[3]{2744}=14$$
Additionally, it's not hard to prove that if
$$a=\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}$$
and
$$b=\sqrt{y\sqrt{x\sqrt{y\sqrt{x\sqrt{...}}}}}$$
then $ab=xy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that $ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$
If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that
$$ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$$
Hint: Rewrite the defining equation as $f(x,y) = h(r(x,y), \theta(x,y))$, with $r(x,y) = \sqrt{x^2+y^2}$ and $\theta (x,y) = tan^{-1} (\frac{y}{x})$, and differentiate with respect to $x$ and $y$.
Defining equation as $$f(x,y) = h \left[r(x,y), \theta(x,y) \right]$$, with $$r(x,y) = \sqrt{x^2+y^2} \\ \theta (x,y) = tan^{-1} (\frac{y}{x})$$
(1)
$$f_x = \frac{\partial h }{\partial r } \frac{\partial r }{\partial x } +\frac{\partial h }{\partial \theta } \frac{\partial \theta }{\partial x } =
\frac{\partial h }{\partial r } \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial h }{\partial \theta } \frac{y}{x^2+y^2} \tag{A}
$$
$$f_{xx} = \frac{\partial}{\partial x} \left( \frac{\partial h }{\partial r } \frac{y }{\sqrt{y^2+x^2} } - \frac{\partial h }{\partial \theta } \frac{y}{x^2+y^2} \right) = \frac{\partial^2 h }{\partial x \partial r } \frac{y^2 }{(y^2+x^2)^{\frac{3}{2}} } - \frac{\partial^2 h }{\partial x \partial \theta } \frac{2yx}{(x^2+y^2)^2} \tag{B}
$$
.
Also from tag A, substituting $\left( \frac{\partial h}{\partial r} \right)$ for $(h)$
$$\frac{\partial}{\partial x}(h) = \frac{\partial}{\partial r } (h) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial }{\partial \theta } (h) \frac{y}{x^2+y^2}$$
$$ \implies \frac{\partial^2 h }{\partial x \partial r } = \frac{\partial}{\partial x} \left( \frac{\partial h}{\partial r} \right) = \frac{\partial}{\partial r } \left( \frac{\partial h}{\partial r} \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial }{\partial \theta } \left( \frac{\partial h}{\partial r} \right) \frac{y}{x^2+y^2} $$ $$
\implies \frac{\partial^2 h }{\partial x \partial r }=
\left( \frac{\partial^2 h}{\partial r^2} \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \left( \frac{\partial^2 h}{\partial \theta .\partial r} \right) \frac{y}{x^2+y^2} \tag{C} $$
Second Substitution
$$ \implies \frac{\partial^2 h }{\partial x \partial \theta } = \left( \frac{\partial^2 h}{\partial r \partial \theta} \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \left( \frac{\partial^2 h}{\partial \theta^2.} \right) \frac{y}{x^2+y^2} \tag{D} $$
It follows, placing (C) and (D) in (B):
$$f_{xx} = \left[ \left( \frac{\partial^2 h}{\partial r^2} \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \left( \frac{\partial^2 h}{\partial \theta .\partial r} \right) \frac{y}{x^2+y^2} \right] \frac{y^2 }{(y^2+x^2)^{\frac{3}{2}} } - \left( \left( \frac{\partial^2 h}{\partial r \partial \theta} \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \left( \frac{\partial^2 h}{\partial \theta^2.} \right) \frac{y}{x^2+y^2} \right) \frac{2yx}{(x^2+y^2)^2}
$$
.
(2) I would then proceed similarly with $f_y$
$$f_y = \frac{\partial h }{\partial r } \frac{\partial r }{\partial y } +\frac{\partial h }{\partial \theta } \frac{\partial \theta }{\partial y } =
\frac{\partial h }{\partial r } \frac{y }{(y^2+x^2) ^{\frac{1}{2}} }+\frac{\partial h }{\partial \theta } \frac{x}{x^2+y^2}
$$
...
(3) After Adding $f_{xx}$ and $f_{yy}$ , and simplifying, Hopefully, I should get an expression such as $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$.
.
.
The rest of my result does not seem to lead to the desired outcome $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$; there is no $h_r$ to be seen in $f_{xx}+f_{yy}$.
EDIT- Is the approach, eventhough long, I undertook would normally let one able to show that $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$? Is there an alternative approach to the one of Mr. Surd shown as answer involving only manipulation of partials (The chapter where the problem is does not have any concept related to the one shown by Mr. Surd).
Much appreciated for your input/help.
| Let $f(x,y)=f(r\cos\theta ,r\sin\theta )=h(r,\theta ).$
Then, $$\nabla _{(r,\theta )}h=\begin{pmatrix}\cos\theta &\sin\theta \\ -r\sin\theta &r\cos\theta \end{pmatrix}\nabla f\implies \nabla f=\frac{1}{r}\begin{pmatrix}r\cos\theta &-\sin\theta \\r\sin\theta &\cos\theta \end{pmatrix}\nabla _{(r,\theta )}h=:\begin{pmatrix}h_1(r,\theta )\\ h_2(r,\theta )\end{pmatrix}.$$
Now, you can observe that $$\Delta f=\nabla \cdot \nabla f,$$ and thus, you get $$\Delta f=\cos\theta \frac{\partial h_1}{\partial r}-\frac{\sin\theta}{r} \frac{\partial h_1}{\partial \theta }+\sin\theta \frac{\partial h_2}{\partial r}+\frac{\cos\theta}{r} \frac{\partial h_2}{\partial \theta }.$$
I let you finishing all the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$x^4-2x^3+mx^2-2x+1=0$, $m \in R$, $x_1=x_2 \in R-\{-1\}$ Given the equation: $x^4-2x^3+mx^2-2x+1=0$, $m \in R$
To which interval does $m$ belong, so that the equation has $x_1=x_2 \in \mathbb R\setminus\{-1\}$, provided that $x_1,x_2,x_3,x_4$ are the roots of the equation.
Since I got this from Vieta's:
$$2x_1+x_3+x_4=2$$
$$x_1^2+2x_1x_3+2x_1x_4+x_3x_4=m$$
$$x_1^2x_3x_4=1$$
$$x_1^2x_3+x_1^2x_4+x_1x_3x_4+x_1x_3x_4=2$$
I figured I could do this:
$$x_1^2(x_3+x_4)+2x_1x_3x_4=2$$
$$x_3+x_4=2-2x_1$$
$$x_1x_3x_4= \frac{1}{x_1}$$
$$x_1^2(2-2x_1)+\frac{2}{x_1}=2$$
$$2x_1^4-2x_1^3+2x_1-2=0$$
So I use Horner's on the last line above and get $x_ 1=1$ the only real root which is also $x_2$.
Thus I am able to find $m=2$ which belongs to the interval of the correct answer[which is $(1, \infty )$].
The thing is I am not sure this is the only $m$ that provides a solution to the exercise... Is it the only value of $m$?
| We can get your result also by the following way.
$$(x^4-2x^3+mx^2-2x+1)'=0$$ or
$$4x^3-6x^2+2mx-2=0$$ or
$$2x^4-3x^3+mx^2-x=0$$ or
$$mx^2=-2x^4+3x^3+x,$$ which gives
$$x^4-2x^3+(-2x^4+3x^3+x)-2x+1=0$$ or
$$x^4-x^3+x-1=0$$ or
$$(x^3+1)(x-1)=0,$$
which gives $x=1$ and $m=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?
First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$
and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$
$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$
$\iff \frac{ab}{a+b}=a+b$
$\iff ab=(a+b)^2$
$\iff ab= a^2 + 2ab + b^2$
$\iff 0 = a^2+2ab+b^2-ab$
$\iff 0= a^2+b^2+ab$
Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:
$ 0= a^2+b^2+ab \iff 0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$
$\iff 0 = \frac{a^2+b^2}{ab} + 1$
$\iff -1 = \frac{a^2+b^2}{ab}$
$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$
If a and b have the same sign, then $ab>0$.
So $\frac{a^2+b^2}{ab}$ could be negative only if:
*
*$a>0 \text{ and } b<0$ or
*$a<0 \text{ and } b>0$
Taking into account that $a \neq -b$, any of these two options would be possible only if either:
*
*$|a| > |b|$
*$|b| > |a|$
Assuming that $|a| > |b|$ we have the following:
$|a| > |b| \implies |a|*|a| > |b|*|a|$
$\implies |a|^2 > |ab|$
$\implies \frac{|a|^2}{|ab|}>1$
$\implies |\frac{a^2}{ab}|>1$
And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
Assuming $|b|>|a|$ we arrive at the same conclusion:
$|b| > |a| \implies |b|*|b| > |a|*|b|$
$\implies |b|^2 > |ab|$
$\implies \frac{|b|^2}{|ab|}>1$
$\implies |\frac{b^2}{ab}|>1$
And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
$\blacksquare$
Is this correct??
| If $a$ and $b$ have the same sign, the magnitude of $\frac 1{a+b}$ is less than either $\frac 1a$ or $\frac 1b$ so they must have opposite signs. By symmetry we can demand $a$ be positive. If $a \gt -b, \frac 1{a+b}$ is positive while $\frac 1a +\frac 1b$ is negative. The opposite happens if $a \lt -b$ so there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate the following complex integrals using parameterisation $\int \limits_{\beta} f(z) dz$ Evaluate the following complex integrals using parameterisation
$$\int \limits_{\beta} f(z) dz$$
where $\beta$ represents the line segment from $-i$ to $2+5i$, followed by the line segment from $2+5i$ to $5i$, and $f(x+yi)=iy+x^2$
What i tried
I tried using parameterisation such as letting $z=t+it$ but i couldnt find a suitable one. Also for the line segment from $-i$ to $2+5i$ can i split it to become $-i$ to $5i$ and then from $5i$ to $2+5i$ . And then parameterise each one individually? Is there a general approach to solving this type of questions. Could anyone please explain. Thanks
| So you have an integral $\int_\beta f(z)\, dz$ where $\beta$ represents two line segments. First, let's parameterize each one:
$C_1$ goes from $-i$, which means the point is $(0,-1)$, to $2+5i$, which means the point is $(2,5)$, so:
\begin{align}
C_1(t) &= (0,1) + t[(2, 5)-(0,-1)] \\
C_1(t) &= (2t, 6t-1).
\end{align}
$C_2$ goes from $2+5i$, which means the point is $(2,-5)$, to $5i$, which means the point is $(0,5)$, so:
\begin{align}
C_2(t) &= (2,5) + t[(0, 5)-(2,5)] \\
C_2(t) &= (2-2t, 5).
