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Evaluating $\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$ How can we evaluate:
$$\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$$
I tried to use integration by parts, but the positive sign of the $x^2$ in the square root doesn't cancel with the integral of $\sin^{-1}(x)$, so how can I proceed ?
|
$\int\dfrac{x\sin^{-1}x}{\sqrt{1+x^2}}dx$
$=\int\dfrac{\sin^{-1}x}{2\sqrt{1+x^2}}d(x^2)$
$=\int\sin^{-1}x~d(\sqrt{1+x^2})$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sqrt{1+x^2}~d(\sin^{-1}x)$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+\sin^2u}}{\sqrt{1-\sin^2u}}d(\sin u)$ $(\text{Let}~x=\sin u)$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sqrt{1+\sin^2u}~du$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}du$
$=\sqrt{1+x^2}\sin^{-1}x-\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}\right)du$
For $\int\sin^{2n}u~du$ , where $n$ is any natural number,
$\int\sin^{2n}u~du=\dfrac{(2n)!u}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts from the formula of http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions.
$\therefore\sqrt{1+x^2}\sin^{-1}x-\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}\right)du$
$=\sqrt{1+x^2}\sin^{-1}x-u-\sum\limits_{n=1}^\infty\dfrac{(-1)^n((2n)!)^2u}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2\sin^{2k-1}u\cos u}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
$=\sqrt{1+x^2}\sin^{-1}x-\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2u}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2\sin^{2k-1}u\cos u}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
$=\sqrt{1+x^2}\sin^{-1}x-\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2\sin^{-1}x}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2x^{2k-1}\sqrt{1-x^2}}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
|
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|
find general solution to the Differential equation Find the general solution to the differential equation
\begin{equation}
\frac{dy}{dx}= 3x^2 y^2 - y^2
\end{equation}
I get
\begin{equation}
y=6xy^2 + 6x^2 y\frac{dy}{dx} - 2y\frac{dy}{dx}
\end{equation}
rearrange the equation
\begin{equation}
\frac{dy}{dx} = \frac{y-6xy^2}{6x^2 y - 2y}
\end{equation}
simplified the equation
\begin{equation}
\frac{dy}{dx} = \frac{1-6xy}{6x^2 - 2}
\end{equation}
how can I solve this equation to get a general solution of $y=ax+c$
|
You might use separating variable method: you have: $\frac{1}{y^2}dy = (3x^2 -1)dx$, and you integrate both sides to get the answer.
|
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|
If $a,b,c$ are positive integers and $a^2+b^2=c^2$ and $a$ is a prime, what can we conclude about primeness of b and c? Let $a,b,c$ be positive integers and they satisfy $a^2+b^2=c^2$, and if $a$ is prime, can we conclude whether $b$ and $c$, are both prime, composite or neither? If yes, why, if not why not?
I can conclude that $b$ and $c$ have to be one odd and the other one even using $a^2 = c^2-b^2=(c-b)(c+b)$. But I couldn't conclude anything about their primeness. Can anyone show me some ideas or its reasoning and maybe any useful and related theorems?
Thanks very much!
|
$c$ is prime and $b$ isn't :
$$3^2+4^2=5^2$$
both are composite :
$$7^2+24^2=25^2$$
However, $b$ can never be a prime. Proof :
$a=2$ or $b=2$ is impossible (first pythagorean triplet is $3,4,5$)
if both $a$ and $b$ are odd prime numbers, then $a^2\equiv b^2\equiv 1 \mod 4$ (this is true for all odd numbers), and therefore $c^2\equiv 2\mod 4$ (impossible)
|
{
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"url": "https://math.stackexchange.com/questions/621246",
"timestamp": "2023-03-29T00:00:00",
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|
How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$
My try: since
$$\Longleftrightarrow\sum_{sym}\left(\dfrac{a-b}{a+2b+c}+\dfrac{1}{2}\right)\ge\dfrac{5}{2}$$
$$\Longleftrightarrow \sum_{sym}\left(\dfrac{3a+c}{a+2b+c}\right)\ge 5$$
use Cauchy-Schwarz inequality,we have
$$\sum_{sym}\dfrac{3a+c}{a+2b+c}\sum_{sym}((3a+c)(a+2b+c))\ge\left(\sum_{sym}(3a+c)\right)^2$$
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$
becasue this is not hold
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$for $a,b,c,d,e>0$
and this method is from this simaler inequality :
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$
then I can't,Thank you
|
I'll solve the three variable case (the $a,b,c,d,e$ case is similar).
Assume without loss of generality $a \geq b \geq c$. Multiplying by the denominators we obtain
$$
a^3+a^2 c+a b^2-6 a b c+b^3+b c^2+c^3 \geq 0
$$
Now use the rearrangment inequality twice. We can split the inequality in two parts:
$$
a^2c+b^2a+c^2b \geq 3abc
$$
and
$$
a^3+b^3+c^3 \geq 3abc
$$
For the first one, use the rearrangment inequality to the sequences $ab, ac, bc$ and $a,b,c$ (which are listed in increasing order).
As for the second one, you can use the extension of the rearrangement inequality (also explained in the link) with three equal sequences $a,b,c$. Sum up the two inequalities and you are done.
|
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|
Solve in $\mathbb Z$ the equation: $x^5 +15xy + y^5=1$ Solve in $\mathbb Z$ the equation: $x^5 +15xy + y^5=1$
I tried: $x(15y+x^4)+y^5=1$
But don't have much ideas on how to continue, thanks!
|
Let's find all pairs $(x,y) \in \mathbb{Z} \times \mathbb{Z} $ that satisfies
$$\tag{*}\label{main-poly} x^5 +15xy + y^5=1$$
It's obvious that $(1,0)$ and $(0,1)$ are solutions of $\eqref{main-poly}$ , because if one of $x,y$ equals $0$, then the other one must be equal to $1$.
It is also obvious, that if $x,y \neq 0$, then one of them must be positive, and the other one negative.
Without loss of generality (because of symmetry), we can assume that $x>0$ and $y<0$. Also, if the pair $(x,y)$ is a solution, the pair $(y,x)$ will also be the solution.
For $ x = 2 $ we got:
$$ 30y + y^5 = -31 $$
which is true only for $y = -1$, because for every integer $y<-1$ the left side of equation will be always less than $-31$.
Therefore $(-1,2)$ and $(2,-1)$ safisfies $\eqref{main-poly}$.
For $ x=3 $ there is no solution, because we get
\begin{gather}
243 + 45y + y^5 = 1 & \Longleftrightarrow & y(y^4+45) = -242 = -2 \cdot 121 = -2 \cdot (45 + 76)
\end{gather}
and there is no such integer $y$ satisfying this.
Let's see what happens for $ x \ge 4$.
Firstly, let's assume that $ \quad |y| \ge x \quad \Longleftrightarrow \quad y \le -x . \quad $ Then we have no solutions, because
$$ x^5 + 15xy + y^5 \ \le \ x^5 + 15xy - x^5 \ = \ 15xy \ < \ 0 \ < \ 1. $$
In the other case, that is for $ \ \ \ |y| < x \ \Longleftrightarrow \ y > -x \Longleftrightarrow \ x > -y , \ $ we have:
$$ x^5 + 15xy + y \cdot y^4 > x^5 + 15xy + yx^4 = x^4 (x+y) + 15xy $$
because $ \ y^4 < x^4 \ $ and $ \ y<0, \ $ thus $ \ y^5 > yx^4. \ $
Next, because $ \ x \ge 4 \ $ and $ \ |y| \le x-1, \ $ we have
$$ x^5 + 15xy + y^5 \ > \ x^4 (x+y) + 15xy \ \ge \ x^4 (x+(-(x-1))) + 15xy \ = \ x^4 + 15xy = \\ = \ x(x^2 \cdot x+15y) \ \ge \ x(4^2 x + 15y) \ = \ x(x + 15x + 15 y) \ > \ x(x + 15(-y) + 15 y) \ = \\ = x^2 \ \ge \ 16 \ > \ 1 $$
so there cannot be $ \ x^5 + 15xy + y^5 \ = \ 1$.
So, the only possible solutions of $(*)$ are: $ (1,0)$, $ (0,1)$, $ (-1,2)$, $ (2,-1)$,
|
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|
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series.
EDIT: I corrected a mistake in the formula I wanted to use.
|
You can also use that
$$2\sin(x)\Bigl[\cos 2x+\cos 4x + \cos 6x+\cos 8x\Bigr] = \sin 9x-\sin x = 2\cos 5x\sin 4x$$
so that inserting $x=\tfrac\pi5=\pi-4\tfrac\pi5$ yields the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometry Identity Problem Prove that:
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{ \cos A}$$
I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/cos A and then simplifying but it didn't work.
|
You may also simplify the expression by using the definition of the tangent and the secant function:
\begin{align}
&\dfrac{\tan A +\sec A-1}{\tan A - \sec A +1}
\\=&\dfrac{\dfrac{\sin A}{\cos A} +\dfrac1{\cos A}-1}{\dfrac{\sin A}{\cos A} - \dfrac1{\cos A} +1}
\\=&\dfrac{\sin A -\cos A+1}{\sin A+\cos A-1}
\\=&\dfrac{(\sin A -\cos A+1)\cdot(\sin A+\cos A+1)}{(\sin A+\cos A-1)\cdot(\sin A+\cos A+1)}=\dfrac{(\sin A +1)^2-\cos^2 A}{(\sin A+\cos A)^2-1}
\\=&\dfrac{\sin^2 A+(1-\cos^2 A) +2\sin A}{2\sin A\cos A}\end{align}
The rest is fairly straightforward.
|
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|
Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$ Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$
Solution :
$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$
and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$
$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$
Can we have other option available so that we will be able to find the solution more quicker way,
Since this is a conic then the given lines let $L_1 =2y-x-3=0 $ and $L_2 = 2x+y-1=0$ are perpendicular to each other .. Can we use this somehow please suggest... thanks..
|
As the Rotation of the axes does not change the eccentricity of a curve
set $2y-x-3=X,2x+y−1 =Y$ so that the given equation becomes $$4X^2-9Y^2=80\iff \frac{X^2}{\frac{80}4}-\frac{Y^2}{\frac{80}9}=1$$
Now we know, $$b^2=a^2(e^2-1)$$ where $a,b\le a$ are the semi-major & semi-minor axes and $e$ is the eccentricity of the hyperbola
|
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|
Prove the series identity
Prove an identity:
$$\sum_{n=2}^{ \infty } \frac{2}{(n^3-n)3^n}=- \frac{1}{2}+ \frac{4}{3} \cdot \sum_{n=1}^{ \infty } \frac{1}{n \cdot 3^n}$$
I've checked that the left-hand-side of this identity is convergent absolutely, hence I can write it as:
$$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)} \cdot \frac{1}{n3^n}$$
I've also calculated the sum $$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)}= \frac{3}{2}
$$
But now, I don't see what can I do with the right-hand-side and what to do with $$\sum_{n=2}^{ \infty }\frac{1}{n3^n}$$
|
$$\sum_{n=2}^\infty{2\over(n+1)(n-1)n3^n}=\sum_{n=2}^\infty \left[{1\over (n-1)n}-{1\over (n+1)n}\right]{1\over 3^n}=\sum_{n=2}^\infty\left[\color{red}{{1\over (n-1)}}-{2\over n}+\color{blue}{1\over (n+1)}\right]{1\over 3^n}=\sum_{n=2}^\infty\left[\color{red}{{1\over 3(n-1)3^{n-1}}}-{2\over n3^n}+\color{blue}{3\over (n+1)3^{n+1}}\right]=\color{red}{{1\over 3}\sum_{n=1}^\infty{1\over n3^n}}-2\sum_{n=2}^\infty{1\over n3^n}+\color{blue}{{3}\sum_{n=3}^\infty{1\over n3^n}}=\left({1\over 3}-2+{3}\right)\sum_{n=1}^\infty{1\over n3^n}+{2\over 3}-\color{blue}{{3}\times\left({1\over 3}+{1\over 2\times3^2}\right)}=-{1\over 2}+{4\over 3}\sum_{n=1}^\infty{1\over n3^n}$$
|
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|
Probability the three points on a circle will be on the same semi-circle Three points are chosen at random on a circle. What is the probability that they are on the same semi circle?