\end{align}
Perfect, you have parametrized $\beta$, now the question is: how do you include this in the integral? You have this function $f(z) = x+iy$ and you have to convert it to $z$ notation, such as this:
\begin{align}
f(z) &= x^2 +iy \\
f(z) &= x -x + x^2 + iy \\
f(z) &= x + iy -x + x^2 \\
f(z) &= z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2.
\end{align}
And now the integral is:
\begin{align}
\int_\beta f(z) \, dz &= \int_\beta \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\
&= \int_{C_1} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\
&+ \int_{C_2} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz.
\end{align}
Before you put $C_1(t)$ and $C_2(t)$, you have to convert them to $z$ notation, which might be strange but it is easy:
\begin{align}
C_1(t) &= z_1(t) = 2t + i(6t-1) \\
C_2(t) &= z_2(t) = 2 -2t + 5i.
\end{align}
Let's do the first intregal:
\begin{align}
\int_{C_1} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &= \int_0^1 \left(2t + i(6t-1) -2t + 4t^2 \right) (2+i6)\,dt \\
&= -\dfrac{28}{3} + 12i.
\end{align}
By the way, $dz_1 = (2+6i)dt$. The second integral is:
\begin{align}
\int_{C_2} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &= \int_0^1 \left(2 -2t + 5i -(2-2t) + (2-2t)^2 \right) (-2)\,dt \\
&= -\dfrac{8}{3} - 10i.
\end{align}
By the way, $dz_2 = -2 dt$. Therefore, the result is $-12 +2i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer. Given
$$(1+x)^n= \binom {n}{0} + \binom{n}{1} x+ \binom{n}{2} x^2+ \cdots + \binom {n}{n} x^n.$$
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer.
I tried to use negative $x$ and even $i$ but could not eliminate $\binom{n}{2} $ term.
| This answer repeatedly takes advantage of one fact,
$$g(k)=\frac{(-1)^k+1}{2}$$
Returns $1$ for a nonnegative even integer, and $0$ for a nonnegative odd integer.
Our sum is,
$$\sum_{k=0,\text{even}}^{n/2} {n \choose 2k}$$
$$=\sum_{k=0}^{n/2} \frac{1+(-1)^k}{2}{n \choose 2k}$$
$$=\frac{1}{2}\left(\sum_{k=0}^{n/2} {n \choose 2k}+\sum_{k=0}^{n/2} (-1)^k {n \choose 2k} \right)$$
Now notice,
$$f(x):=\sum_{k=0}^{n/2} {n \choose 2k}x^{2k}$$
$$=\sum_{k=0,\text{even}}^{n} {n \choose k} x^k$$
$$=\sum_{k=0}^{n} \frac{1+(-1)^k}{2}{n \choose k}x^k$$
$$=\frac{1}{2}\left((1+x)^n+(1+(-x))^n\right)$$
$$=\frac{1}{2}\left((1+x)^n+(1-x)^n \right)$$
But
$$f(i)=\sum_{k=0}^{n/2} {n \choose 2k} (-1)^k$$
And so our sum is,
$$\frac{1}{2}(f(1)+f(i))$$
$$=\frac{1}{4}\left(2^n+(1+i)^n+(1-i)^n \right)$$
A simpler way: Notice $1,-1,i,-i$ are the $4$ (fourth) roots of unity. Then considering, $\frac{1+(-1)^k+i^k+(-i)^k}{4}=1$ when $k$ a negative integer is divisible by $4$ but equal to zero when $k$ is not divisible by $4$, we easily get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Mathematical Olympiad Treasures Problem 1.92 how does this relate to geometric sequence You are supposed to show that if
$a,b,c$ are nonzero real numbers
and
$(ab+bc+ca)^3 =abc(a+b+c)^3$
$a, b, c$ are terms of a geometric sequence.
One given solution is that:
$(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc) (b^2 −ac) (c^2 −ab)$
How does the new factorization imply that $a,b,c$ are part of geometric sequence?
| You got $(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc) (b^2 −ac) (c^2 −ab)$. Now since $(ab+bc+ca)^3 =abc(a+b+c)^3$, it implies $(ab+bc+ca)^3 −abc(a+b+c)^3=0$. Then;
$$(a^2 −bc) (b^2 −ac) (c^2 −ab)=0$$
This implies one of $(a^2 −bc),(b^2 −ac),(c^2 −ab)$ must be zero. Then we have $a^2 =bc$ or $b^2 =ac$ or $c^2 =ab$.
If $a^2=bc$, then $b,a,c$ or $c,a,b$ in G.P. Similar for other cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Demonstrate that all integers of a certain form are divisible by primes of a certain other form I wish to demonstrate that for any $n\in\mathbb{N}$ all prime divisors of a number $n^2 + n + 1$ are either equal to $3$ or of the form $3k + 1$ with positive integer $k$.
Thus far, I have tried checking cases. If $n$ is of the form $3m+1$, we have that
$n^2+n+1=(3m+1)^2+(3m+1)+1=9m^2+9m+3=3(3m^2+3m+1)$, which is clearly divisible by 3.
If $n$ is of the form $3m$, we have
$n^2+n+1=(3m)^2+(3m)+1=9m^2+3m+1=3(3m^2+m)+1$
If $n$ is of the form $3m+2$, we have
$n^2+n+1=(3m+2)^2+(3m+2)+1=9m^2+15m+7=3(3m^2+5m+2)+1$
The only problem is that since $(3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1$, the numbers being of the form $3k+1$ does not prove that their divisors must be of the form $3k+1$, so this does not really prove anything.
Any input is highly appreciated.
| Disclaimer: The following answer is based on finite fields, where the polynomial $x^2+x+1$ should ring some bells.
We have
$$n^3-1 = (n-1)(n^2+n+1).$$
Now let $p$ be a prime of the form $p = 3k+2$ for some positive $k$ and assume that $p$ divides $n^2+n+1$ and thus also divides $n^3-1$. Then we have that
$$n^3 \equiv 1 \mod{p}.$$
Looking at the finite field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$, we thus get that either $$n \equiv 1 \mod{p}$$ or $n$ is an element of order three in the unit group of $\mathbb{F}_p$. But this unit group has order $3k+1$ and can thus not contain elements of order three.
Therefore, the second case is not possible and we conclude $n \equiv 1 \mod{p}$, meaning that $p$ divides $n-1$.
Now, all that is left is to compute $gcd(n-1,n^2+n+1)$ to get that $p$ does not divide $n^2+n+1$.
(Hint: That gcd will be either three or one, but never include $p > 3$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
| It is ok. You could have also put $b=1-a$ and minimize a quadratic function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
For $08$.
For $0<x<\dfrac{\pi}{4}$ prove that
$$\frac{\cos x}{\sin^2x(\cos x-\sin x)}>8$$
I am trying to use the following for x $\in$ (0,$\frac{\pi}{4}$)
Cos x $\in$ ($\frac{1}{√2}$,1)
(sinx)^2 $\in$ (0,$\frac{1}{2}$)
(cosx-sinx) $\in (0,1)$
| Let me try. We have $$\frac{\cos x}{\sin^2 x (\cos x - \sin x)} = \frac{1+\tan^2 x}{\tan^2 x(1-\tan x)}.$$
Note that $0 < x < \dfrac{\pi}{4}$, so $0 < \tan x < 1$,
$$\tan x(1-\tan x) \leq \frac{1}{4}(\tan x + 1-\tan x)^2 = \frac{1}{4}.$$
So we have $$\text{LHS} \geq 4\frac{1+\tan^2x}{\tan x} \geq 8.$$
The equality happens when $\tan x = \dfrac{1}{2}$ for first and $\tan x = 1$ for second, so $\text{LHS} > 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written without assuming any prior calculus knowledge so I can't prove this by finding the minimum and the limits as x approaches $\infty \text{ and} -\infty $.
So there has to be a simple algebraic proof involving neither the quadratic formula nor calculus but I'm stuck.
Here are some things I thought:
Method 1:
$1+x+x^2 = 0 \iff 1+x+x^2+x = x$
$\iff x^2+2x+1 = x$
$\iff (x+1)^2 = x $
And here maybe prove that there is no x such that $(x+1)^2 = x$ ???
Method 2:
$1+x+x^2 = 0$
$\iff x^2+1 = -x$
By the trichotomy law only one of these propositions hold: $x=0$ or
$x>0$ or $x<0$.
Assuming $x=0$:
$x^2+1= 0^2+1 = 0 +1 = 1$
$-x = - 0 = 0$
And $1\neq 0$
Assuming $x>0$:
$x>0 \implies -x < 0$
And $x^2+1 \ge 1 \text{ } \forall x$
With this method I have trouble proving the case $x<0$:
I thought maybe something like this could help but I'm not sure:
$x<0 \implies -x=|x|$
$x^2 = |x|^2$
And then prove that there is no x such that $|x|^2 + 1 = |x|$??
Can anyone please help me? Remember: No calculus or quadratic formula allowed.
| If
$x^2 + x + 1 = 0, \tag 1$
then
$x^2 + x + \dfrac{1}{4} = -\dfrac{3}{4}; \tag 2$
but
$(x + \dfrac{1}{2})^2 = x^2 + x + \dfrac{1}{4}, \tag 3$
so
$(x + \dfrac{1}{2})^2 = -\dfrac{3}{4}; \tag 4$
but no real has a negaitive square, so . . .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 14,
"answer_id": 1
} |
Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $Q_1$, $Q_2$, $Q_3$ so that
$$(x^2 + y^2 + z^2)^2 - 3( x^3 y + y^3 z + z^3 x) = Q_1^2 + Q_2^2 + Q_3^2$$
I would like to find an explicit writing of the quartic forms, with rational quadratic forms $Q_i$. Maybe more than $3$ terms are necessary.
| $$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)=\frac{1}{2}\sum_{cyc}(x^2-y^2-xy-xz+2yz)^2=$$
$$=\frac{1}{6}\sum_{cyc}(x^2-2y^2+z^2-3xz+3yz)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Monotonicity of $f(x) =\sin(\ln(x))-\cos(\ln(x))$ Find the interval in which $f(x) =\sin(\ln(x))-\cos(\ln(x))$ is increasing.
After differentiating we get $$f'(x) = \frac{\cos\left(\ln(x)\right)}{x} +\frac{\sin\left(\ln(x)\right)}{x}$$
Now how do we analyze this expression?
| $$f'(x)=\frac{\sin \log x+\cos \log x}{x}$$
$\log x$ is an increasing function. Substitute $u=\log x$. Denominator is positive because $x$ is argument of logarithm.