If I have two portions $x$ and $y$, then $x+y= \pi r$...if the projected angles are $c_1$ and $c_2$. then it will imply that $c_1+c_2=\pi$...I have assumed uniform distribtuion so that $f(c_1)=\frac{\pi}{2}$...to calculate $P(c_1+c_2= \pi)$ I have integrated $c_2$ from $0$ to $\pi-c_1$ and $c_1$ from $0$ to $\pi$..but not arriving at the answer of $\frac 3 4$
|
Without loss of generality, we can assume the circumference of the circle to be equal to $1$. Cut the circle at the first point $A$ and spread it out as a line. Let the other two points $B$ and $C$ be located at distances of $x$ and $y$ from $A$. The midpoint of this line is $M$ and $0 \leq x,y \leq 1$.
Points $B$ and $C$ can be both located on same side of $M$ or on either side of $M$. If both points lie on the same side of $M$, then all $3$ points lie on the same semi-circle.
$$
P(\text{same side}) = 2 * \frac{1}{4} = \frac{1}{2}
$$
We are multiplying by 2 as the points can lie on both sides of $M$.
They can also be located on the either side of $M$ and in this case, the conditions for all 3 points to lie on the same semi-circle are
$$
x + 1 - y < 0.5 \Rightarrow y > x + 0.5 \text{ if } y > x\\
y + 1 - x < 0.5 \Rightarrow x > y + 0.5 \text{ if } x > y
$$
Representing these two regions graphically, we get the following area:
The area of the shaded region is $\frac{1}{4}$. Combining all results, we get the probability that $A$, $B$ and $C$ all lie on the same semi-circle is
$$\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$.
|
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|
Evaluate $\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$ Find $$\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$$ by hand. I'm aware Mathematics gives $\frac{\pi^2}{4}$, but I need to learn this without the aid of Mathematica. I tried using substitutions like $u= \tan \frac{x}{2}$ and some trig identities, but I still can't work this out.
|
I prove lemma that says that $$\int_0^\pi xf(\sin x)dx=\frac{\pi}2\int_0^\pi f(\sin x)dx$$
let $t=\pi-x$,so
\begin{align*}
\int_0^\pi xf(\sin x)dx&=\int_\pi^0(\pi-t)f(\sin(\pi-t))(-dt)=\int_0^\pi(\pi-t)f(\sin t)dt\\
&=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi tf(\sin t)dt
\end{align*}
so
$$\int_0^\pi xf(\sin x)dx=\dfrac{\pi}{2}\int_0^\pi f(\sin x)dx$$
Finally
\begin{align*}
\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx&=\dfrac{\pi}{2}\int_0^\pi\frac{ \sin x}{1+\cos ^2 x}dx=-\dfrac{\pi}{2}\int_0^{\pi}\frac{1}{1+\cos ^2 x}d\cos x\\
&=-\dfrac{\pi}{2}\arctan\cos x\Bigg|_0^\pi=\dfrac{\pi^2}{4}
\end{align*}
|
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|
Divisibility Of Positve Integers Suppose a,b and c are three positive integers which satisfy the condition that ($a$2+$b$2+$c$2) is divisible by $(a+b+c)$.
Prove that there exists infinitely many positive integers $n$ for which ($a$n+$b$n+$c$n) is also divisible by$(a+b+c)$.
|
We have that $(a^{2^n}+b^{2^n}+c^{2^n})$ is divisible by $(a+b+c)$ for each $n\geq 0$.
The proof is by induction on $n$. The base cases $n=0$ and $n=1$ are given to us; suppose that it is true for $n$ and $n-1$. Then $$2(a^{2^{n-1}}b^{2^{n-1}}+a^{2^{n-1}}c^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}) = (a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2 - (a^{2^n}+b^{2^n}+c^{2^n})$$ is also divisible by $(a+b+c)$.
We then have $$(a^{2^{n+1}}+b^{2^{n+1}} + c^{2^{n+1}}) = (a^{2^n}+b^{2^n}+c^{2^n})^2 - 2(a^{2^{n-1}}b^{2^{n-1}}+a^{2^{n-1}}c^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}})^2 + 4(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})(a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}),$$
and each term on the right hand side is divisible by $(a+b+c)$ by the above and the inductive hypothesis.
|
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|
Finding relatives of the series $\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}$. Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$
$$
\begin{align*}
\varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\
\varphi &=\frac{2}{1}+\sum_{k=0}^{\infty}\cdots&(2)\\
\varphi &=\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}&(3)\\
\varphi &=\frac{5}{3}+\sum_{k=0}^{\infty}\cdots&(4)\\
\varphi &=\frac{8}{5}+\sum_{k=0}^{\infty}\cdots&(5)\\
\varphi &=\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2(k+1))!}{((k+1)+1)!(k+1)!2^{4(k+1)+3}}&(6)\\
\vdots&\\
\end{align*}
$$
When looking at the leading terms of $(3)$ and $(6)$ $\;\frac{3}{2}$ and $\frac{13}{8}$ respectively, one is tempted to conjecture that there are similar formulas to fill the holes in the above table.
I'd like to know if such family of formulas exist.
Thanks.
EDIT:
Note that both formulas connect the Golden Ratio $\varphi$ to Catalan Numbers
$$
C_{k}=\frac{(2k)!}{(k+1)!k!}
$$
so for $(3)$ we have
$$
\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}
$$
and for $(6)$ we have
$$
\varphi =\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4(k+1)+3}}
$$
So, maybe this could be used, somehow, to find the other formulas.
|
Yes, you are right. There is a family. A little experimentation shows that,
$$\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}$$
$$\varphi =\frac{13}{2^3}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}$$
$$\varphi =\frac{207}{2^7}+\sum_{k=0}^{\infty}(-1)^{k+2}\frac{C_{k+2}}{2^{4k+11}}$$
$$\varphi =\frac{1657}{2^{10}}+\sum_{k=0}^{\infty}(-1)^{k+3}\frac{C_{k+3}}{2^{4k+15}}$$
$$\varphi =\frac{53019}{2^{15}}+\sum_{k=0}^{\infty}(-1)^{k+4}\frac{C_{k+4}}{2^{4k+19}}$$
and so on. Sorry, but the numerators of the fractions are not Fibonacci numbers. It seems hard to predict what they will be.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $ I can't figure out how to prove the following inequality:
$$
1/2 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2}
$$
Thanks
|
Since $\cos\frac{\pi}{7}$ is a root of the Chebyshev polynomial $U_6(x)$
$$U_6(x) = 64x^6-80x^4+24x^2-1$$
we have
$$\frac{1}{4\cos^2\frac{\pi}{7}}=16\cos^4\frac{\pi}{7}-20\cos^2\frac{\pi}{7}+6,$$
so:
$$4\sin^2\frac{\pi}{14}+\frac{1}{\cos^2\frac{\pi}{7}}=4-2\cos\frac{\pi}{7}-2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7},$$
or:
$$4\sin^2\frac{\pi}{14}+\frac{1}{4\cos^2\frac{\pi}{7}}=4-2\sum_{j=1}^{3}\cos\frac{\pi j}{7}=5-2\cdot\frac{\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}}=7-8\cos^2\frac{\pi}{7}=3-4\cos\frac{2\pi}{7}.$$
Now $\cos\frac{2\pi}{7}$ is a root of the third-degree polynomial
$$p(x)=U_6\left(\sqrt{\frac{x+1}{2}}\right)=8x^3+4x^2-4x-1,$$
so $3-4\cos\frac{2\pi}{7}$ is a root of
$$q(x)=-8\cdot p\left(\frac{3-x}{4}\right)=x^3-11x^2+31x-13.$$
Now it is easy to check that $q(1/2)=-1/8<0$ and $q(\xi)>0$, where $\xi$ is the smallest positive root of $11x^2-31x+13$. Since $3-4\cos\frac{4\pi}{7}$ and $3-4\cos\frac{6\pi}{7}$, the other roots of $q(x)$, are bigger than $3$, and all the roots of $q(x)$ are positive, we have found:
$$4\sin^2\frac{\pi}{14}+\frac{1}{4\cos^2\frac{\pi}{7}}=3-4\cos\frac{2\pi}{7}\in\left(\frac{1}{2},\frac{20}{39}\right)=I.$$
Since $q(x)$ concave over $I$, we can further improve this bound with a step of the Newton's method with starting point $x=1/2$ and a step of the secant method with endpoints $\frac{1}{2},\frac{20}{39}$:
$$3-4\cos\frac{2\pi}{7}\in\left(\frac{42}{83},\frac{63512}{125503}\right).$$
The difference between the upper and lower bound is now just $3.552\cdot 10^{-5}$, so the first four figures of the LHS are $0.506$.
|
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|
Find the sum of the series. I need to find the following sum:
$$\sum_{s=0}^{n+1}{(-1)}^{n-s}4^s\binom{n+s+1}{2s}$$
First I tried to simplify this:
$$\begin{split}
\sum_{s=0}^{n+1}{(-1)}^{n-s}4^s\binom{n+s+1}{2s}
&= {(-1)}^n\sum_{s=0}^{n+1}{(-1)}^{s}2^{2s}\binom{n+s+1}{2s} \\
&= \left[{(-1)}^{m-1}\sum_{s=0}^m{(-1)}^{s}x^{2s}\binom{m+s}{2s}\right](2)
\end{split}
$$
Now I reduced the problem to the following:
"Find generating function for the following sequence"
$$\sum_{s=0}^m{(-1)}^{s}x^{2s}\binom{m+s}{2s}$$
Does anyone have any ideas how to solve this problem? Because if you put it to the Wolfram|Alpha result is terryfing and I hope that generating function produced by wolfram is too generalized (for any values of x and m).
UPD: I put the wrong sequece to Wolfram|Alpha, here is the correct one.
So, Wolphram|Alpha says now, that:
$$\sum_{s=0}^m{(-1)}^{s}x^{2s}\binom{m+s}{2s} = \frac{2\cos\left((2m+1)\arcsin\left(\frac2x\right)\right)}{\sqrt{4-x^2}}$$
Unfortunately, it is undefined for $x=2$. While when we set $x=2$ for initial query (Sum[(-1)^s*2^(2s)*Binom(m+s,2s),{s,0,m}]) the answer is following:
$$\sum_{s=0}^m{(-1)}^{s}2^{2s}\binom{m+s}{2s} = {(-1)}^m(2m+1)$$
And I still wondering, how to prove that?
|
Suppose we seek to evaluate
$$\sum_{q=0}^{n+1} (-1)^{n-q} 4^q {n+q+1\choose 2q}
= (-1)^n \sum_{q=0}^{n+1} (-1)^{q} 4^q {n+q+1\choose n+1-q}.$$
We use the integral
$${n+q+1\choose n+1-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+q+1}}{z^{n-q+2}} \; dz.$$
This has the property that it is zero when $q\gt n+1$ so we may extend
$q$ to infinity to get
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\sum_{q\ge 0} (-1)^{q} 4^q (1+z)^q z^q\; dz.$$
This yields
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\frac{1}{1+4(1+z)z} \; dz
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\frac{1}{(2z+1)^2} \; dz.$$
Extracting the residue we get
$$(-1)^n \sum_{q=0}^{n+1} {n+1\choose n+1-q} \times
(-1)^q \times (q+1) \times 2^q
\\ = (-1)^n \sum_{q=0}^{n+1} {n+1\choose q} \times
(-1)^q \times (q+1) \times 2^q.$$
Now observe that
$$(x(1+x)^n)' = \sum_{r=0}^n {n\choose r} (r+1) x^r
= (1+x)^n + nx (1+x)^{n-1}.$$
This gives two components for the sum, the first is
$$(-1)^n (-1)^{n+1} = -1,$$
and the second is
$$(-1)^n \times (n+1) \times (-2) \times (-1)^{n}$$
for a final answer of
$$-1- 2(n+1) = -2n-3.$$
|
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|
Help with limit calculation Can anyone help me with this limit please:
I have been trying to solve this for 2 hours with no success:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$
|
We don't need to make all the detailed calculations as the highest exponent dominates when $n\to\infty$
$$1+4+7+...+(3n-2)=\frac n2\left(1+3n-2\right)=\frac{3n^2-n}2$$
$$\implies \left(1+4+7+...+(3n-2)\right)^2=\left(\frac{3n^2-n}2\right)^2=\frac{9n^4}4+O(n^3)$$
$$1^3+4^3+7^3+...+(3n-2)^3=\sum_{1\le r\le n}(3r-2)^3=\sum_{1\le r\le n}(27r^3-54r^2+36r-8)$$
$$=27\sum_{1\le r\le n}r^3-54\sum_{1\le r\le n}r^2+36\sum_{1\le r\le n}r-\sum_{1\le r\le n}8=27\left(\frac{n(n+1)}2\right)^2+O(n^3)=\frac{27n^4}4+O(n^3)$$
So, the limit$(F)$ will be $$\lim_{n\to\infty}\frac{\frac{9n^4}4+O(n^3)}{\frac{27n^4}4+O(n^3)}$$
Dividing the numerator & the denominator by $n^4,$
$$F=\lim_{n\to\infty}\frac{\frac{27}4+O\left(\frac1n\right)}{\frac94+O\left(\frac1n\right)}=\frac{\frac{27}4}{\frac94}$$
|
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|
Prove $\frac{c_n(a_1,\ldots,a_n)}{c_{n-1}(a_2,\ldots,a_n)}=a_1 + \frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_{n-1}+\frac{1}{a_n}}}}$ For $n>0$ and $a_1,...,a_n \in K$ let $c_n(a_1,...,a_n)$ be the determinant of the matrix
$$
\begin{pmatrix}
a_1 & 1 & 0 & \cdots & 0 \\
-1 & a_2 & \ddots & \ddots & \vdots \\
0 & \ddots & \ddots & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & 1 \\
0 & \cdots & 0 & -1 & a_n
\end{pmatrix}
$$
Show for $n \ge 2$ that following holds:
$$
\frac{c_n(a_1,...,a_n)}{c_{n-1}(a_2,...,a_n)}
=
a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_{n-1}+\frac{1}{a_n}}}}
$$
I want to show it using induction over n but I already fail at the initial step:
For $n = 2$ I have to show:
$$
\frac{c_2(a_1,a_2)}{c_{2-1}(a_2)}
=
a_1 + \frac{1}{a_2}
$$
which is
$$
\frac{a_1a_2 + 1}{a_2} \neq a_1 + \frac{1}{a_2}
$$
My also have no idea for the induction step where I have to show:
$$
\frac{c_{n+1}(a_1,...,a_{n+1})}{c_n(a_2,...,a_{n+1})}
=
a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}}
$$
So first I calculate $c_{n+1}(...)$ by developing after the first column which gives me:
$$
a_1 \cdot \det
\begin{pmatrix}
a_2 & 1 & & & & \\
-1 & a_3 & 1 & & & \\
& -1 & \ddots & & & \\
& & & \ddots & & \\
& & & & & 1 \\
& & & & -1 & a_{n+1}
\end{pmatrix}
+
\det
\begin{pmatrix}
1 & 0 & & & & \\
-1 & a_3 & 1 & & & \\
& -1 & a_4 & & & \\
& & & \ddots & & \\
& & & & & 0 \\
& & & & 0 & a_{n+1}
\end{pmatrix}
$$
which after developing after the first row gives me
$$
a_1 \cdot c_n(a_2,...,a_{n+1}) + c_{n-2}(a_3,...,a_{n+1}).
$$
Analog for $c_n(...)$:
$$
c_n(...) = a_2 c_{n-1}(...) + c_{n-3}(...)
$$
So I have
$$
\frac{c_{n+1}(...)}{c_n(...)} = \frac{a_1c_n(...) + c_{n-2}(...)}{a_2c_{n-1}(...) + c_{n-3}(...)}
$$
written in another way it is
$$
\frac{a_1c_n(...)}{c_n(...)} + \frac{c_{n-2}}{a_2c_{n-1}(...) + c_{n-3}(...)}
$$
I write it as
$$
a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{c_{n-3}(...)}{c_{n-2}(...)}}
$$
and then
$$
a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{1}{\frac{c_{n-2}(...)}{c_{n-3}(...)}}}
$$
so
$$
a_1 + \frac{1}{a_2 \cdot \left( a_3 + \cfrac{1}{a_4 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right) + \frac{1}{\left( a_4 + \cfrac{1}{a_5 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right)}}
$$
I dont know how to preced from here, any help or simpler solutions are appreciated!
|
but I already fail at the initial step
Not really. You have
$$\frac{c_2(a_1,a_2)}{c_1(a_2)} = \frac{a_1a_2+1}{a_2} = \frac{a_1a_2}{a_2} + \frac{1}{a_2} = a_1 + \frac{1}{a_2}.$$
In your induction step, you also have the necessary ingredients,
$$c_{n+1}(a_1,\dotsc,a_{n+1}) = a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1}),$$
you just have made a dimension error and thought it was $c_{n-2}(a_3,\dotsc,a_{n+1})$. The dimension of that matrix is $\bigl((n+1)-2\bigr)^2$, not $(n-2)^2$.
And then you rename, calling $b_k = a_{k+1}$ for $k = 1,\dotsc, n$, to get
$$\begin{align}
\frac{c_{n+1}(a_1,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})} &= \frac{a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\
&= a_1 + \frac{c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\
&= a_1 + \frac{1}{\frac{c_n(b_1,\dotsc,b_n)}{c_{n-1}(b_2,\dotsc,b_n)}}\\
&= a_1 + \cfrac{1}{b_1 + \cfrac{1}{b_2 + \cfrac{1}{\ddots + \cfrac{1}{b_{n-1}+ \frac{1}{b_n}}}}}\\
&= a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots + \cfrac{1}{a_{n}+ \frac{1}{a_{n+1}}}}}}
\end{align}$$
using the induction hypothesis on the sequence $b_1,\dotsc,b_n$.
|
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|
How find this postive integer $a$ such $a(x^2+y^2)=x^2y^2$ always have roots Find all postive integer numbers of $a$,such this equation
$$a(x^2+y^2)=x^2y^2,xy\neq0$$
always have integer roots $(x,y)$
my try: since
$$\dfrac{x^2y^2}{x^2+y^2}\in N$$
and I can't
Thank you
|
Take any two positive coprime integers $u$ and $v$, and a third positive integer $t$.
Let $q$ be the square-free part of $u^2 + v^2$: thus $u^2 + v^2 = qs^2$ for some positive integer $s$.
Let $a = u^2 v^2 t^2 q$.
Then the equation $a(x^2+y^2) = x^2y^2$ has the non-trivial solution $x = tsqu$ and $y = tsqv$ because
$a(x^2+y^2) = u^2v^2t^2q(tsq)^2 (u^2+v^2) = u^2v^2t^2(tsq)^2q^2s^2 = (tsq)^4u^2v^2 = x^2y^2.$
Conversely, if there is a non-trivial solution to this equation for some $a$, then $a$ must necessarily have the form given above.
To see this, let $z = \gcd(x,y)$ and write $x = zu$ and $y = zv$ for some coprime positive integers $u,v$. Then $a(u^2+v^2) = z^2u^2v^2$. Since $u^2v^2$ and $u^2+v^2$ cannot share a prime factor, $a = bu^2v^2$ for some positive integer $b$. Let $q$ be the square-free part of $u^2+v^2$ and write $u^2 + v^2 = qs^2$ for some positive integer $s$. Then $b q s^2 = z^2$ forces $q$ to divide $b$. So $b = qw$ and $w (qs)^2 = z^2$; hence $w = t^2$ is a perfect square. Putting everything together gives $a = u^2v^2t^2q$ as claimed.
|
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|
Singular Value Decomposition: Nonsquare matrix, third column of U My textbook says the singular value decomposition of the below matrix:
$A = \left( \begin{matrix} -2&1\\ 1&2 \\ 1 & 2 \end{matrix} \right)$
is given by:
$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \sqrt{10}&0\\0&\sqrt{5}\\0&0\end{matrix}\right)\left( \begin{matrix} \frac{1}{5}&\frac{2}{5}\\-\frac{2}{5}&1\end{matrix}\right)=U\Sigma V^T$
However, I don't see how to get the third column of U, not only because A only yields 2 singular values, but also because the third column is multiplied out anyway. How did my textbook come up with this result?
|
It seems like there are some typos in $V^T$, the result should be:
$$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \sqrt{10}&0\\0&\sqrt{5}\\0&0\end{matrix}\right)\left( \begin{matrix} \frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\-\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{matrix}\right)=U\Sigma V^T$$
If we need to step through the process, please give a yell.
|
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|
Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
Show that
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
What I have:
$P(1)$:
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
Replace n with 1
$$\frac{1}{1} \le 2 - \frac{1}{1}$$
Conclusion
$$1 \le 1$$
Prove:
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
P(K) Assume
$$\sum_{i=1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$$
$P(K) \implies P(k + 1)$
Performed Steps:
Working the LHS to match RHS
$$2 - \frac{1}{k} + \frac{1}{(k+1)^2}$$
Edit: Fixed error on regrouping
$$2 - \left[\frac{1}{k} - \frac{1}{(k+1)^2}\right]$$
Work the fractions
$$2 - \left[\frac{1}{k} \frac{(k+1)^2}{(k+1)^2} - \frac{1}{(k+1)^2} \frac{k}{k} \right]$$
$$2 - \left[\frac{(k+1)^2 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + 2k + 1 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + k + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1) + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1)}{k(k+1)^2} + \frac{1}{k(k+1)^2} \right]$$
$$2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}$$
EDIT: I fixed my mistake of my regrouping and signs; had completely missed the regrouping.
This is the final step I got to. I am hung on where to go from here. The answers given have been really helpful and I'm happy with them. I'd just like to know the mistake I made or next step I should take.
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
Thanks for the help
|
You want to prove that
$$P(k):\qquad \sum_{i=1}^k{1\over i^2}\leq 2-{1\over k}$$
implies
$$P(k+1):\qquad \sum_{i=1}^{k+1}{1\over i^2}\leq 2-{1\over k+1}\ .$$
Therefore we have to prove that
$$\left(2-{1\over k}\right)+{1\over (k+1)^2}\leq 2-{1\over k+1}\ ,$$
which is the same as
$${1\over (k+1)^2}\leq {1\over k}-{1\over k+1}\quad\left(={1\over k(k+1)}\right)\ .$$
But this is obvious.
|
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|
What is completing the square? Why is it called, completing the square? Is square metaphorical in this sense? How do you complete the square and what is it used for?