$\sin u + \cos u > 0\to \sin u > - \cos u$
$\sin u = -\cos u$ when $\tan u=-1$ when $u=\dfrac{3\pi}{4}+k\pi,\forall k\in\mathbb{Z}$
therefore $\sin u > - \cos u$ for
$2k\pi<u<\dfrac{3\pi}{4}+2k\pi\lor \dfrac{5\pi}{4}+2k\pi<u<2(k+1)\pi$
that is
$2k\pi<\log x<\dfrac{3\pi}{4}+2k\pi\lor \dfrac{5\pi}{4}+2k\pi<\log x<2(k+1)\pi$
and finally
$e^{2k\pi}< x<e^{\frac{3\pi}{4}+2k\pi}\lor e^{\frac{5\pi}{4}+2k\pi}<x<e^{2(k+1)\pi}$
these intervals are very huge, for instance for $k=3$ the derivative is positive in
$(1.5\times 10^8,1.62009\times 10^9)\cup (7.79343\times 10^9,8.22263\times 10^{10})$
and graph is pretty weird, too
Hope this helps
edit
$f''(x)=-\dfrac{2 \sin (\log (x))}{x^2}$
Thus $x=e^{2 \pi k}$ are inflexion points because $f''(x)=0$
$x=e^{2 \pi k+\frac{3 \pi }{4}}$ are maxima because $f''(x)<0$
$x=e^{2 \pi k+\frac{5 \pi }{4}}$ are minima
as $k\to -\infty$ maxima and minima oscillate in intervals exponentially smaller and smaller when $x\to 0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Factoring a polynomial with possibly repeated root I've been trying to factor $x^3 - 8x^2 + 17x - 4$. One of factors is $x - 4$ so I got $(x - 4)(x^2 -4x + 1)$, but I don't know how to factor the second part $x^2 -4x + 1$.
I also have a similar problem with $x^3(x^3 - 3x + 2)$. I'm supposed to get $x^3(x - 1)^2 (x + 2)$, but I don't know how to get here. Is there a method to factor repeated roots?
| $$x^2-4x+1=x^2-4x+4-3=(x-2)^2-(\sqrt3)^2=(x-2-\sqrt3)(x-2+\sqrt3)$$
$$x^3-3x+2=x^3-2x^2+x+2x^2-4x+2=(x^2-2x+1)(x+2)=(x-1)^2(x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
| Famke's answer is the simplest; however, we can use a variational argument, as well.
For all variations that maintain $ab+bc+ca=2$, we have
$$
(b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
To minimize $a^2+b^2+c^2$, we must have
$$
2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2}
$$
for all variations that satisfy $(1)$.
Orthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need
$$
2a=\lambda(b+c),\quad2b=\lambda(a+c),\quad\text{and}\quad2c=\lambda(a+b)\tag{3}
$$
Summing these up, we get that $\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\sqrt{\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
M tosses $7$ fair coins and has $M$ heads. $A$ tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A).$ $M$ tosses $7$ fair coins and has M heads. A tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A)$.
I suppose that both distributions are binominal, but I don't know what to do next.
| The values $6$ and $7$ are sufficiently small that you could calculate this by hand.
The probabilty that $A$ gets zero heads is $ \frac{1}{64}$ and $M$ get one or more heads is $\frac{127}{128}$.
The probabilty that $A$ gets one head is $ \frac{6}{64}$ and $M$ get two or more heads is $\frac{120}{128}$.
...
The probabilty that $A$ gets six heads is $ \frac{1}{64}$ and $M$ gets seven heads is $\frac{1}{128}$.
We have
\begin{eqnarray*}
P= \frac{1}{64} \frac{127}{128}+\frac{6}{64} \frac{120}{128}+\frac{15}{64} \frac{99}{128}+\frac{20}{64} \frac{64}{128}+\frac{15}{64} \frac{29}{128}+\frac{6}{64} \frac{8}{128}+\frac{1}{64} \frac{1}{128}=\frac{4096}{8192}
\end{eqnarray*}
So the probability is $\color{red}{\frac{1}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$
into partial fractions.
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(x-3)^2}$$
but this is apparently incorrect.
According to the text, the decomposition is
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
I discussed this with my friend that the fraction first decomposes into
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{Bx +C}{(x-3)^2}$$
but I can't see how he derived this.
I don't understand how he is correct.
| When you decompose, the degree of the numerator will be less than the degree of the denominator. How much less is yet to be determined. Assume that it is one degree less, and then if you get a zero coefficient, so be it.
If you had something like...$\frac {P(x)}{(x^2 +x + 1)(x-1)}$
your first step would be $\frac {Ax + B}{x^2+ x+ 1} + \frac {C}{(x-1)}$
With each numerator of degree 1 less than each denominator.
And the you might want to break down the first term to make it easier to integrate.
e.g. $\frac {A(x+\frac12)}{x^2 + x + 1} + \frac {B}{(x+\frac 12)^2 + \frac 34}$
With this problem.
$(x-3)^2$ makes for a degree 2 denominator, and you need a degree 1 numerator.
So $\frac {A}{x+1} + \frac {Bx+C}{(x-3)^2}$ would be a good place to start.
but $\frac {B}{(x-3)} + \frac {C}{(x-3)^2}$ is easier to integrate, so you might want to skip the intermediate step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the sum $\sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$ Let $a$ and $b$ be natural numbers such that $a \geq b \geq 1$. How can we evaluate the following sum?
$$ \sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$$
| $$\begin{align}
\sum_{n=1}^b n\binom an\binom bn
&=\sum_{n=1}^b \binom an\cdot n\binom bn\\
&=\sum_{n=1}^b \binom an\cdot b\binom {b-1}{n-1}\\
&=b\sum_{n=1}^b \binom a{a-n}\binom {b-1}{n-1}\\
&=b\binom {a+b-1}{a-1}
&&\text{(Vandermonde)}\\
&=b\binom {a+b-1}{b}\\
&=(a+b-1)\binom {a+b-2}{b-1} \\
&=(a+b-1)\binom {a+b-2}{a-1}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $abcd=1$ then $a^4b+b^4c+c^4d+d^4a\geq a+b+c+d$
I want to prove that for all $a>0$, $b>0$, $c>0$ and $d>0$ with $abcd=1$: $$a^4b+b^4c+c^4d+d^4a\geq a+b+c+d.$$
I think that this should be provable by AM-GM inequality, but I could not manage to prove it. Can you give me a hint?
Best wishes
| The hint:
Write
$$xa^4b+yb^4c+zc^4d+td^4a\geq a^2bcd,$$
where $x$, $y$, $z$ and $t$ are non-negatives such that $x+y+z+t=1$
and after using AM-GM solve the system, which you got.
I got $x=\frac{23}{51}$, $y=\frac{7}{51}$, $z=\frac{11}{51}$ and $t=\frac{10}{51}$,
which gives the following magical solution in one line.
$$\sum_{cyc}a^4b=\frac{1}{51}\sum_{cyc}(23a^4b+7b^4c+11c^4d+10d^4a)\geq$$
$$\geq\frac{1}{51}\sum_{cyc}\left(51\sqrt[51]{a^{23\cdot4+10}b^{23+7\cdot4}c^{7+11\cdot4}d^{11+10\cdot4}}\right)=\sum_{cyc}a^2bcd=\sum_{cyc}a.$$
Also we can use C-S here.
C-S inequality it's the following.
Let $b_i>0$. Prove that:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$
Hence, by C-S we obtain:
$$\sum_{cyc}a^4b=\sum_{cyc}\frac{a^3}{cd}=\sum_{cyc}\frac{(a^2)^2}{acd}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{cyc}acd}=\frac{(a^2+b^2+c^2+d^2)^2}{abc+abd+acd+bcd}.$$
Thus, it remains to prove that
$$(a^2+b^2+c^2+d^2)^2\geq(a+b+c+d)(abc+abd+acd+bcd),$$ which is obvious by Muirhead
because $(4,0,0,0)\succ(2,1,1,0)$, $(2,2,0,0)\succ(2,1,1,0)$ and $(2,2,0,0)\succ(1,1,1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
I want to find all the rational solutions of $x^2+y^3=z^3$ I want to find all the rational solutions of $x^2+y^3=z^3$,
and any help would be appreciated.
My answer is as follows：
$$x^2=z^3-y^3=(z-y)(z^2+zy+y^2)$$
Let $x=ab,a,b\in\Bbb Q$
when
$$ z-y=a^2,z^2+zy+y^2=b^2$$
$$z^2+z(z-a^2)+(z-a^2)^2=b^2$$
Sort it out, and that's a Pell's equation:
$$\left(\frac{2b}{a^2}\right)^2-3\left(\frac{2z}{a^2}-1\right)^2=1$$
Let $s=\frac{2b}{a^2},t=\frac{2z}{a^2}-1$
We obtained
$$s^2-3t^2=1$$
The least $s_0=2,t_0=1$,other solutions can be obtained from:
$$s_{n+1}=2s_n+3t_n,t_{n+1}=s_n+2t_n(n=0,1,2,...)$$
Last we obtained:
$$x=ab=\frac{a^3}{2}s_n,z=\frac{a^2}{2}(t_n+1),y=\frac{a^2}{2}(t_n-1)$$
Example:Let $a=\frac{1}{3},s_1=7,t_1=4$,then $x=\frac{7}{54},y=\frac{1}{6},z=\frac{5}{18}$.
I don't know what the other solutions are,I'd be glad of some help with this,thank you for you help.
| EDIT: Let's first consider the more symmetric equation
$$x^2=y^3+z^3\tag1.$$If $y=0$, we see that gives the solutions $$(x,y,z)=(t^3,0,t^2)$$ for rational $t$. If $y\neq0$, we can set $x=ry$ and $z=ty$ with rational $r,t$. That gives $r^2=y(1+t^3)$, i.e. $r=0$, giving $t=-1$ and thus the solutions $$(x,y,z)=(0,y,-y),$$ or $y=r^2/(1+t^3)$, giving the solutions $$(x,y,z)=\left(\frac{r^3}{1+t^3},\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$
That should take care of all possible cases.
To return from (1) to the original equation, we just have to change the sign of $y$, so the original equation has the solutions
$$(x,y,z)=(t^3,0,t^2),$$
$$(x,y,z)=(0,y,y),$$
$$(x,y,z)=\left(\frac{r^3}{1+t^3},-\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sequence : $f(f(n))+f(n+1)=3n$ Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))+f(n+1)=3n$, $\forall n \in \mathbb{Z}^+$ ?
My attempt :
Substitute $n=1$, $f(f(1))+f(2)=3$ so $f(2) \in \{1$ or $2\}$
If $f(2)=1, f(f(1))=2$
Substitute $n=2$, $f(f(2))+f(3)=6$ so $f(1)+f(3)=6$
Substitute $n=3$, $f(f(3))+f(4)=9$
If $f(2)=2$, $f(f(1))=1$ so $f(1)\not= 2$
| We will show that there is no such function.