Thank you, regards.
|
Consider the quadratic term $x^2+2x-11$. The goal of completing the square is to write this term in a form where $x$ appears only once and the idea behind this is to use the binomial formula $(a+b)^2=a^2+2ab+b^2$ in opposite direction. So we set $a=x$ and then it looks like $2ab$ should be $2x$ and since $a=x$ we get $2xb=2x$ which gives us $b=1$. Then
$$(x+1)^2=x^2+2x+1^2=x^2+2x+1.$$
But wait a minute, we want $x^2+2x-11$. What do we do? We add the missing $1$ to complete the square:
\begin{align}
x^2+2x-11 &= x^2+2x+1-1-11 \\
&= (x+1)^2-1-11 \\
&= (x+1)^2-12.
\end{align}
|
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|
Find the signature of the quadratic form Very simple question but something doesn't make sense to me.
We are given a quadratic form (bilinear map but on the same vector twice):
$Q(v) = v^t *\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}*v$
and we are asked to find the signature....But this matrix isn't symmetric. Not only is it not symmetric, it's not diagonlizable. Is the signature defined well in this case?
|
Think I got it.
Suppose $u=\begin{pmatrix} x \\ y \\ z\end{pmatrix}$.
$Q(u)=\begin{pmatrix} x & y & z\end{pmatrix} \begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\1 & 1 & 0\end{pmatrix} \begin{pmatrix} x\\y\\z\end{pmatrix} = \begin{pmatrix}x+z \\ x+y+z \\y\end{pmatrix} \begin{pmatrix} x \\y\\z\end{pmatrix}=x^2+xz+xy+y^2+2yz$ which corresponds to the matrix:
$\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& 1 & 1\\ \frac{1}{2} &1 &0\end{pmatrix}$
Where the first column is x and row is x, second column and row is y, third column and row is z. And the signature of that matrix is (2,1,0) and that's the answer.
|
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Quadratic congruence relation Is there a general formula for solving quadratic congruence relation like $ax^2-bx+c = 0$ in $\mathbb{Z}_{n}$ where $n \in \mathbb{N}$?
For example, I am trying to solve $x^2-3x+2 = 0$ in $\mathbb{Z}_{200}$. What kind of tools I need to solve it?
|
There are techniques, but no general formula. In this case there is a method, based on the factorization $(x-1)(x-2)$.
For any $x$, the numbers $x-1$ and $x-2$ are relatively prime. Note that $200=2^3\cdot 5^2$, and work modulo $8$ and $25$ separately.
Since $x-1$ and $x-2$ are relatively prime, we have $(x-1)(x-2)\equiv 0\pmod{8}$ if and only if $x\equiv 1\pmod{8}$ or $x\equiv 2\pmod{8}$.
Similarly, we have $(x-1)(x-2)\equiv 0\pmod{25}$ if and only if $x\equiv 1\pmod{25}$ or $x\equiv 2\pmod{25}$.
That gives $4$ possibilities:
(i) $x\equiv 1\pmod{8}$, $x\equiv 1\pmod{25}$;
(i) $x\equiv 2\pmod{8}$, $x\equiv 2\pmod{25}$;
(iii) $x\equiv 1\pmod{8}$, $x\equiv 2\pmod{25}$;
(iv) $x\equiv 2\pmod{8}$, $x\equiv 1\pmod{25}$.
Using the Chinese Remainder Theorem, or inspection, we can then write down the $4$ solutions modulo $200$.
The solutions to (i) and (ii) are trivial. For (iii), the solution is (by inspection) $177$ (modulo $200$. We leave (iv) to you.
|
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conditions on $\{a_n\}$ that imply convergence of $\sum_{n=1}^{\infty} a_n$ (NBHM 2011) Question is :
For a sequence $\{a_n\}$ of positive terms, Pick out the cases which imply convergence of $\sum_{n=1}^{\infty} a_n$.
*
*$\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$
*$\sum_{n=1}^{\infty} n^2a_n^2<\infty$
*$\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$
For the first case $\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$ :
Suppose $a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n\in \mathbb{N}$ then we would have :
$n^{\frac{3}{2}}a_n > \frac{3}{2}.n^{\frac{3}{2}} (\frac{1}{n})^{\frac{3}{2}}=\frac{3}{2}$ but then $n^{\frac{3}{2}}a_n$ would not converge to $\frac{3}{2}$
So, for large $n$ we should have $$a_n \leq\frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}\Rightarrow \sum_{n=1}^{\infty}a_n \leq \sum_{n=1}^{\infty} \frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}=\frac{3}{2}\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$$
Right hand side converges and by comparison test $\sum_{n=1}^{\infty}a_n$ should converge.
For the second case $\sum_{n=1}^{\infty} n^2a_n^2<\infty$ :
Suppose that $na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ then we would have $$n^2a_n^2> \dfrac{1}{n}$$ but then it is given that $\sum_{n=1}^{\infty} n^2a_n^2$ which would imply that $\sum_{n=1}^{\infty}\dfrac{1}{n}$ converges which is a contradiction.
Thus we should have $na_n\leq \dfrac{1}{\sqrt{n}}$ i.e., $a_n\leq \frac{1}{n\sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}$ for all $n$ large.
So, We have $$\sum_{n=1}^{\infty}a_n\leq\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$
right hand side is convergent so by comparison test $\sum_{n=1}^{\infty}a_n$ converges.
For third case $\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$ :
we would have $\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n} <\lim_{n\rightarrow \infty} (\frac{n}{n+1})^2=1$
Thus, by ration test, $\sum_{n=1}^{\infty} a_n$ Converges absolutely and so is convergent.
My choice of bounds :
*
*$a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n$ in first case
*$na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ in second case
All these came to me just by choice and i would like to know if there is some sense behind this choice.
Please help me to clear this and make this solution a bit more clear.
Please do not give an alternative solution
until this problem is fully verified.
Thank you.
|
For (a) , if we take $$\{b_n\}=\frac{1}{n^{\frac{3}{2}}}$$, then $$\lim_{
n\to \infty}\frac{a_n}{b_n}=lim_{n\to \infty}n^{\frac{3}{2}}a_n=\frac{3}{2}$$ which is finite . Hence by limit comparison test $\{a_n\}$ converges.
For (b) since $$\sum({na_n})(\frac{1}{n})\le \{\sum n^2a_n^2\}^{\frac{1}{2}}\{ \sum\frac{1}{n^2}\}^{\frac{1}{2}}$$.
Hence $\{a_n\}$ is convergent.
|
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Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? And how? What method is required? Thanks.
|
$\int_1^\infty \frac{1}{x^3+x}dx$
$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}$
Therfore, $A+B=0, C=0, A=1$, then $B=-1$.
$\int \frac{1}{x^3+x}dx=\int \frac{dx}{x}-\int \frac{x dx}{x^2+1}=\ln x -\frac{1}{2}\ln (x^2+1)=\ln\frac{x}{\sqrt{x^2+1}}$
For $\int_1^\infty \frac{1}{x^3+x}dx=\lim_{x \to \infty}\ln\frac{x}{\sqrt{x^2+1}}-\ln\frac{1}{\sqrt{1^2+1}}=0-\ln\frac{1}{\sqrt{2}}<\infty.$
Integral is convergent, therefore the series too.
|
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How prove this $\prod\limits_{cyc}(a^3+2b+\frac{2}{a^2+1})\ge 64$ let $a,b,c>0$ and such
$$abc\ge 1$$
show that
$$\left(a^3+2b+\dfrac{2}{a^2+1}\right)\left(b^3+2c+\dfrac{2}{b^2+1}\right)\left(c^3+2a+\dfrac{2}{c^2+1}\right)\ge 64$$
my try:
$$\sum_{cyc}\ln{\left(a^3+2b+\dfrac{2}{a^2+1}\right)}\ge 6\ln{2}$$
Then I can't,Thank you
|
Hint: first prove that $a^3+\frac{2}{a^2+1} \geq 2a$
|
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Prove the given inequality $$\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $$
$$\implies \frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
Given if $A>B>C$ and $A+B+C=180^\circ$
Is that implication correct if not then please correct it otherwise try to solve the first inequality.
This is not the original problem, but this problem arose when I was solving another trigonometric equation.
Now if anybody prove or disprove the above inequality then that will also be the solution of my problem.
|
The original inequality is $\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $
And we have to prove this $\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$.
Now we have to interchange only the positions of $tan(B-C)$ and $sin^{2}B$. $sin^{2}B$ is always positive since it is a square. We need to care only about $tan(B-C)$.
If $tan(B-C)<0$ then $(B-C)>90^\circ \implies B>90^\circ+C$.
$C$ is a non-zero angle. Therefore $B>90^\circ$ and $A>B>90^\circ$.
Consider the fact that $A+B+C=180^\circ$. This means that the angles A, B , C form the angles of a triangle. But we know that there cannot exist two obtuse angles in a triangle. Therefore our supposition is wrong and hence $B-C<90^\circ$ which leads to $tan(B-C)>0$.
Now we can interchange $tan(B-C)$ and $sin^{2}B$ without changing the inequality sign. Hence we have the given inequality $$\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
|
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How to solve this trigonometric identity? I have tried it several times and I do not know what I am doing wrong. This is the identity:
$$\csc^2x - \csc x = \frac{\cot^2 x}{1 + \sin x}$$
Trying with RHS, I get:
$$\frac{\cos^2 x }{ \sin^2 x} + \frac{\cos^2 x }{ \sin x}$$
With LHS:
$$\frac{1 }{ \sin^2 x} - \frac{\sin x }{ \sin^2 x}$$
But I do not know how to continue after that. Am I missing something?
|
$$\frac{1}{sin^2x}-\frac{1}{sinx}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}-\frac{sinx}{sin^2x}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}(1-{sinx})=\frac{1}{sin^2x}(\frac{cos^2x}{1+sinx})$$
$$\frac{1}{sin^2x}(1-{sinx})(1+sinx)=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(1-{sin^2x})=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(cos^2x)=\frac{1}{sin^2x}{cos^2x}$$
And we're done.
|
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|
Determine appropriate $c$ and $x_0$ for Big-O proofs. "Prove that $f(x)$ is $O(x^2)$:"
$$f(x) = \frac{x^4+2x-7}{2x^2-x-1}$$
Let $c=10$ (addition of coefficients of the numerator less the addition of coefficients of the denominator), and $x_0 = 1$ (the lowest coefficient among all elements in the function).
$$\frac{x^4+2x-7}{2x^2-x-1} \leq \frac{x^4+2x^4-7x^4}{2x^4-x^4-x^4} \leq 10x^2$$
Am I going on the right track? For any $x \geq 1$, the second fraction becomes troublesome, as you then get a division by zero. I'm not sure how to approach these questions if someone could assist me.
|
Note that
$$\frac{x^4 + 2x -7}{2x^2 - x - 1} \leq \frac{x^4 + 2x^4 -7}{2x^2 - x - 1} \leq \frac{3x^4}{2x^2 - x - 1} \leq \frac{3x^4}{x^2} = 3x^2.$$
The last inequality holds true when
$$2x^2 - x - 1 \geq x^2, $$
that is, for all $x \geq \frac{1 + \sqrt{5}}{2}$.
|
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|
Sum of a power series $n x^n$ I would like to know:
How come that
$$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$
Why isn't it infinity?
|
Using the ratio test, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)x^{n+1}}{nx^n} = \dfrac{(n+1)x}{n}$. So, the series converges when $|x|<1$.
$$F(x) = \sum_{n=1}^\infty n x^n = x + 2x^2 + 3x^3 + ...$$
$$xF(x) = \sum_{n=1}^\infty n x^{n+1} = x^2 + 2x^3 + 3x^4 + ...$$
$$F(x) - xF(x) = x + x^2 + x^3 + x^4... = \sum_{n=1}^\infty x^n = \dfrac{x}{1 - x}$$
It's a geometric series, defined when $|x| < 1$.
$$F(x)(1 - x) = \dfrac{x}{1-x}$$
$$F(x) = \dfrac{x}{(1-x)^2}$$
|
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|
Determine $2$ missing digits for modulo $11$ An account number verification system works as follows:
All digits in a 10-digit account number all multiplied by following weights:
$$(6 \ 3 \ 7 \ 9 \ 10 \ 5 \ 8 \ 4 \ 2 \ 1)$$
Resulting numbers are summed and divided by $11$. The result must be $0$.