Since $f(2)+ff(1)=3$, we have two case:
$f(2)=2$:
$ff(2)+f(3)=6\implies f(3)=4$
$ff(3)+f(4)=9\implies 2f(4)=9$.
The last one is impossible hence $f(2)=1$.
$f(2)=1:$
$$
ff(2)+f(3)=6\implies f(3)=6-f(1)\\
ff(3)+f(4)=9\implies f(6-f(1))+f(4)=9.
$$
First of all $f(1)\neq 1,2$.
If $f(1)=5$ then :
$$
ff(1)=2\implies f(5)=2\\
f(3)=1\implies f(4)=4\\
ff(4)+f(5)=4+f(5)=4+2\neq 3\times 4.
$$
So $f(1)\neq 5$.
If $f(1)=3$:
$$
ff(1)=f(3)=2\\
f(3)=6-f(1)=3
$$
So again $f(1)\neq 3$.
If $f(1)=4$ then $f(3)=2$:
$$
f(f(1))=f(4)=2\\
ff(3)+f(4)=9\implies 1+f(4)=9.
$$
So $f(1)\neq 4$.
Therefore there is no function satisfying this condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Need help with level 3 Calculus problem
Two curves $$C_1: ([f(y)]^{2/3} + ([f(x)]^{1/3}) = 0\quad\mbox{and}\quad C_2: [f(y)]^{2/3}+ [f(x)]^{2/3} = 12, $$ satisfying the relation $$ (x-y)f(x+y)-(x+y)f(x-y) = 4xy(x^2-y^2).$$
1.) Evaluate the area bounded by $C_1$ and $C_2$.
2.) Evaluate the area bounded by $C_2$ and $|x|+|y|=\sqrt{12}$.
3.) Evaluate the area bounded by $C_1$ and $x+y+2=0$.
It was hard to write math here so:
Have a look at this pic for clear question
| According to Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
the general solution of the function equation
$$(x-y)f(x+y)-(x+y)f(x-y) = 4xy(x^2-y^2)$$
is $f(x)=x^3+kx$ with $k\in\mathbb{R}$. So the problem seems to be ambiguous because $f$ (and therefore also the curves $C_1$ and $C_2$) is not uniquely determined.
Note that if we let $k=0$ then $f(x)=x^3$ and
$$C_1: \; y^2+x=0\quad \mbox{and}\quad C_2:x^2+y^2=12.$$
In this case the area bounded by $C_1$ and $C_2$ is
$$2\int_{-\sqrt{12}}^{-3}\sqrt{12-x^2}dx+2\int_{-3}^0\sqrt{-x}dx=
2\pi+\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$ My attempts:
I cannot apply the Eisenstein's criteria here, because there is no prime number that divides the constant term i.e. $1$ Taking a translation of the form $x \rightarrow x+a$ does not solve this issue either.
Next, I tried the mod tests: $\operatorname{mod}2$ doesn't work since $x^4-x^2+1=(x^2+x+1)(x^2-x+1)$, similarly in $\operatorname{mod}3$ $x^4-x^2+1=(x^2+1)^2$. Now I can go on and maybe eventually find a $\operatorname{mod}p$ that works, but that is very time consuming, specially in examinations.
So, I'll use the rational root test. The possibilities for roots are $\pm 1$ and it is easy to see that neither is a root.
The only possibility left then are quadratic factors, say, $(x^2+ax+b)(x^2+cx+d)=x^4-x^2+1$
This gives me a set of equations $bd=1, a+c=0, b+d+ac=-1$. So either $b=d=1$, in which case $a=\pm \sqrt3 \notin \mathbb{Q}$, or $b=d=-1$, which gives $a=\pm i \notin \mathbb{Q}$.
So such factorization is not possible and hence the given polynomial is irreducible.
Is this solution correct? Also, is there an easier way to solve this? Thank you.
| There are no rational roots, so no linear factors.
If $p(x)$ is a factor of $x^4-x^2+1$ then $p(-x)$ is, too.
If $x^4-x^2+1 = (x^2+a)(x^2+b)$ then $x^2-x+1=(x+a)(x+b)$. Show that $x^2-x+1$ is irreducible.
On the other hand, you'd have to have $x^4-x^2+1=(x^2+ax+b)(x^2-ax+b)$ where $b^2=1$ and $a\neq 0$. This means that $x^4+(2b-a^2)x^2+b^2 = x^4-x^2+1$.
So you need $2b-a^2=-1$. If $b=-1,$ then this means $a^2=-1$, and if $b=1$ then $a^2=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$
So
$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$
So
$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$
where
$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$
I got stuck up to integrate $J$.
| HINT: set $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}
$$
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
your Integrand is given by $$-1/2\,{\frac { \left( {t}^{2}+2\,t-1 \right) \left( {t}^{2}-2\,t-1
\right) }{ \left( 2\,t+1 \right) \left( t-2 \right) \left( {t}^{2}+
1 \right) }}
$$ with $t=\tan(x/2)$
and don't Forget $$dx=\frac{2dt}{1+t^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$
Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that
$$\det \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = (a^2+b^2+c^2+d^2)^2$$
I'm looking for a smart way to solve this problem. If we denote
$$A = \begin{bmatrix}
a & -b \\
b & a \\
\end{bmatrix}$$
and
$$B = \begin{bmatrix}
-c & -d \\
-d & c \\
\end{bmatrix}$$
we have that
$$ \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} $$
So it's sufficient to proof that
$$ \det \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} = (\det A - \det B)^2. $$
Help?
| Generally， if $B$ is symmetric such that $A^TB=BA$, then
$$ \det \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} = \det (AA^T +BB^T). $$
In fact
$$ \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}\begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}^T=\begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}\begin{bmatrix}
A^T & -B^T \\
B^T & A^T \\
\end{bmatrix} = \begin{bmatrix}
AA^T+BB^T & -AB^T+BA^T \\
-BA^T+AB^T & AA^T+BB^T \\
\end{bmatrix}=\begin{bmatrix}
AA^T+BB^T & 0 \\
0 & AA^T+BB^T \\
\end{bmatrix} $$
and hence
$$ \det\begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}=\sqrt{\det\begin{bmatrix}
AA^T+BB^T & 0 \\
0 & AA^T+BB^T \\
\end{bmatrix}}=\det(AA^T+BB^T).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Prove $\ln(1+x)\geq x-\frac{x^2}{2}$
When $x\geq0$ prove that:
$$\ln(1+x)\geq x-\frac{x^2}{2}.$$
My effort: From the Taylor series:
$$\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
I don't know how to continue from here.
| You want to prove $\ln(1+x)\ge x-\frac{x^2}{2}$ for all $x\ge 0$.
To prove it for $0\le x<1$, we'll use Taylor series. It converges because $-1<x\le 1$. See, e.g., here for more information. Or here.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
$\ln(1+x)\ge x-\frac{x^2}{2}$ is equivalent to $\frac{x^3}{3}-\frac{x^4}{4}+\cdots\ge 0$.
We'll prove for all $k\ge 3$, $k\in\mathbb Z$, $\frac{x^k}{k}-\frac{x^{k+1}}{k+1}\ge 0$. We could prove this using derivatives, but I'll use a simpler method here. The inequality is equivalent to
$\frac{x^k(k+1-kx)}{k(k+1)}\ge 0$
and we have
$\frac{k+1}{k}>1> x$, $x^k\ge 0$ because $x\ge 0$.
Using derivatives:
Proof: let $f(x)=(k+1)x^k-kx^{k+1}$.
Then $f'(x)=k(k+1)x^{k-1}(1-x)\ge 0$.
$f(x)\ge f(0)=0$.
Also notice that $\left(\frac{x^3}{3}-\frac{x^4}{4}\right)+\frac{x^5}{5}\ge 0$, where also $\frac{x^5}{5}\ge 0$, $\frac{x^{2t+1}}{2t+1}\ge 0$ for all $t\ge 2$, $t\in\mathbb Z$.
Therefore, if $0\le x<1$, then $\frac{x^3}{3}-\frac{x^4}{4}+\cdots \ge 0$, i.e. $\ln(1+x)\ge x-\frac{x^2}{2}$.
By the way, notice that when finding the sum $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$, you have to find the limit of partial sums so that you have to sum the terms in the exact same order $\frac{x^3}{3}$, $\frac{x^3}{3}-\frac{x^4}{4}$, $\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$, etc. because see the Riemann series theorem: summing the terms in a different order could result in a different number or diverge if the series is conditionally convergent. But in this case, $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ is not conditionally convergent when $-1<x<1$, because if $-1<x<1$, then it converges to $\ln(1+x)-x+\frac{x^2}{2}$ and $\frac{x^3}{3}+\frac{x^4}{4}+\cdots$ converges to $-\ln(1-x)-x-\frac{x^2}{2}$.
Here's a proof for $x\ge 1$. If $x>1$, then the Taylor series diverges, so we can't use it. $$x-\frac{x^2}{2}=\frac{2x-x^2}{2}=\frac{1}{2}-\frac{(x-1)^2}{2}\le $$
$$\le \frac{1}{2}<\ln 2=\ln(1+1)\le \ln(1+x)$$
because $1=\ln e<\ln 4=\ln 2^2$, i.e. $1<2\ln 2$, because $e<4$ and $\ln x$ is strictly increasing when $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Find $\frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3}$ for $ax^3 + bx^2 + cx + d$ Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\ &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{\left (-\frac{d}{a} \right)^3}\end{align}$$
But then I don't know how to substitute the numerator.
| If $x \neq 0$ is a solution to $at^3+bt^2+ct+d=0$ then since $a+b(\frac{1}{x})+c(\frac{1}{x})^2+d(\frac{1}{x})^3=0$, $\frac{1}{x}$ is a solution to $dt^3+ct^2+bt+a=0$. Thus if we have $\frac{1}{x_i}=y_i$ for $i=1,2,3$,
\begin{align*}
\frac{1}{x_1^3}+\frac{1}{x_2^3}+\frac{1}{x_3^3}&=y_1^3+y_2^3+y_3^3\\
&=(y_1+y_2+y_3)\left[(y_1+y_2+y_3)^2-3(y_1y_2+y_2y_3+y_3y_1)\right]+3y_1y_2y_3\\
&=(-\frac{c}{d})\left[(-\frac{c}{d})^2-3\cdot\frac{b}{d}\right]-3\cdot\frac{a}{d}\\
&=\frac{-c^3+3bcd-3ad^2}{d^3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why the minimum of this multivariate polynomial is not a critical point? I have this multivariate polynomial function
$$\begin{align*} f(L,x,y,z) &= L^2 + (1 - L x)^2 + (1 - L y)^2 + (1 - L z)^2 \\
&+ (2 - L x y)^2 + (2 - L x z)^2 + (2 - L y z)^2 + (3 - L x y z)^2
\end{align*}$$
To compute its minimum an idea is to computes its critical points and then verify which of them minimizes $f$. This is a differentiable function, so all maxima and minima of $f$ are critical points, right?
Solving the equation $\nabla f(L,x,y,z) = (0,0,0,0)$ we can find its critical points. Supposing $L \neq 0$, we have the following polynomial system.
$$ \begin{cases}
\begin{align*} X &+ Y + Z + 2 X Y + 2 X Z + 2 Y Z + 3 X Y Z \\
&-L(1 + X^2 + Y^2 + Z^2 + X^2 Y^2 + X^2 Z^2 + Y^2 Z^2 + X^2 Y^2 Z^2) = 0
\end{align*}\\
-1 + Y - Y^2 + Z + 2 Y Z - Z^2 - Y^2 Z^2 = 0\\
-1 + X - X^2 + Z + 2 X Z - Z^2 - X^2 Z^2 = 0\\
-1 + X - X^2 + Y + 2 X Y - Y^2 - X^2 Y^2 = 0\\
\end{cases}$$
I tried to solve this system using Mathematica and PHCPack. Both programs solved the system without a problem, but both missed the actually minimum. I had to use a minimizer routine in Mathematica to get the minimum $(L,x,y,z) = \left( \frac{1+\sqrt{17}}{9}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4} \right)$. This point minimizes $f$ but is not a solution of the system
above.
My question is, why this minimum is not a critical point? How is that possible?
| Reduce[Grad[L^2+(1-L x)^2+(1-L y)^2+(2-L x y)^2+(1-L z)^2+(2-L x z)^2+(2-L y z)^2+
(3-L x y z)^2,{L,x,y,z}]=={0,0,0,0}&&L!=0,{L,x,y,z}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2y^2=2\Longleftrightarrow a^2\cos^2{t}+2b^2\sin^2{t}=2.$$
Trig identities gives
$$a^2(1-\sin^2{t})+2b^2\sin^2{t}=a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
Solving for $\sin{t}$ gives
$$\sin{t}=\pm\sqrt{\frac{a^2-2}{a^2+2b^2}}=\pm f(a,b).$$
Since $$-1\leq\sin{t}\leq1\Longleftrightarrow -1\leq\pm f(a,b)\leq1.$$
Answer a) gives $f(2,1)=\frac{\sqrt{3}}{3} \in[-1,1]$. OK!
Answer b) gives $f(1,2)=\frac{i}{3}\in \mathbb{C}$. Disregard.
Answer c) gives $f(1,\sqrt{2}) = \frac{i\sqrt{5}}{5}\in \mathbb{C}$. Disregard.
So according to me, the correct answer should be a). But correct answer is d). Where did I go wrong?
| Putting $t=0$ and $t=\frac{\pi}{2}$ into the parametric equation $(x,y) = (a\cos t, b\sin t)$, you'll see that both $(a,0)$ and $(0,b)$ have to be points on the ellipse. Since these points have to satisfy $x^2+2y^2=2$, plugging in these coordinates tells you that $a^2=2$ and $2b^2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Diagonalization of a quadratic form I have to diagonalize the following quadratic form:
$$
Q(x,y,z,t)=x^2+y^2+2xy+2xz+2xt+2yz
$$
I want to complete the squares, so I observe that the first three terms are the square $(x+y)^2$. Then I sum and subtract the quantities $2x^2$, $2z^2$, $y^2$ and $t^2$. Now, how do I have to continue? Which is the associated matrix?
| Consume all the terms containing $x$ first ... etc ...
\begin{eqnarray*}
(x+y+z+t)^2-(y+z+t)^2+(y+z)^2-z^2
\end{eqnarray*}
Edit:
\begin{eqnarray*}
\begin{pmatrix}1 & 1 & 1&1 \\0 & 1 & 1&1 \\0 & 1& 1&0 \\0&0&1&0 \\\end{pmatrix}^{T}
\begin{pmatrix}1 & 0 & 0&0 \\0 & -1 & 0&0 \\0 & 0& 1&0 \\0&0&0&-1 \\\end{pmatrix}
\begin{pmatrix}1 & 1 & 1&1 \\0 & 1 & 1&1 \\0 & 1& 1&0 \\0&0&1&0 \\\end{pmatrix}=
\begin{pmatrix}1 & 1 & 1&1 \\1 & 1 & 1&0 \\1 & 1& 0&0 \\1&0&0&0 \\\end{pmatrix}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$.
From the assumption we get that $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}$.
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
|
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
Squaring and clearing denominators yields $(2k+1)^2 (3k+4) \leq (2k+2)^2 (3k+1)$. Expanding yields $12k^3+28k^2+19k+4 \leq 12k^3+28k^2+20k+4$. This simplifies to $19 k \leq 20k$, which is trivially true for positive $k$. Thus proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$.
If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $$\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$$
My Attempt:
$$p(q-r)x^2+q(r-p)x+r(p-q)=0$$
Comapring with $ax^2+bx+c=0$, we get:
$$a=p(q-r)$$
$$b=q(r-p)$$
$$c=r(p-q)$$.
Since the roots are equal:
$$b^2-4ac=0$$
$$(q(r-p))^2-4p(q-r)r(p-q)=0$$
Multipying and simplfying a bit;
$$q^2r^2-2pq^2r+p^2q^2-4p^2qr+4pq^2r+4p^2r^2-4pr^2q=0$$.
| Dividing by $p^2q^2r^2$
$$=\frac{1}{p^2} +\frac{2}{pr}+\frac{1}{r^2}-\frac{4}{qr}+\frac{4}{q^2}-\frac{4}{pq}$$
$$=\Bigl(\frac{1}{p}\Bigl)^2+\Bigl(\frac{-2}{q}\Bigl)^2+\Bigl(\frac{1}{r}\Bigl)^2$$
$$2\Bigl( \frac{1}{p}\Bigl)\Bigl(\frac{1}{r}\Bigl)+2\Bigl(\frac{1}{p}\Bigl)\Bigl(\frac{-2}{q}\Bigl)$$
$$+2\Bigl(\frac{1}{r}\Bigl)\Bigl(\frac{-2}{q}\Bigl)$$
Using
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$
$$=\Bigl(\frac{1}{p}+\frac{1}{r}-\frac{2}{q}\Bigl)^2$$
You can carry on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find limit of the function $$\lim_{x\to 0^+} \frac { \sqrt {x+bx^2}-\sqrt x}{bx^{3/2}}$$
I tried taking x common and all but it doesn't work.Are there are any shortcuts for finding the answer?Any help is greatly appreciated
Thanks in advance
| First of all $b$ cannot be zero.
By multiplying both numerator and denominator by $\sqrt{x+bx^2}+\sqrt{x}$
the function becomes:
$$\frac{(\sqrt{x+bx^2}+\sqrt{x})(\sqrt{x+bx^2}-\sqrt{x})}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}=\frac{bx^2}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}$$ $$=\frac{\sqrt{x}}{\sqrt{x+bx^2}+\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x(bx+1)}+\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x}(\sqrt{1+bx}+1)}=\frac{1}{\sqrt{1+bx}+1} \to \frac{1}{2}$$ as $x \to 0^+$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$ by using Mean Value Theorem? Given Question is Prove that
$$\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$$
when $0<a<b<1$ using Mean Value Thoerem.
I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and used LMVT and established that $\frac{b}{\sqrt{a-c^2}}=\sin^{-1}b$ where $c \in (0,b)$.
But I don't know how to proceed after this. Please guide.
| According to MVT, there exists $c$, $a\leq c\leq b$ such that
$$f(b)-f(a)=f'(c)(b-a)$$
so with $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$ we see
$$\sin^{-1}b-\sin^{-1}a=\dfrac{1}{\sqrt{1-c^2}}(b-a)$$
step by step
\begin{align}
a\leq &c\leq b\\
a^2\leq &c^2\leq b^2\\
1-b^2\leq &1-c^2\leq 1-a^2\\
\dfrac{1}{\sqrt{1-a^2}}\leq &\dfrac{1}{\sqrt{1-c^2}}\leq \dfrac{1}{\sqrt{1-b^2}}\\
\dfrac{b-a}{\sqrt{1-a^2}}\leq &\dfrac{b-a}{\sqrt{1-c^2}}\leq \dfrac{b-a}{\sqrt{1-b^2}}\\
\dfrac{b-a}{\sqrt{1-a^2}}\leq &\sin^{-1}b-\sin^{-1}a\leq \dfrac{b-a}{\sqrt{1-b^2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that the deltoid is a plane algebraic curve of degree $4$?
How can I show that a deltoid is a plane algebraic curve of degree 4?
I have searched that the parametric equation for deltoid is given by
$$\begin{align}
x&= 2 \cos t + \cos 2t \\
y&=2 \sin t - \sin 2t
\end{align}$$
So, by using some trigonometric properties, we can write
$$\begin{align}
x = 2 \cos t + 2 \cos^2 t - 1 &\quad\to\quad x^2 = \phantom{-}4 \cos^4t + 8\cos^3 t - 4 \cos t + 1 \\
y =2 \sin t -2\sin t \cos t\;\, &\quad\to\quad y^2=-4 \cos^4t+ 8 \cos^3t -8\cos t+4
\end{align}$$
| Write the equations as
$
x=2c^2+2c-1\\
y^2=-4c^4+8c^3-8c+4
$
where $c=\cos t$.
Feeding
Eliminate[{x==2c+2c2-1,y2=-4c^4+8c^3-8c+4},c]
to WA
gives
$$
x^4 - 8 x^3 + x^2 (2 y^2 + 18) + 24 x y^2 = -y^4 - 18 y^2 + 27
$$
You can do this by hand by computing the resultant of $2c^2+2c-1-x$ and $-4c^4+ 8 c^3 -8c+4-y^2$, seen as polynomials in $c$.
The resultant is the determinant of this $6 \times 6$ Sylvester matrix:
$$
\pmatrix{
2 & 2 & -x-1 & 0 & 0 & 0 \\
0 & 2 & 2 & -x-1 & 0 & 0 \\
0 & 0 & 2 & 2 & -x-1 & 0 \\
0 & 0 & 0 & 2 & 2 & -x-1 \\
-4 & 8 & 0 & -8 & 4-y^2 & 0 \\
0 & -4 & 8 & 0 & -8 & 4-y^2 \\
}
$$
It will take some work to compute this determinant by hand but it's not impossible.