Example:
Account number: $1111111111$
Sum: $1 \cdot 6 + 1 \cdot 3 + 1 \cdot 7 + 1 \cdot 9 + 1 \cdot 10 + 1 \cdot 5 + 1 \cdot 8 + 1 \cdot 4 + 1 \cdot 2 + 1 \cdot 1$
Result: $55$
$55 \bmod{11} = 0$: Check OK
Goal:
I need to determine $2$ out of $10$ digits ($8$ are known) on positions $6$ and $7$.
Example:
$21942xy925$
where $x$ can only be $8$ or $9$ and $y$ can be any integer.
How can I calculate the correct values for $x$, $y$ without resorting to brute-force methods?
|
It amounts to solving $\ \color{#c00}8y \equiv -k - 5x\!\pmod{11}\,$ for $\,x =8\,$ or $\,9.\,$ and $\,k=$ sum from other digits.
Notice that $\,{\rm mod}\ 11\!:\,\ 1/\color{#c00}8 \equiv 12/{-}3 = -4,\,$ so $\, y \equiv 4k+20x\equiv 4k-2x \equiv \smash[b]{\underbrace{4k+6}_{\large x=8},\ \underbrace{4k+4}_{\large x=9}} $
The same technique works for other unknown digits.
|
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|
Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$ $a$,$b$,$c$ are non-negative real numbers such that $~x^{2}+y^{2}+z^{2}=1$
show that $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
|
By C-S we obtain:
$$\sum_{cyc}\left(\frac{1}{y^2+z^2}+\frac{1}{1-yz}\right)=\sum_{cyc}\left(\frac{(3x+y+z)^2}{(3x+y+z)^2(1-x^2)}+\frac{(y+z)^2}{(y+z)^2(1-yz)}\right)\geq$$
$$\geq\frac{25(x+y+z)^2}{\sum\limits_{cyc}(3x+y+z)^2(1-x^2)}+\frac{4(x+y+z)^2}{\sum\limits_{cyc}(y+z)^2(1-yz)}.$$
Thus, it remains to prove that
$$\frac{25(x+y+z)^2}{\sum\limits_{cyc}(3x+y+z)^2(1-x^2)}+\frac{4(x+y+z)^2}{\sum\limits_{cyc}(y+z)^2(1-yz)}\geq9.$$
The last inequality we can prove by the $uvw$'s technique.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$\frac{25u^2}{\sum\limits_{cyc}(3u+2x)^2(9u^2-6v^2-x^2)}+\frac{4u^2}{\sum\limits_{cyc}(3u-x)^2(9u^2-6v^2-yz)}\geq\frac{1}{9u^2-6v^2},$$
which is eighth degree, which says that it's a quadratic inequality of $w^3$
and the rest is smooth, but a lot of work.
The starting inequality is a known unsolved problem and you can be first, which ends a proof.
Good luck!
|
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Given $x^2 + y^2 + z^2 = 3$ prove that $x/\sqrt{x^2+y+z} + y/\sqrt{y^2+x+z} + z/\sqrt{z^2+x+z} \le \sqrt3$ Given $x^2 + y^2 + z^2 = 3$
Then prove that $${x\over\sqrt{x^2+y+z}} + {y\over\sqrt{y^2+x+z}} + {z\over\sqrt{z^2+x+y}} \le \sqrt 3$$
I tried using the Cauchy-Schwarz inequality but the inequality is coming in opposite direction.
|
Using the Cauchy-Schwarz inequality, we have
$$(1+1+1)(x^2+y^2+z^2)\ge (x+y+z)^2\Longrightarrow x^2+y^2+z^2\ge x+y+z$$
and using the Cauchy-Schwarz inequality, we have
$$(x^2+y+z)(1+y+z)\ge (x+y+z)^2\Longrightarrow \dfrac{\sqrt{1+y+z}}{x+y+z}\ge\dfrac{1}{\sqrt{x^2+y+z}}$$
so
$$\Longleftrightarrow\dfrac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z}\le\sqrt{3}$$
By the Cauchy-Schwarz inequality, we have
\begin{align*}&(x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+z+x})^2\le[x+y+z][(x+y+z)+2(xy+yz+xz)]\\
&\le(x+y+z)[x^2+y^2+z^2+2(xy+yz+xz)]=(x+y+z)^3
\end{align*}
so
$$\dfrac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z}\le\sqrt{x+y+z}$$
and
$$\sqrt{x+y+z}\le\sqrt{x^2+y^2+z^2}=\sqrt{3}$$
|
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|
Eigenvalues in orthogonal matrices Let $A \in M_n(\Bbb R)$.
How can I prove, that
1) if $ \forall {b \in \Bbb R^n}, b^{t}Ab>0$, then all eigenvalues $>0$.
2) if $A$ is orthogonal, then all eigenvalues are equal to $-1$ or $1$
|
The first part of the problem is well solved above, so I want to emphasize on the second part, which was partially solved.
An orthogonal transformation is either a rotation or a reflection. I will focus on 3D which has lots of practical use. Let us then assume that $A$ is an orthonormal matrix in $\mathbb{R}^3 \times \mathbb{R}^3$.
We find that its eigenvalues are either $1, \text{e}^{\pm i \theta}$ for a rotation or $\pm 1$ for a reflection.
*
*For a rotation:
We have the following sequence of equalities (since $\det A = 1$)
\begin{eqnarray*} \det(I -A) &=& \det(A) \det (I-A) \\ &=&
\det(A^T) \det(I-A) \\ &=& \det(A^T - I) \\ &=& \det(A - I)
\\ &=& -\det(I-A) \quad , \text{since 3 is odd}, \end{eqnarray*}
So $\det(I-A)=0$ and $\lambda_1=1$ is an eigenvalue of $A$.
Now, let $u_1$ the unit eigenvector of $\lambda_1$, so $A u_1 =
u_1$. We show that the matrix $A$ is a rotation of an angle
$\theta$ around this axis $u_1$. Let us form a new coordinate
system using $u_1, u_2, u_1 \times u_2$, where $u_2$ is a vector
orthogonal to $u_1$, so the new system is right handed (has
determinant = 1). The transformation between the old system $\{e_1,
e_2, e_3\}$ and the new system is given by a matrix $B$ with column
vectors $u_1, u_2$, and $u_3$. So we have that $B e_i = u_i$. Let
us call the coordinate transformation (similarity) matrix $C =
B^{-1}AB$. Then
\begin{eqnarray*}
C e_1 = B^{-1} A B e_1 = B^{-1} A u_1 = B^{-1} u_1 = e_1. \end{eqnarray*} The fact that $C e_1 = e_1$ means two things:
*
*The first column of $C$ is $(1,0,0)^T$.
*The first raw of $C$ is $(1,0,0)$.
Then $C$ is a matrix of the type
\begin{eqnarray*}
C = \left ( \begin{array}{ccc}
1 & 0 & 0 \\
0 & a & b \\
0 & c & d \\ \end{array} \right ) \end{eqnarray*} Since $A$ is orthogonal $C$ is orthogonal and so the vectors $(a,c)^T$ and
$(b,d)^T$ are orthogonal and since \begin{eqnarray*}
1 = \theta A = \det C = ad - bc \end{eqnarray*} we have that the minor matrix with entries $a,b,c,d$ is a rotation (orthogonal
with determinat 1). Rotations in 2D are of the form
\begin{eqnarray*}
\left (
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right )
=
\left (
\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}
\right ) \end{eqnarray*} then \begin{eqnarray*} B^{-1}A B = \left ( \begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta \end{array} \right ) \end{eqnarray*} But since the eigenvalues of $B^{-1}AB$ are the same
as those of $A$ we find the eigenvalues of this matrix and those
would be the eigenvalues of $A$.
For the eigenvalues of this matrix we have that
\begin{eqnarray*} \det ( C - \lambda I)= 0 \implies (\cos \theta
- \lambda)^2 + \sin^2 \theta = 0. \end{eqnarray*} That is,
\begin{eqnarray*} 1 -2 \lambda \cos \theta + \lambda^2 = 0
\end{eqnarray*} Then
\begin{eqnarray*} \lambda = \frac{2 \cos \theta \pm \sqrt{4 \cos^2
\theta - 4}}{2} = \cos \theta \pm i \sin \theta = \text{e}^{\pm i
\theta}.
\end{eqnarray*}
This is the interesting result by Euler where he claimed that all
rigid body rotations can be written as rotation about one vector
(here $u_1$ by some angle $\theta$).
*Eigenvalues of a reflection. If we repeat all the steps above for a
reflection we find that now
\begin{eqnarray*} B^{-1}A B = \left ( \begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos 2 \theta & \sin 2 \theta \\
0 & \sin 2 \theta & -\cos 2 \theta \end{array} \right ) \end{eqnarray*}
Now
\begin{eqnarray*}
\det ( C - \lambda I)= 0 \implies -(\cos
2 \theta - \lambda)(\cos 2 \theta + \lambda ) - \sin^2 2 \theta =
0.
\end{eqnarray*}
and then
\begin{eqnarray*} -\sin^2 2 \theta - \cos^2 2 \theta + \lambda^2 =0
\end{eqnarray*} and from here
\begin{eqnarray*} \lambda = \pm 1. \end{eqnarray*}
.
Please see the following reference which I used for this answer:
Orthogonal Matrices
|
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|
Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere.
Hints are also welcome.
|
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=a^2b+abc+c^2b\leq a^2b+2abc+c^2b=b(a+c)^2=4\cdot b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(a+b+c)^3=\frac{4}{27}(x+y+z)^3.$$
|
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|
Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also for $\sqrt {x}-\sqrt {y}=1 $)
|
Square both sides:
$$x + 2 \sqrt{xy} + y = 1.$$
Isolate the radical and square:
$$xy = \left(\frac{1-x-y}{2}\right)^2.$$
Rotate the axes by $\pi/4$ to new axes ${X,Y}$:
$$\frac{Y^2-X^2}{2} = \left(\frac{1 - \sqrt{2}Y}{2}\right)^2.$$
Simplfying gives:
$$X^2 = 2\sqrt{2}Y - 1,$$
which is the equation for a parabola.
|
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|
$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the derivatives: $f^{(n)}(x) = \dfrac{1}{2i} \cdot ((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x})$
Now, I use that: $$ (1+i)^n = {\sqrt{2}}^n \cdot \left(\cos\dfrac{n \pi}{4} + i \sin\dfrac{n \pi}{4}\right) \\(1 - i)^n = \sqrt{2}^n \cdot \left( \cos \dfrac{-n \pi}{4} + i \sin \dfrac{-n \pi}{4} \right)$$
Plugging that mess, I get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\left(\cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4}\right) e^{ix} - \left(\cos \dfrac{-n \pi}{4} + i \sin \dfrac{- n \pi}{4}\right) e^{-ix} \right)$$
But, $e^{ix} = \cos x + i \sin x$, and using Moivre's theorem, that makes: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\cos \left( x + \frac{n \pi}{4}\right) + i \sin \left( x + \frac{n \pi}{4}\right) - \left(\cos \left( - x - \frac{n \pi}{4}\right) + i \sin \left( -x -\frac{ n \pi}{4}\right)\right)\right)$$
and since $\cos$ is an even function, and $\sin$ is odd, we get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot 2i \sin \left(x + \dfrac{n \pi}{4}\right)$$
Simplifying, the answer would be $f^{(n)}(x) = e^x \cdot \sqrt{2}^n \cdot \sin\left(x + \dfrac{n \pi}{4}\right)$. I'm almost positive that this is it, but I just want to be sure. Thank you in advance!