However, you don't need to compute the determinant to prove that the curve is algebraic, because the determinant is clearly a polynomial in $x$ and $y$. It is also clear that its is degree $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
express the union and intersections of the circles How to express the union and intersections of the circles $ \ x^2+(y+1)^2=9$ and $x^2+(y-1)^2=9 \ $ in terms of set comprehension notation.
Answer:
Let ,
$ A=\{(x,y): \ x^2+(y+1)^2=9 \} \ \ and \ \ B=\{(x,y): \ x^2+(y-1)^2=9 \} \ $
Thus,
$A \cup B \ =\{(x,y): x=\pm \sqrt{9-(y+1)^2} \ \cup \ x=\pm \sqrt{9-(y-1)^2} \} $
Am I right ?
If there is more suitable way ?
| $x=\pm \sqrt{9-(y+1)^2} \ \cup \ x=\pm \sqrt{9-(y-1)^2}$
is meaningless, because the $\cup$ operator is an operator on sets, not equations. If anything, it should say
$x=\pm \sqrt{9-(y+1)^2} \ \lor \ x=\pm \sqrt{9-(y-1)^2}$
since $\lor$ is the symbol for "or".
Other than that, I see no easy way to describe the set, except maybe just lose the square roots and $\pm$ by writing
$$A\cup B=\{(x,y): x^2+(y+1)^2=9\lor x^2+(y-1)^2=9\}$$
or, shorter, but less informative:
$$A\cup B = \{(x,y): x^2+(y\pm 1)^2=9\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^2}{t} \times \frac{t}{1 - t^2}$$
$$\frac{x}{y} = t \tag{3}$$
Substitute $(3)$ into $(1)$
$$x = \frac{y}{x} - \frac{x}{y}$$
$$x = \frac{y^2 - x^2}{xy}$$
$$x^2y = y^2 - x^2$$
$$x^2y + x^2 = y^2$$
$$x^2(y + 1) = y^2$$
$$x^2 = \frac{y^2}{y + 1}\tag{4}$$
$$x = \sqrt{\frac{y^2}{y + 1}}$$
From $(4):$
$$x^2y + x^2 = y^2$$
$$y^2 - x^2y = x^2$$
$$y(y-x^2) = x^2$$
I have no idea how to progress from that point. I've tried different methods in paper to make $y$ the subject of the formula, but none of them were fruitful.
How do I make $y$ the subject of the formula in $(4)$?
| Alternatively:
$$x=\frac{1}{t}-t \Rightarrow x^2=\frac{1}{t^2}-2+t^2=(y+1)-2+\frac{1}{y+1} \Rightarrow $$
$$y^2-x^2y-x^2=0 \Rightarrow y=\frac{x^2\pm \sqrt{x^4+4x^2}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Are there multiple ways to translate a permutation $\alpha$ to a permutation $\beta$? Are there multiple ways to translate a permutation $\alpha$ to a permutation $\beta$?
I came across the question below and found multiple solutions for $\gamma$. My approach to finding solutions is not very rigorous, but I did check my answers and they seem to be correct.
Is my work correct, and how would I go about finding all possible $\gamma$?
$
\alpha = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
1 & 7 & 8 & 6 & 5 & 4 & 3 & 2 & 9
\end{pmatrix}
$,
$
\ \ \beta = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
1 & 2 & 5 & 4 & 6 & 7 & 3 & 9 & 8
\end{pmatrix}
$
Find $\gamma$ such that $\alpha = \gamma^{-1}\beta\gamma$.
$$
\alpha = \gamma^{-1}\beta\gamma \iff \gamma\alpha\gamma^{-1} = \beta \iff \gamma \text{ translates } \alpha \text{ to } \beta
$$
\begin{align*}
\alpha &= (1)(2\ 7\ 3\ 8)(4\ 6)(5)(9) \\
\beta &= (1)(2)(3\ 5\ 6\ 7)(4)(8\ 9) \\
\gamma &= (1)(2\ 6\ 9\ 4\ 8\ 5)(3)(7)
\end{align*}
$$
\text{This result is obtained by noting the similarities between the cycles mean it's likely that } \\
\gamma \text{ translates } 1 \to 1, \ \ 3 \to 3, \ \ 7 \to 7 \ \implies \ 2 \to 6, \ \ 8 \to 5 \\\ \\
\text{After that, there are multiple ways of translating }(4\ 6) \to (8\ 9) \text{ and } (5),\ (9)\ \to \ (8),\ (4),\ e.g.: \\\ \\
(2\ 6\ 9\ 4\ 8\ 5), \ \ (2\ 6\ 8\ 5\ 4\ 9), \ \ (2\ 6\ 8\ 5)(4\ 9)
$$
| Yes, if two permutations are conjugate they are conjugate in multiple ways. The cycle structure is preserved by conjugation but beyond that there are two kinds of freedom in the choice of the conjugating permutation: first, the cycles may be "spun" arbitrarily. And equal cycles may be permuted. In your case $\alpha$ and $\beta$ has cycle counts $c_1=3,$ $ c_2=1,$ $ c_4=1,$ (and the rest of the $c_i=0$ so you have $2\times 4$ ways to spin the 2 and 4 cycles and $3!=6$ ways to permute the 1 cycles, making $48=\prod_i (c_i!)i^{c_i}$ different conjugating maps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $, by I.H.
I am stuck here. how do I show that this expression is $ < 4^{n+1}$?
| You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into very complicated expressions, and are most likely wrong. Hints or solutions are welcomed. Sorry if this is a duplicate.
| As Mr. Chip suggested there’s no need to drag around $a$ and $b$ here.
So I'm gonna proceed in this fashion:
$$(1+x)^7-1^7-x^7=7x(1+x)(1+x+x^2)^2$$
Now we divide by $(1+x)$ or we can factor it out:
Note that various divisions like that may call into question the rigor here. But it should be as easy as pie for you (I'm leaving it for you).
Well, we go on
$$(1+x)^6-(1-x+x^2-x^3+x^4-x^5+x^6)=7x(1+x+x^2)^2$$
$$7x+14x^2+21x^3+14x^4+7x^5=7x(1+x^2+x^4+2x+2x^2+2x^3)$$
$$7x(1+2x+3x^2+2x^3+x^4)=7x(1+2x+3x^2+2x^3+x^4)$$
So we got our result slightly faster this way.
My point is you should be very proficient even when dealing with complicated expressions including the knoweledge of Multinomial theorem, which I used here on autopilot for getting the square of the trinomial. And the formula for $1\pm x^n$ and all the related subtleties are supposed to be easy when we talk about Olympiads. Seeing here $1^7+x^7$ should ring a bell that we can factor out $1+x\,.\,$ Hope my advice might be of some use.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solve identity : $\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$ $$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$$
I am doing this identity for about an hour and I can't get to the result. I use a common denominator but that just makes a mess that takes forever to do and gets me nowhere. Surely there is an easier way that I am not seeing? Some hint or help please?
| Use $a^3-b^3=(a-b)(a^2+ab+b^2)$
Therefore we have,
$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x \cos x)$$
Your expression in LHS
$$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x} $$
Simplifies to
\begin{align}
\text{LHS}
&=\frac{1}{\cos x - \sin x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{2\cos x}{\cos^2 x - \sin^2 x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{2 \cos x+\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{\sin^2 x-1}{\cos^2x-\sin^2x}\\
&= \frac{cos^2 x}{\sin ^2x-\cos^2x}\\
&= \frac{1}{\tan^2x-1}= \text{RHS}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\to a} (a-x) \tan \frac {\pi x}{2a}$
Evaluate:
$$\lim_{x\to a} (a-x) \tan \dfrac {\pi x}{2a}.$$
My attempts:
$$=\lim_{x\to a} (a-x) \dfrac {\sin \left(\dfrac {\pi x}{2a}\right)}{\cos \left(\dfrac {\pi x}{2a}\right)}=1\cdot\lim_{x\to a} \dfrac {a-x}{\cos \left(\dfrac {\pi x}{2a}\right)}.$$
| Let $y = \frac{\pi x}{2a}$
\begin{align} \lim_{x \to a} \frac{a-x}{\cos \left( \frac{\pi x}{2a}\right)} &= \lim_{y \to \frac{\pi}{2}} \frac{\pi a-(2ay)}{\pi \cos(y)}\\
&=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos(y)}\\
&=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos\left(y-\frac{\pi}{2} + \frac{\pi}{2}\right)}\\
&= \frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{y-\frac{\pi}2}{\sin\left(y-\frac{\pi}{2} \right)}\\
&=\frac{2a}{\pi}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding all non-negative integers solutions to $x_1+x_2+x_3+...+x_6=20$ such that $x_{2n+1} \le x_{2n+2}$ for $0 \le n \le 2$ Solve $x_1+x_2+x_3+x_4+x_5+x_6=20$ such that $x_{2n+1}\leq x_{2n+2}, 0\leq n \leq2$
Edit : I solved it.
Let $0 \le a,b,c \le 20$ such that $0 \le a+b+c \le 20$
$x_1+x_2+x_3+x_4+x_5+x_6=20 \;and\ x_{2n+1}\leq x_{2n+2}\\
\implies x_2=x_1+a,\;x_4=x_3+b,\;x_6=x_5+c\\
\implies 2(x_1+x_2+x_3)+a+b+c=20\\
\implies 2(x_1+x_2+x_3)=20-a-b-c\;\;\;(1)$
So now we need for a fixed values of $a,b,c$ to find the number of non-negative integers solutions to the equation at $(1)$
Now, for all $0 \le i \le 20 $
Let $A_i$ be the set of all non negative solutions for $a+b+c=i$
Let $X_i$ be the set of all non negative solutions for $2(x_1+x_2+x_3)=20-i$
Now,
$2(x_1+x_2+x_3)=20-i \implies (x_1+x_2+x_3)= \frac{20-i}{2}$
And since we are interesting in non-negative integers we can say that
$20-i\nmid 2 \implies X_i=\emptyset \implies |X_i|=0\;\;\;(2)$
Otherwise,
If $i$ is even we will want to compute
$\left|A_i\right|\cdot \left|X_i\right|$ because for any solution a+b+c=i in $A_i$ there's the corresponding solution in $X_i$
Therefore we have:
$\sum _{i=0}^{20}\:\left(\left|A_i\right|\cdot \left|X_i\right|\right)=\sum _{i=0}^{20}\:\begin{pmatrix}i+3-1\\ \:i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:10-1\end{pmatrix}\\ \text{and from (2)}\\\sum _{i=0}^{20}\:\begin{pmatrix}i+3-1\\ \:i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:10-1\end{pmatrix}= \sum \:_{i=0}^{10}\:\begin{pmatrix}2i+3-1\\ \:\:2i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:\:10-i\end{pmatrix}=\\ \sum _{i=0}^{10}\:\left(\frac{\left(2i+2\right)\left(2i+1\right)}{2}\right)\left(\frac{\left(12-i\right)\left(11-i\right)}{2}\right)=9009$
And we've got a palindrome.
| I verified your solution a different way:
Let $N(i)$ denote the number of ways to have $x_{1}+x_{2}=i$ with $x_{1}\leq x_{2}$. It is easy to show that $N(i)=\lfloor\frac{i}{2}\rfloor+1$, for $0\leq i$.
Your problem now reduces to computing the following sum:
$\sum_{(k_{1},k_{2},k_{3})\in K}N(k_{1})\cdot N(k_{2})\cdot N(k_{3})$
where $K=\{(k_{1},k_{2},k_{3}):k_{1}+k_{2}+k_{3}=20,\ k_{i}\geq 0\}$, i.e. the set of ordered pairs of three non-negative integers summing to 20.
After a bit of sum-manipulating we get the equivalent sum:
$\sum_{k_{1}=0}^{20}N(k_{1})\sum_{k_{2}=0}^{20-k_{1}}N(k_{2})\cdot N(20-(k_{1}+k_{2}))$
Now throw elegance to the curb, enter the sum into Mathematica, and get 9009 $\checkmark$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$T(n)=2T(n-1)+n,T(1)=1,n\ge2$
Solve the recurrence $T(n)=2T(n-1)+n,T(1)=1,n\ge2$
My approach:
$$T(n)=2T(n-1)+n$$
$$T(n)=2(2T(n-2)+(n-1))+n$$
$$T(n)=4T(n-2)+2(n-1)+n\dots$$
$$T(n)=2^kT(n-k)+2^{k-1}(n-k+1)+\dots+2^{k-k}n$$
From initial conditions, $k=n-1$:
$$T(n)=2^{n-1}1+2^{n-2}2+\cdot\cdot\cdot\cdot+2^{n-n}n$$
Now what?
| $$f(x)=x2^{n-1}+x^22^{n-2}+...+x^n=2^n\left(\frac{x}{2}+\left(\frac{x}{2}\right)^2+...+\left(\frac{x}{2}\right)^n\right)=$$
$$=2^n\cdot\frac{\frac{x}{2}\left(\left(\frac{x}{2}\right)^n-1\right)}{\frac{x}{2}-1}=\frac{x^{n+1}-x2^n}{x-2}.$$
Now, calculate $f'(1).$
I got $$1\cdot2^{n-1}+2\cdot2^{n-2}+...+n\cdot2^0=f'(1)=2^{n+1}-n-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine all real $x$ such that $\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$ I just need you guys to check this solution and tell me where I can improve. The key questions to be answered are:
*
*Are things clear?
*Is there any unnecessary redundancy?
*Any logical fallacies?
*Quality of stringency and mathematical language?
Problem: Determine all real $x$ such that $$\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$$
Attempt: Let $f(x)=\arccos{\frac{1-x^2}{1+x^2}}$ and $g(x)=-2\arctan{x}.$ By the definition of the inverse trigonometric functions it now follows that
$$\left\{
\begin{array}{rcr}
D_{f} & ?=? & [-1,1] \\
V_{f} & = & [0,\pi] \\
\end{array}
\right. \quad \text{and} \quad \left\{
\begin{array}{rcr}
D_{g} & = \mathbb{R}\\
V_{g} & = & 2\left(-\frac{\pi}{2},\frac{\pi}{2}\right)=(-\pi,\pi)\\
\end{array}
\right.$$
Where $V$ denotes ranges and $D$ denotes domains. By $D_{f}$ it follows that $-1\leq \frac{1-x^2}{1+x^2}\leq1.$ Upon examination of the inequalities $-1\leq\frac{1-x^2}{1+x^2}$ and $\frac{1-x^2}{1+x^2} \leq 1$ one finds that they get satisfied $\forall x\in\mathbb{R}.$ This implies that $D_{f}=D_{g}=\mathbb{R}.$ The the function values can only be the same in the intersection of their respective ranges; $[0,\pi]\cap(-\pi,\pi)=[0,\pi).$ So $f(x)\rightarrow\pi$ when $x\rightarrow -\infty$ and $g(x)\rightarrow\pi$ when $x\rightarrow -\infty.$ We also note that both $f(x)$ and $g(x)\rightarrow 0$ when $x\rightarrow0^{-}.$ This means that the solutions, if they exist, should be in the interval $(-\infty,0]$. Lets find them:
Taking cosine of both sides we get $$\frac{1-x^2}{1+x^2}=\cos{(2\arctan{x})}=2\cos^2{(\arctan{x})}-1.$$ Using the fact that $\cos{\arctan{x}}=\frac{1}{\sqrt{1+x^2}}$ from
So $$\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}-1 = \frac{1-x^2}{1+x^2} \Longleftrightarrow x=x.$$
This means that the solutions are all reals, but according to our earlier conclusion the solutions must exist in $(-\infty,0].$ Thus the solution set to the original equation is $$x\in(-\infty,0]\cap(-\infty,\infty)=(-\infty,0].$$
| Your solution seems good. Here's an alternative way.
It is easy to see that
$$
-1\le\frac{1-x^2}{1+x^2}\le1
$$
for every $x$.
The derivative of
$$
f(x)=\arccos\frac{1-x^2}{1+x^2}
$$
is
\begin{align}
f'(x)&=-\frac{1}{\sqrt{1-\dfrac{(1-x^2)^2}{(1+x^2)^2}}}
\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}
\\[6px]
&=\frac{1+x^2}{|2x|}\frac{4x}{(1+x^2)^2}
\\[6px]
&=\frac{x}{|x|}\frac{2}{1+x^2}
\end{align}
for $x\ne0$.
This means that
$$
\arccos\frac{1-x^2}{1+x^2}=
\begin{cases}
a+2\arctan x & \text{for $x>0$} \\[6px]
b-2\arctan x & \text{for $x<0$}
\end{cases}
$$
for some constants $a$ and $b$. When $x=1$, we have
$$
\arccos0=\frac{\pi}{2}
\qquad
a+2\arctan 1=a+2\frac{\pi}{4}
$$
so we conclude $a=0$; similarly, for $x=-1$ we have
$$
b-2\arctan(-1)=b+\frac{\pi}{2}
$$
hence also $b=0$.
Your equation is thus satisfied for every $x<0$ and also for $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| The structure of the equations indicate that $7$, $49$, $133$, $406$, ... is a sequence that is defined as a homogeneous linear difference equation of order $2$ with the characteristic equation $\lambda^2+c_1\lambda +c_0=0$ with the roots $x$ and $y$. Therefore we have
\begin{eqnarray*}
133 + \;\;49c_1 + \;\;7c_0 &=& 0 \\
406 + 133c_1 + 49c_0 &=& 0
\end{eqnarray*}
with $c_1 = -(x+y)$ and $c_0=xy$. From this, we easily get $x+y$ and $xy.$ Then we can find $x$ and $y$ and use the first two equations to figure out $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Infinite sum of $\cot^{-1} (n^2 + 3/4)$. I am trying to find the infinite sum
$$\sum_{n=1}^\infty \cot^{-1} (n^2 + ( \frac{3}{4})),$$
I tried to get a telescopic series but I couldn't find one.
| As $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$
$$\dfrac{4n^2+3}4=1+\dfrac{4n^2-1}4=1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2$$
Again,$\dfrac{2n+1}2-\dfrac{2n-1}2=1$
So, we can write $\cot^{-1}\left(n^2+\dfrac34\right)=\cot^{-1}\left(\dfrac{1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2}{\dfrac{2n+1}2-\dfrac{2n-1}2}\right)$
$=\cot^{-1}\left(\dfrac{2n-1}2\right)-\cot^{-1}\left(\dfrac{2n+1}2\right)$
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/2440420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| Two solutions of $82=x^4+(x+2)^4$ are given by $x=1$ and $x=-3$. It follows that
$$ x^4+(x+2)^4-82 = (x-1)(x+3) q(x) $$
where $q(x)$ is a quadratic polynomial. Find it, solve $q(x)=0$ and you have the complete set of solutions ($q(x)=2\left[(x+1)^2+10\right]$, so the only real solutions are the trivial ones).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Proving an identity in floor function For every positive integer $n$ ,show that
$$\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor=\lfloor{\sqrt{4n+1}}\rfloor=\lfloor{\sqrt{4n+2}}\rfloor=\lfloor{\sqrt{4n+3}}\rfloor$$
My Attempt:
$$\left(\sqrt{n}+\sqrt{n+1}\right)^2=2n+1+2\sqrt{n(n+1)}$$
Now, $\sqrt{n(n+1)}$ is geometric mean of $n$ and $n+1$, therefore
$$n<\sqrt{n(n+1)}<n+1$$
$$2n<2\sqrt{n(n+1)}<2(n+1)$$
$$4n+1<2n+1+2\sqrt{n(n+1)}<4n+3$$
$$4n+1<\left(\sqrt{n}+\sqrt{n+1}\right)^2<4n+3$$
But after this not able to get anywhere
| Note that for an integer, $x$, $$x^2 \equiv 0 \pmod 4$$ or
$$x^2 \equiv 1 \pmod 4$$
Hence there is no complete square between $4n+1$ and $4n+3$.
Hence $\lfloor \sqrt{4n+1} \rfloor = \lfloor \sqrt{4n+3} \rfloor$, the overall task can be completed by your earlier result of $4n+1 < (\sqrt{n} + \sqrt{n+1})^2 < 4n+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$ $\ge$1 Prove your assertion. Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$$\ge$1 Prove your assertion.
I guess $m$$=$1 because if $m$ divides $n^3-n$ $\Rightarrow$ $m$ divides $n(n-1)(n+1)$ then $m$ divides at least one of them, but $\forall$$n$ natural number, $n$,$n-1$,$n+1$ are coprime, so $m$$=$1
Can anybody help correct or improve the answer?
| We have $n^3 -n = n\cdot\left(n^2 - 1\right) = n\cdot(n-1)\cdot(n+1)$. These are three consecutive numbers so at least one of them has to be divisible by $3$ and one of them has to be divisible by $2$. Hence, $m$ has to be at least $2\cdot 3 = 6$.