|
There is a nice point that tells $$D^n\{e^{kx}f(x)\}=e^{kx}(D+k)^nf(x)$$ It can be proved by an inductive approach on $n$. By using it we get: $$D\left(e^x\sin x\right)=e^x(D+1)\sin x=e^x(\cos x+\sin x)=\sqrt{2}e^x\sin(x+\pi/4)\\D^2\left(e^x\sin x\right)=D(\sqrt{2}e^x\sin(x+\pi/4))=\sqrt{2}e^x(D+1)\sin(x+\pi/4)=\sqrt{2}e^x[\cos(x+\pi/4)+\sin(x+\pi/4)]=(\sqrt{2})^2e^x\sin(x+2\pi/4)$$ So we can guess that $D^n(e^x\sin x)=(\sqrt{2})^ne^x\sin(x+n\pi/4)$. What is remained is to prove it by induction on $n$. If $D^k(e^x\sin x)=(\sqrt{2})^ke^x\sin(x+k\pi/4)$ then $$D^{k+1}(e^x\sin x)=D[(\sqrt{2})^ke^x\sin(x+k\pi/4)]=(\sqrt{2})^ke^x(D+1)\sin(x+k\pi/4)=\cdots=(\sqrt{2})^{k+1}e^x\sin(x+(k+1)\pi/4)$$
|
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|
Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$. I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is?
Problem: Let $F$ be a field with $2^n$ elements, with $n$ odd. Show that for $a,b \in F$ that $a^2+ab+b^2=0$ implies that $a=0$ and $b=0$.
Proof:
Suppose $a,b \in F$ and $a^2+ab+b^2=0$.
$\implies a^2+2ab+b^2 = ab$
$\implies \frac{2^n}{2}(a^2+2ab+b^2) = \frac{2^n}{2}ab$
$\implies \frac{2^n}{2}a^2+ 2^nab+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$
$\implies \frac{2^n}{2}a^2+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$ (since F is a group under addition, then every element to the $|F|$ multiple is the identity thus $2^n(ab) = 0$)
$\implies \frac{2^n}{2}(a^2+b^2) = \frac{2^n}{n}ab$
$\implies a^2+b^2 = ab$
$\implies a^2-ab+b^2 = 0 = a^2+ab+b^2$
$\implies -ab = ab \implies 2ab=0 \implies ab=0$.
Thus, $a=0$ or $b=0$. However, if just one of them is zero, then so is the other ($a=0 \implies a^2+ab+b^2 = 0 \implies b^2 = 0 \implies b=0$). Thus, $a=0$ and $b=0$.
QED
Anyways, if there is something wrong with this proof, could someone give me a subtle hint perhaps? I've been stuck on this seemingly simple problem for awhile now.
|
Hint $\ $ You cannot divide by $\,2\,$ since $\,2 = 0\,$ in $\,\Bbb F_{2^n}.\,$ Instead, notice that
$\, 0 = (a-b)(a^2+ab+b^2) = a^3-b^3.\,$ If $\,a\,$ or $\,b\neq 0,\,$ wlog. $\,b\ne 0,\,$ then $\,c^3 = 1\,$ for $\,c = a/b.\,$ If $\,c \ne 1\,$ then $\,c\,$ has order $= 3,\,$ so Lagrange's Theorem $\Rightarrow 3\,$ divides $\,|\Bbb F_{2^n}^{*}| = 2^n-1,\,$ contra $\,{\rm mod}\ 3\!:\ 2^n\! =(-1)^{2k+1}\!\equiv -1.\,$ So $\,c = 1,\,$ so $\,a\!=\!b,\,$ so $\,a^2\!+\!ab\!+\!b^2\!=3b^2 = 0,\,$ so $\,b = 0.$
|
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|
Is there a shortcut for taking this matrix to a certain power? One can take a diagonal matrix to a certain power by just taking diagonal elements that power. There is a similar polynomial-time (in respect to the matrix dimensions) shortcut for triangular matrices.
Is there a shortcut for exponentiating a square matrix of the form
$$
\left( \begin{array}{cccc}
a & b & c & d \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \end{array} \right)
$$
, where $a$, $b$, $c$, and $d$ are constants?
|
By using Cayley Hamilton Theorem. If you are doing the calculations by hand, then the problem is simple if the characteristic polynomial has simple roots.
The general approach is to divide $x^n$ by $x^4-a x^3 - b x^2 - c x - d$ find the remainder. Thus suppose
$$
x^n = Q(x) (x^4-a x^3 - b x^2 - c x - d) + { \alpha x^3 + \beta x^2 + \gamma x + \delta } \tag 1
$$
Here we do not need the quotient $Q(x)$ but just the remainder $\alpha x^3 + \beta x^2 + \gamma x + \delta $
Then, replacing $x$ by $A$ we have
$$
A^n = Q(A) (A^4-a A^3 - b A^2 - c A- d I) + { \alpha A^3 + \beta A^2 + \gamma A + \delta I } \tag 2
$$
In (2) we have $$(A^4-a A^3 - b A^2 - c A- d I) =0 ~\text{Verify this by direct calculation!}$$
so
$$
A^n = { \alpha A^3 + \beta A^2 + \gamma A + \delta I }$$
One can easily calculate
$$A^2=\pmatrix{b+a^2&c+a\,b&d+a\,c&a\,d\cr a&b&c&d\cr 1&0&0&0\cr 0&1&
0&0\cr }$$
$$A^3=\pmatrix{c+2\,a\,b+a^3&d+a\,c+b^2+a^2\,b&a\,d+b\,c+a^2\,c&
\left(b+a^2\right)\,d\cr b+a^2&c+a\,b&d+a\,c&a\,d\cr a&b&c&d\cr 1&0&
0&0\cr }$$
The key to the answer is finding the remainder. One can use recursion (synthetic division) or remainder theorem to find the remainder.
Here is a simple example on finding the remainder when the characteristic equation has simple roots.
Example: Let
$$a=2 , b=1 , c=-2 , d=0 $$
Then the roots of $x^4-a x^3 - b x^2 - c x - d$ are $x=0$, $x=1$, $x=-1$ and $x=-2$.
Now let
$$x^n = Q(x) (x^4-a x^3 - b x^2 - c x - d) + \alpha x^3 + \beta x^2 + \gamma x + \delta$$
We substitute different roots we got as
Set $x=0$
$$ 0 = \delta \tag 3$$
Set $x=1$:
$$
1 = \alpha+\beta+\gamma+\delta \tag 4$$
Set $x=-1$
$$
(-1)^n = -\alpha+\beta-\gamma+\delta \tag 5$$
Set $x = 2$
$$
2^n = 8 \alpha + 4 \beta + 2 \gamma + \delta \tag6$$
We can solve to get
$$ \alpha={{2^{n}-\left(-1\right)^{n}-3}\over{6}} , \beta={{
\left(-1\right)^{n}+1}\over{2}} , \gamma=-{{2^{n}+2\,\left(-1\right)
^{n}-6}\over{6}} , \delta=0 $$
So we have
$$
A^n =
{{2^{n}-\left(-1\right)^{n}-3}\over{6}} A^3 +
{{
\left(-1\right)^{n}+1}\over{2}} A^2 + {{2^{n}+2\,\left(-1\right)
^{n}-6}\over{6}} A$$
|
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|
Evaluate of number represented by the infinite series $\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$. Evaluate of number represented by the infinite series
$$\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$$
|
$n > 0:$
Let $x = \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}}$
Assuming convergence:
$$\begin{align*} x = \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}} & \implies x^2 = n + \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}} = n + x \\ & \implies x^2 - x - n = 0\end{align*}$$
This is a simple quadratic to solve (take the positive root as $n > 0$).
|
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|
How to prove this inequality $\sqrt{\prod\limits_{cyc}(a+\sqrt{\frac{bcd}{a}})}+2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd$ Let $a,b,c,d>0$. Show that
$$\sqrt{\left(a+\sqrt{\dfrac{bcd}{a}}\right)\left(b+\sqrt{\dfrac{acd}{b}}\right)\left(c+\sqrt{\dfrac{abd}{c}}\right)\left(d+\sqrt{\dfrac{abc}{d}}\right)}+2\sqrt{abcd}\ge ab+bc+cd+da+ac+bd$$
My idea: make use of
$$a+\sqrt{\dfrac{bcd}{a}}=a+\dfrac{\sqrt{abcd}}{a}\ge 2\sqrt{\sqrt{abcd}}$$
It is said this inequality can be proved using AM-GM( Cauchy-Schwarz) inequality. But I can't.
Thank you for you help in proving it.
|
WLOG, assume $abcd = 1$. The desired inequality is written as
$$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} + 2 \ge ab + bc + cd + da + ac + bd.$$
Since $ab + bc + cd + da + ac + bd \ge 6$, it suffices to prove that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1) - (ab + bc + cd + da + ac + bd - 2)^2 \ge 0.$$
Using $d = \frac{1}{abc}$, we have
$$\mathrm{LHS} = \frac{(ab - 1)^2(bc-1)^2(ca-1)^2}{a^2b^2c^2}.$$
We are done.
|
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|
Why these integrals are evaluated differently? $\cos(x)/(a-b\cos(x))$ In attempt to solve electrostatics problem I came up to this integral that I am trying to integrate:
$$\int_0^{2\pi}\frac{\cos(x)}{a-b\cos(x)} \, dx$$ where $a>b$ and both are real numbers.
For the domain $[0, 2\pi]$ this function is symmetric at $\pi$ (Plot: 1)
So it is expected that:
$$\int_0^{2\pi}\frac{\cos(x)}{a-b\cos(x)}dx = 2\int_0^{\pi}\frac{\cos(x)}{a-b\cos(x)}dx$$
I could not integrate it by hand (maybe some of you guys can?). But Wolfram Mathematica gives weird answers for both of these integrals - however if evaluated they lead to same answer!
This is what Mathematica gives me:
$$\int_0^{2 \pi } \frac{\cos (x)}{a-b \cos (x)} \, dx = \frac{2 \pi \left(a \left(\sqrt{\frac{a+b}{a-b}}-1\right)-b\right)}{b (a+b)}$$
$$\int_0^{\pi } \frac{\cos (x)}{a-b \cos (x)} \, dx = -\frac{\frac{a \log \left(\frac{a+b}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}-\frac{a \log \left(-\frac{a+b}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}+\pi }{b}$$
If I add these logarithms it gives log(-1) which is imaginary, how come these integrals are the same?
|
These integrals are very similar to the ones for which Euler introduced the tangent half-angle substitution, sometimes erroneously called the "Weierstrass substitution" (I hope the quotation marks are intimidating . . . . .)
\begin{align}
t & = \tan\frac x 2 \\[8pt]
2\arctan(t) & = x \\[8pt]
\frac{2\,dt}{1+t^2} & = dx \\[8pt]
\cos x & = \cos(2\arctan t) = 2\cos^2(\arctan t) - 1 \\[8pt]
& = 2\left(\frac{1}{\sqrt{1+t^2}}\right)^2 - 1 = \frac{1-t^2}{1+t^2}
\end{align}
(and $\sin x$ can be found similarly, but we don't need it here, and neither did Euler in the integrals he did this way that I am aware of).
So we have
$$
\begin{align}
\int_0^{2\pi} \frac{\cos x}{a-b\cos x}\,dx & = \int_{-\infty}^\infty \frac{\frac{1-t^2}{1+t^2}}{a-b\frac{1-t^2}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} \\[8pt]
& = \int_{-\infty}^\infty \frac{1-t^2}{a(1+t^2)^2- b(1+t^2)(1-t^2)} 2\,dt
\end{align}
$$
One of the factors of the denominator is $1+t^2$. Another depends on $a$ and $b$.
Now go on to partial fractions. If $a>b>0$, you'll have two irreducible quadratic factors in the denominators, so you'll get some arctangents and maybe also some logarithms.
|
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|
Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin{C}$
These are my steps for doing this:
$\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \sin{\theta} $
Let $\mathbf{a} = \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix}$ and let $\mathbf{b} = \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix}$
$\therefore \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix} = (5\sqrt{3})(15)\sin{\theta} $
$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$
If I substitute these values into the general formula:
area $= \frac{1}{2}ab\sin{C}$
I get:
area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$
$\therefore$ area $= -\dfrac{75}{2}$
However this isn't right, the area should be $\dfrac{75}{\sqrt{2}}$
I feel I'm missing something really obvious but I can't spot it, can anyone help?
Thank you.
|
Alternative solution
$a = |OA| = \sqrt{1^2 + 5^2 + 7^2} = \sqrt{75} = 3 \sqrt{5}$
$b = |AB| = \sqrt{(10-1)^2 + (10--5)^2 + (5--7)^2} = \sqrt{450} = 15\sqrt{2}$
$c = |BO| = \sqrt{10^2+10^2+5^2} = 15.$
Now you can calculate the semiperimeter $s$ which is just $\frac{1}{2} \left(3 \sqrt{5} + 15 \sqrt2 + 15 \right)$, and use Heron's formula to find the area:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
|
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|
Show that $e^x > 1 + x + x^2/2! + \cdots + x^n/n!$ for $n \geq 0$, $x > 0$ by induction Show that if $n \geq 0$ and $x>0$, then
$$ e^x > 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}.$$
Not sure where to get started with this induction proof.
|
Base case: $e^x > 1$ for $x > 0$.
Induction: suppose we are given $k$ such that for all $x > 0$
$$
e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!}
$$
Change variables to $t$ and integrate both sides:
$$
\int_0^x e^t \; dt > \int_0^x \left( 1 + t + \frac{t^2}{2!} + \cdots + \frac{t^k}{k!} \right) \; dt
$$
$$
e^x - 1 > x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$$
$$
e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$$
Note that if $f > g$ on $(a,b)$ and both are integrable, then $\int_a^b f > \int_a^b g$ (not just $\ge$).
|
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|
Alternative ways to solve $\frac{\sin{\theta}+1}{\cos{\theta}}\leq 1$? The inequality is:
$$ \frac{\sin{\theta}+1}{\cos{\theta}}\leq 1\quad \text{ with } \cos{\theta}\neq0 \land 0\leq \theta\lt 2\pi$$
I've tried splitting it up into cases of $\theta$ that make $\cos{\theta}$ positive so: $0\leq\theta<\frac{\pi}{2}$ and $\frac{3\pi}{2} \lt\theta\lt 2\pi$ thus allowing the inequality to simply to:
$$ \sin{\theta}-\cos{\theta}\leq -1 $$
which after using the fact that $\sin{\theta}-\cos{\theta}=\sqrt{2}\sin{(\theta-\frac{\pi}{4})}$ and thus solving this inequality$$\sqrt{2}\sin{(\theta-\frac{\pi}{4})}\leq -1 \text{ for } 0\leq\theta<\frac{\pi}{2} \text{ or }\frac{3\pi}{2} \lt\theta\lt 2\pi $$
and this inequality
$$ \sqrt{2}\sin{(\theta-\frac{\pi}{4})}\geq -1 \text{ for } \frac{\pi}{2}\lt\theta<\frac{3\pi}{2} $$
which give the solutions:
$$ \theta=0 \text{ and } \frac{\pi }{2}<\theta<\frac{3 \pi }{2}\text{ and } \frac{3 \pi }{2}<\theta<2 \pi $$
I'm just curious if there's another way to do it that isn't quite so case based since it seems that I'm just solving it by plugging in values and testing. So I'm wondering if there's a better more insightful or efficient way to to this?
|
\begin{align}
1 + \sin \theta &= ( \cos (\theta /2) + \sin (\theta /2))^2 \\
\cos \theta &= ( \cos ^2 (\theta /2) - \sin ^2 (\theta /2) \\
\frac{1 + \sin \theta }{\cos \theta } &= \frac{ \cos (\theta /2) + \sin (\theta /2) }{ \cos (\theta /2) - \sin (\theta /2)} = \frac{ 1+ \tan (\theta /2)}{1 - \tan (\theta /2)}
\end{align}
Therefore the given inequation reduces to $\frac{2 \tan (\theta /2)}{1 - \tan (\theta /2)} \leq 0$ whose solution is $\tan (\theta /2) \leq 0$ or $\tan (\theta /2) \geq 1$.
|
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|
solving $\int x^7\sqrt{3+2x^4}dx$ I'm trying to solve $\int x^7\sqrt{3+2x^4}dx$
All I have so far is
Let $u$ = $3+2x^4$
$du$ = $8x^3$ $dx$
$\frac{du}{8x^3}$ = $dx$
Therefore,
$\int x^7\sqrt{u}$ $\frac{du}{8x^3}$
$\frac{1}{8}$$\int x^4\sqrt{u}$ ${du}$
Since there is still a $x$ variable in the integral, I'm not sure where to go from here. Any ideas?
|
Given $\int x^7 \sqrt{3+2x^4} dx$. Assume that $3+2x^4= u$ then $8x^3 dx=du$. Hence the givenm integral becomes $\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}$ which is \begin{align*}
&\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}\\
=&\frac{1}{16}\int (u^{3/2}-3u^{1/2})du\\
=&\frac{1}{16}(\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2})+c\\
=&\frac{1}{16}(\frac{2}{5}(3+2x^4)^{5/2}-2(3+2x^4)^{3/2})+c
\end{align*}
where $c$ is constant of integration.
|
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|
Find the surface area obtained by rotating $y=1+3x^2$ from $x=0$ to $x=2$ about the y-axis. Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.
Having trouble evaluating the integral:
Solved for $x$:
*
*$x=0, y=1$
*$x=2, y=13$
$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\sqrt\frac{y-1}3'}^2\,dy$$
I got stuck at 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+(1/12)+(1/(y-1))}
any help would be great thanks
|
Note that $((\sqrt{\frac{y-1}{3}})')^2=(\frac{1}{2\sqrt{\frac{y-1}{3}}} (1/3))^2=\frac{1}{4(y-1)}$
$$=\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\frac{1}{4(y-1)}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y-1}3+\frac{1}{12}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y}3-\frac{1}{4}}dy
\\= 6\pi\{(\frac{y}3-\frac{1}{4})^\frac{3}{2}\}_1^{13}$$
|
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|
Trigonometric identity for $a\sin 2x + b\sin(x+\alpha )$ The following is a known trigonometric identity,
$$a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,$$
where
$$c=\sqrt {a^2+b^2+2ab\cos \alpha },\,$$
and
$$\beta =\arctan \left(\frac {b\sin \alpha }{a+b\cos \alpha }\right) + \begin{cases} 0 & \text{if } a+b \cos \alpha \geq 0,\\
\pi & \text{if } a+b\cos \alpha <0.\end{cases}$$
Is there an identity for the following ?
$$a\sin 2x + b\sin(x+\alpha )$$
Alternatively, can you find the values of $x$ that satisfy the following equation?
$$a\sin 2x - b\sin(x+\alpha )=0$$
|
As for the last question:
\begin{align*}
a\sin2x-b\sin(x+\alpha)&=0 \\
2a\sin x\cos x - b(\sin x\cos\alpha + \cos x\sin\alpha)&=0\\
-b\cos\alpha\sin x &= \cos x \cdot(b\sin\alpha-2a\sin x) \\
\frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} &= \cos x \\
\bigg( \frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} \bigg)^2 &= \cos^2 x = 1-\sin^2x
\end{align*}
gives you an equation of degree $4$ in $\sin x$ that you can try to solve your favorite way. (You'll need to check any solutions against the original equation, since squaring both sides will introduce extraneous solutions.)
|
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|
Arc length of parametric equations Find the arc length of the parametrized path $x(t) = t^2/2$ and $y(t) = t^3/3$ for $1 \le t \le 3$.
I try to do the standard integral of $\sqrt{\mathrm dx^2 + \mathrm dy^2}$, but get stuck when I have to integrate $\sqrt{t^2 + t^4}$.
Any ideas?
|
$$
\int\mathrm{d}t \sqrt{t^2+t^4} = \int\mathrm{d}t\ \ t\sqrt{1+t^2}
$$
for positive t. Now let $\sqrt{1+t^2}=y-t$:
$$
1+t^2=y^2+t^2-2yt
$$
$$
t=\frac{y^2-1}{2y}
$$
$$
\mathrm{d}t=\mathrm{d}y \frac{y^2+1}{2y^2}
$$
and
$$
\sqrt{1+t^2}=\frac{y^2+1}{2y}.
$$
Substituting yields:
$$
\int \mathrm{d}y \frac{y^2+1}{2y^2} \frac{y^2-1}{2y} \frac{y^2+1}{2y} = \frac{1}{8}\int \mathrm{d}y \frac{(y^4-1)(y^2+1)}{y^4}= \frac{1}{8}\int \mathrm{d}y \frac{y^6+y^4-y^2-1}{y^4}=\frac{1}{8}\int \mathrm{d}y\left( y^2+1-y^{-2}-y^{-4}\right)
$$
which is no big deal!
|
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|
Adding fractions is not at all obvious Why does $\frac{5}{4} + \frac{2}{3}$ need to be rewritten as $\frac{15}{12} + \frac{8}{12}$ to be added? It's not obvious.
I'm looking towards the fact that any integer can be rewritten as $x=qy$ but these work for rational numbers as well.
Can anyone clarify why exactly fraction addition works only when you find a common denominator?
|
First, we have to define: what is a fraction? For now, we will just say it is an ordered pair $(a, b)$, where $b \ne 0$. We identify $(a, 1)$ with the integer $a$. (The notation doesn't matter, I just chose this so that we don't accidentally use identities we already know)
Here is what we want:
*
*$(a, 1) + (b, 1) = (a + b, 1)$
*$(a, 1)(b, 1) = (ab, 1)$
*$(a, 1)(1, b) = (a, b)$
*$(ac, bc) = (a, b)$
*Multiplication to distribute over addition.
*Multiplication and addition to be distributive and commutative.
This is sufficient to define everything! First, we show that $(a, 1)(1, a) = (1, 1)$:
$$(a, 1)(1, a) = (a, a) = (a1, a1) = (1, 1)$$
Next, we show $(1, a)(1, b) = (1, ab)$:
$$(ab, 1) \cdot (1, a) \cdot (1, b) = (ab, a) \cdot (1, b) = (b, 1) \cdot (1, b) = (1, 1)$$
Because $(ab, 1) \cdot \left( (1, a) \cdot (1, b) \right) = (1, 1)$, we know that $(1, a) \cdot (1, b) = (1, ab)$.