However, for each prime $p \in \mathbb P$ with $p > 3$, we can find an $n \in \mathbb N$, so that $n^3 - n = (n-1)n(n+1)$ is not divisible by $p$, such as $n = p + 2$, since then the expression becomes $(p+1)(p+2)(p+3)$ and none of these factors can be divisible by $p$. We also cannot have $n^3 - n$ always being divisible by bigger powers of $2$ or $3$, any $n = 6k$ is a counterexample of that.
Therefore, $m$ has to be equal to $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral of irrational function $\int \frac{dx}{x\left(\sqrt[3]{x}+\sqrt[5]{x^2}\right)}$
$$\int \frac{dx}{x\left(\sqrt[3]{x}+\sqrt[5]{x^2}\right)}$$
I have tried to solve the integral using, substitution but it doesn’t lead to the correct result.
Do you folks have any ideas about which technique I should use to solve the given integral?
| Plug $x=u^{15}$ so $dx=15u^{14}$
$\sqrt[3]{x}=u^5;\;\sqrt[5]{x^2}=u^6$
Integral becomes
$$15\int \frac{du}{u^7+u^6} =15\int \frac{du}{u^6 (u+1)}=15\int\left[\frac{1}{u^6}-\frac{1}{u^5}+\frac{1}{u^4}-\frac{1}{u^3}+\frac{1}{u^2}+\frac{1}{u+1}-\frac{1}{u}\right]\,du$$
Indeed with partial fraction we have:
$$\frac{a}{u}+\frac{b}{u^2}+\frac{c}{u^3}+\frac{d}{u^4}+\frac{e}{u^5}+\frac{f}{u^6}+\frac{g}{u+1}=\frac{1}{u^6 (u+1)}$$
Numerator is
$u^6 (a+g)+u^5 (a+b)+u^4 (b+c)+u^3 (c+d)+u^2 (d+e)+u (e+f)+f$
which must be identical to $1$ so $f=1,\;e=-1,\; a = -1,\;b = 1,\;c = -1,\;d = 1,\;g = 1$
So the integral gives
$$15 \left(-\frac{1}{5 u^5}+\frac{1}{4 u^4}-\frac{1}{3 u^3}+\frac{1}{2 u^2}-\frac{1}{u}-\log (u)+\log (u+1)\right)+C=$$
$$=15 \left(\frac{1}{2 \sqrt[2]{x^{15}}}+\frac{1}{4 \sqrt[4]{x^{15}}}-\frac{1}{5 \sqrt[3]{x}}-\frac{1}{3 \sqrt[5]{x}}-\frac{1}{\sqrt[15]{x}}-\log \sqrt[15]{x}+\log \left(\sqrt[15]{x}+1\right)\right)+C=$$
$$=\frac{15}{2 \sqrt[2]{x^{15}}}+\frac{15}{4 \sqrt[4]{x^{15}}}-\frac{3}{\sqrt[3]{x}}-\frac{5}{\sqrt[5]{x}}-\frac{15}{\sqrt[15]{x}}+15 \log \left(\sqrt[15]{x}+1\right)-\log x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$ Can you provide a proof of this identity using only calculus?
$$\ln x + 1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln x)^n}{x}$$
By the way, here is how I arrived at it:
There is string of length $x$ units. Select a point on the string uniformly at random and cut the string at that point. Repeat the process with the string on the left side of the cut until the string you have is shorter than $1$ unit. The problem is to figure out the expected number of cuts.
Here is how I did it:
Let $E(x)$ denote the expected number of cuts to be made on a string of length $x$. If $x<1$, clearly, $E(x)=0$. If $x>1$, we have:
\begin{align} E(x) &= 1 + \int_0^x E(u) \cdot \frac{du}{x} \\
&= 1 + \frac 1x \int_1^x E(u) \ du
\end{align}
Multiplying by $x$ and differentiating (applying the Fundamental Theorem of Calculus),
\begin{align} xE'(x) + E(x) &= 1 + E(x) \\
\Rightarrow E(x) &= \ln x + C \end{align}
Clearly for $x=1, \ E(x)=1$ thus $E(x) = \ln x + 1$
However, we can also calculate $E(x)$ in a different way:
Let $P(n,\ x)$ denote the probability of exactly $n$ cuts being made on a string of length $x$.
If $x<1$, $P(n,\ x) = 0$; if $n=1$ and $x>1$ $P(n,\ x)=\frac 1x$; if $n>1$ and $x>1$:
\begin{align} P(n,\ x) &= \int_0^x P(n-1,\ u) \cdot \frac{du}{x} \\
&= \frac 1x \int_1^x P(n-1,\ u) \ du \end{align}
I calculated that $P(1,\ x) = \frac 1x,\ P(2,\ x) = \frac{\ln x}{x},\ P(3,\ x) = \frac{(\ln x)^2}{2x},\ P(4,\ x) = \frac{(\ln x)^3}{6x}$
This led me to hypothesize that $P(n,\ x)=\frac{(\ln x)^{n-1}}{x(n-1)!}$, which can be proven by induction:
\begin{align} P(n,\ x) &= \frac 1x \int_1^x P(n-1,\ u) \ du \\
&= \frac 1x \int_1^x \frac{(\ln u)^{n-2}}{u(n-2)!} \ du \\ &= \frac 1{x(n-2)!} \int_1^x (\ln u)^{n-2}\ d(\ln u) \\
&= \frac 1{x(n-2)!} \left[\frac {(\ln u)^{n-1}}{n-1} \right]_1^x \\ &= \frac{(\ln x)^{n-1}}{x(n-1)!} \end{align}
But then $E(x)$ can be written as follows:
\begin{align} E(x) &= \sum_{n=1}^{\infty} n \cdot P(n,\ x) \\
&= \sum_{n=1}^{\infty} n \cdot \frac{(\ln x)^{n-1}}{x(n-1)!} \\ &= \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}
\\ \\
\therefore \ \ln x + 1 &= \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x} \ \blacksquare \end{align}
So can you prove this result by finding an appropriate Taylor series (this would be especially appreciated) or if not using Taylor series then using just the methods of calculus?
| For, $x>0$, $\text{ln}x =z \in (-\infty, \infty)$.
So, we can write the r.h.s as ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$.
Now, ${\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$
$=\frac{d (\sum_{n=0}^{\infty} \frac{z^{n+1}}{n!})}{dz}$
$=\frac{d {e^{z}}z}{dz}= e^{z}+ze^{z}$
So, ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}={e^{-z}}(e^z+ze^{z})=z+1=\text{ln}x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to solve :$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$ $$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$
Why can't I figure this one out??
Is it possible to cancel out the arcsins?
I know from graphing on my calculator that the answers are $-1, 0$, and $1$, but want help getting there by hand.
Thank you!
| Use $\sin(A-B)$ identity and the fact that $f(f^{-1}(x)) = x$:
$$x\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-x^2} = \frac{x\sqrt 3}{2}$$
So $x=0$ is one solution. Now
$$\sqrt{4-x^2}-\sqrt{1-x^2} = \sqrt{3}$$
Now let $1-x^2 = t$, so that our equation becomes:
$$\sqrt{3+t}-\sqrt{t} = \sqrt{3}$$
Upon squaring, we get:
$$3+t+t - 2\sqrt{(3+t)t} = 3 \\
t^2 = (3+t)(t)$$
From here, we get $t = 0$. Thus $1-x^2 = 0$, so that $x = \pm 1$
So we have three solutions, $x = -1, 0, 1$, which satisfy the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Guess explicit formula for sequence and then prove it is correct just trying to work through the following problem but have come to point where I don't know what to do.
Info: Let $a_0, a_1, a_2, ...$ be the sequence defined recursively as $a_0 = 0, a_k = k + a_{k-1}$ for each integer $k\ge 1$.
Question is: Using this information write out the first 6 terms of the sequence, guess the explicit formula and then prove your guess is correct.
So I've done as follows:
First 6 terms: $a_0 = 0, a_1 = 1, a_2 = 3, a_3 = 6, a_4 = 10, a_5 = 15, a_6 = 21$
Explicit formula guess: $ a_n = \frac {n^2 + n}{2}$
Proof:
Using the guessed explicit formula $a_n = \frac {n^2 + n}{2}$ for each integer $n \ge 0$ we see that
$a_0 = \frac {0^2 + 0}{2} = 0$ so we have the correct initial condition.
Let $a_0, a_1, a_2, ...$ be the sequence defined as $a_0 = 0$ and $a_k = k + a_{k-1}$ for each integer $k \ge 1$
Then, let P(n) be the predicate $a_n = \frac {n^2 + n}{2}$ for each integer $n \ge 0$
Basis step: $a_0 = 0$ from the recursive definition and $\frac {n^2 + n}{2} = 0$, so P(0) is true.
Inductive Hypothesis: Suppose that P(k) is true for some integer $k \ge 0$.
Then $a_{k+1} = (k + 1) + a_k$
= $(k + 1) + \frac {k^2 + k}{2}$
= $\frac{2(k + 1) + (k^2 + k)}{2}$
= $\frac{2k + 2 + k^2 + k}{2}$
= $\frac{k^2 + 3k + 2}{2}$
= $\frac{k^2 + 3k}{2} + 1$??? I'm not sure what to do after the previous step. Any help would be greatly appreciated.
| Induction isn't actually needed here. Notice that each term is a partial sum, in particular:
$$a_n = \sum_{k=0}^n k.$$
You guessed that $a_n = \frac{n^2 +n}{2}$, which is correct, so you can just show that
$$\sum_{k=0}^n k = \frac{n^2 +n}{2}.$$
The usual way is to add the sum here to itself but write it backwards once, i.e.,
\begin{align}2 \sum_{k=0}^n k & = \hspace{.05in} \sum_{k=0}^n k + \sum_{k=0}^n k\\
& =\hspace{.2in}1 +\hspace{.2in} 2 \hspace{.2in} + ...+(n-1) + n \\
& \hspace{.2in}+ n + (n-1) +... + \hspace{.2in}2 \hspace{.2in}+ 1 \\
& =\hspace{.05in} n(n+1)\\
& =\hspace{.05in} n^2 +n .
\end{align}
In the second to last step we added vertically to get $n+1$, $n$ times. Then just divide by two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$ without induction I found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.
Prove for odd $n$ that
$$\sum_{k=1}^{(n+1)/2}\prod_{j=0}^{k-1} \left(\frac{n-2j-1}{n-2j}\right)=\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$
| We need to prove that
$$\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$ or
$$1+\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n}{3}$$ or
$$\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-3}{3}$$ or
$$1+\frac{n-5}{n-4} +\frac{n-5}{n-4}\frac{n-7}{n-6} ... = \frac{n-2}{3}$$ or
$$.$$
$$.$$
$$.$$
$$1+\frac{2}{3}=\frac{5}{3}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.