So now we want to find $(a, b) \cdot (c, d)$. We factor:
$$(a, 1) \cdot (1, b) \cdot (c, 1) \cdot (1, d) = (ac, 1) \cdot (1, bd) = (ac, bd)$$
Now we move on to addition:
$$ \begin{align*}
(a, b) + (c, d) &= (a, 1)(1, b) + (c, 1)(1, d) \\
&= (a, 1)(d, bd) + (c, 1)(b, bd) \\
&= (a, 1)(d, 1)(1, bd) + (c, 1)(b, 1)(1, bd) \\
&= (ad, 1)(1, bd) + (bc, 1)(1, bd) \\
&= [ (ad, 1) + (bc, 1) ] (1, bd) \\
&= (ad + bc, 1)(1, bd) \\
&= (ad + bc, bd)
\end{align*} $$
So, just starting from that bullet list, we have determined that $(a, b) + (c, d)$ must equal $(ad + bc, bd)$. If we change our notation, this gives us what you want.
|
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|
How to solve this trig problem? $\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$ Basic trig problem my brother ask me, but I don't know how to do it:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$
|
Since $\sec(x) = 1/\cos(x)$, let's first consider
$$\cos\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right].$$
Using the addition formula for $\cos$, this is equal to
$$\cos(\sin^{-1}(-5/13)) \cos(\tan^{-1}(4/3)) + \sin(\sin^{-1}(-5/13)) \sin(\tan^{-1}(4/3))$$
Now let's work term-by-term.
The key is to recognise that the arguments of the inverse trig functions are part of Pythagorean triples.
First, $\cos(\sin^{-1}(-5/13))$.
If we write $u = \sin^{-1}(-5/13)$, then $\sin u = -5/13$.
These numbers might remind you of the Pythagorean triple $(5,12,13)$, so if
$$-\sin u = \sin(-u) = \frac{\text{OPP}}{\text{HYP}} = \frac{5}{13},$$
then we must have
$$\cos(-u) = \frac{\text{ADJ}}{\text{HYP}} = \frac{12}{13}.$$
Further, since $\cos(-u) = \cos(u)$ for all real $u$, so our first term is $12/13$.
A similar principle works for our terms with $\tan^{-1}(4/3)$, where we use the Pythagorean triple $(3,4,5)$. Write $w = \tan^{-1}(4/3)$. If $\text{OPP}=4$, $\text{ADJ}=3$ and $\text{HYP} = 5$, then
$$\sin(w) = \frac{4}{5} \text{ and } \cos(w) = \frac{3}{5}.$$
Finally, the $\sin(\sin^{-1}(-5/13))$ term simplifies to $-5/13$ (strictly speaking, this depends on our choice of domain for $\sin^{-1}$).
Plugging this all in, we have
$$\cos\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right]
= \frac{12}{13} \cdot \frac{3}{5} + \frac{-5}{13} \cdot \frac{4}{5}
= \frac{16}{65}.
$$
Then we just flip this fraction to get what you actually want:
$$\sec\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right] = \frac{65}{16}.$$
|
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|
Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
|
Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.
You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$
Now, from, $AM\ge GM$, we have $$\frac {a^4+
a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\right)^{1/4}=a^2bc\tag 1$$
Similarly, $$\frac {a^4+
b^4+b^4+c^4}4\ge \left(a^4\cdot b^4\cdot b^4\cdot c^4\right)^{1/4}=ab^2c\tag 2$$ and also, $$\frac {a^4+
b^4+c^4+c^4}4\ge \left(a^4\cdot b^4\cdot c^4\cdot c^4\right)^{1/4}=abc^2\tag 3$$
Now, summing up $(1),(2),(3)$, we have, $a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$, that is $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c$$
|
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|
modular arithmetic proof Suppose $x$, $y$, and $z$ are integers and $x= 3y^2 -z^2$. Prove that $x\not\equiv1\mod4$.
My thoughts: So I am not sure the route that can prove this. I am trying to just use the simple stuff to set me up. I started with saying if $x\equiv1\mod4$ you would have $4|x-1$ and there will exist an integer, say $t$, where $x-1 = 4t$. I then solved for $x$ just because the other equation was in terms of $x$. So from that we would have $x=4t+1$. Not sure if I am on the right track here or what I can do with what I have.
|
The base is small enough for this problem that we can use proof by exhaustion by letting $y,z \equiv 0,1,2,3 \pmod 4$. And in fact, because $0^2 \equiv 2^2 \pmod 4$ and $1^2 \equiv 3^2 \pmod 4$, you only have to check $4$ different cases:
*
*$y \equiv 0, z \equiv 0 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 0^2 - 0^2 \equiv 0$
*$y \equiv 0, z \equiv 1 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 0^2 - 1^2 \equiv 3$
*$y \equiv 1, z \equiv 0 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 1^2 - 0^2 \equiv 3$
*$y \equiv 1, z \equiv 1 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 1^2 - 1^2 \equiv 2$
And none of these are $1$.
|
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|
Finding intersection points of 2 functions. My method is incomplete. These are the 2 functions :
$y = x^{4}-2x^{2}+1$
$y = 1-x^{2} $
Here's how I solved It :
$x^{4}-2x^{2}+1 = 1-x^{2}$
$x^{4}-x^{2} = 0$
$x^2(x^2-1)=0$
$x^2-1=0$
$x=\pm \sqrt{1} $
Value of $y$ when $x=1$
$y=1-x^2\\y=1-1\\y=0$
Value of $y$ when $x=(-1)$
$y=1-x^2\\y=1-(-1)^2\\y=1-1\\y=0$
So the intersection points of the 2 functions are $(1,0)$ and $(-1,0)$.
The problem
The problem is when I used a graphing calculator to find the intersection points of the above 2 functions it gave me 3 results instead of 2. They were :
$(-1,0),(0,1),(1,0)$
So what am I missing ? Why don't I get 3 intersection points ?
Best Regards !
|
$x^2(x^2-1)=0\Rightarrow \text{either } x^2=0 \text{ or } x^2-1=0$
|
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|
How prove this limit $\lim\limits_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a$ given two other limits related to $f_n$ Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}=ab \quad (|a|<|b|). $$
Prove that:$$\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a $$
I think we must prove $\displaystyle\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}} $ exists,and we prove this limit is $a$,But I can't prove this limit exists.
My idea: since
$$\lim_{n\to\infty}\dfrac{\dfrac{f_{n}}{f_{n+1}}-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=a+b$$
and
$$\lim_{n\to\infty}\dfrac{\left(\dfrac{f_{n}}{f_{n+1}}\right)^2-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=ab$$
But I felt this deal is not useful,
Other idea: I want to take the
Fibonacci sequence to solve this problem, But I can't,
Thank you
|
I'm going to post a failed approach, just so you can see what I tried. This is here only to show an approach and (maybe) generate new ideas.
Let $$x:=a+b\;\;\text{ and }\;\; y:=ab.$$ Assume for now $0<a<b$. Then
$$\frac{x-\sqrt{x^2-4y}}{2}
=\frac{a+b-\sqrt{(a+b)^2-4ab}}{2}
=\frac{a+b-\sqrt{(a-b)^2}}{2}
=\frac{a+b-\vert a-b\vert}{2}
=\frac{a+b-b+a}{2}
=a.$$
Now we replace
$a+b$ with $\lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}$ and
$ab$ with $\lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}$, and due to continuity move the limit outside:
$$\frac{a+b-\sqrt{(a+b)^2-4ab}}{2}
=\lim_{n\to\infty}
\frac{\left(\frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}\right)-\sqrt{\left(\frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}\right)^2
-4\left(\frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}\right)}}{2}
=\lim_{n\to\infty}
\frac{\left({f_{n+1}f_n-f_{n-1}f_{n+2}}\right)
-\sqrt{\left({f_{n+1}f_n-f_{n-1}f_{n+2}}\right)^2
-4\left({f_{n}^2-f_{n-1}f_{n+1}}\right)\left({f_{n+1}^2-f_nf_{n+2}}\right)}}{2\left({f_{n+1}^2-f_nf_{n+2}}\right)}
=\lim_{n\to\infty}
\frac{f_{n+1}f_n-f_{n-1}f_{n+2}
-\sqrt{
f_{n+1}^2f_n^2
-2f_{n+1}f_n
f_{n-1}f_{n+2}
+f_{n-1}^2f_{n+2}^2
-4f_n^2f_{n+1}^2+4f_{n}^2f_nf_{n+2}+4f_{n-1}f_{n+1}f_{n+1}^2-4f_{n-1}f_{n+1}f_nf_{n+2}}
}{2\left({f_{n+1}^2-f_nf_{n+2}}\right)}
$$
At this point there are no useful simplifications to be done inside the radical. It does not seem to agree with the $x$ and $y$ case.
|
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|
If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $(a+b)(a+c)=(b+c)^2$, prove that $(b-c)^2>8(b+c)$.
If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $$(a+b)(a+c)=(b+c)^2$$prove that $$(b-c)^2>8(b+c).$$
The first thing I did after I saw the problem was turning the inequality into this: $$(b+c)^2-8(b+c)-4bc>0$$
After a lot of different approaches I still can't find a way to solve this.
|
$(a+b)(a+c)=(b+c)^2\implies a^2 +a(b+c)+bc-(b+c)^2=0$, which is a quadratic in $a$. Solving for $a=\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}=\frac {-2(b+c)+\sqrt{(b-c)^2+4(b+c)^2}}{2}$, (we take the positive root since $a\ge1$).
or, $\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}\ge1$
or, $(b-c)^2+4(b+c)^2\ge(2+2b+2c)^2$
or, $(b-c)^2 \ge 4(1+b+c)^2-4(b+c)^2=4(1+2b+2c)>8(b+c).$
|
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|
Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$
Then, I get
$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$
If $x = \pi$, then
$$\begin{aligned}
\pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\
\dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n
\end{aligned}$$
I'm stuck. It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.
Any comments or suggestions? By the way, I know that
$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$
I need to know how to get there.
|
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\begin{align}
\pi x\cot\pars{\pi x}&=1 + \sum_{n = 1}^{\infty}{2x^{2} \over x^{2} - n^{2}}
=1 - 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{4}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{6}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
\end{align}
$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}=1 - 2x\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{3}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
$$
For $\verts{z} \sim 0$:
\begin{align}
z\cot\pars{z}&={z \over \tan\pars{z}} \sim {z \over z + z^{3}/3 + 2z^{5}/15}
={1 \over 1 + z^{2}/3 + 2z^{4}/15}
\\[3mm]&\sim 1 - \pars{{z^{2} \over 3} + {2z^{4} \over 15}}
+ \pars{{z^{2} \over 3} + {2z^{4} \over 15}}^{2}
\sim 1 - {z^{2} \over 3} - {2z^{4} \over 15} + {z^{4} \over 9}
=1 - {z^{2} \over 3} - {z^{4} \over 45}
\end{align}
$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}\sim
1 - {\pi^{2} \over 3}\,x - {\pi^{4} \over 45}\,x^{2}
$$
$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n^{4}}}
= -\,\half\,\pars{-\,{\pi^{4} \over 45}}
= \color{#00f}{\large{\pi^{4} \over 90}}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/687676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Finding sums of infinite series as stated in the title the series are infinite and im struggle with how to find the common ratio. and first term (kinda) well I want it to be on the form: ar^k where r is the ratio and a the first term
(3+2^n)/(2^(n+2)) (n=1)
any tips on how to get that series on the form ar^k ?
|
It may help to decompose the series:
$$S = \displaystyle \sum_{n = 1}^{k} \frac{3 + 2^n}{2^{n + 2}} = \displaystyle \sum_{n = 1}^{k} \frac{3}{2^{n + 2}} + \displaystyle \sum_{n = 1}^{k} \frac{1}{4}$$
Going a bit further,
$$S = (\frac{3}{8} + \frac{3}{16} + \frac{3}{32} + \dots + \frac{3}{2^{k+2}}) + (\frac{1}{4} k)$$
So in this case, $a = \frac{3}{8}$ and $r = \frac{1}{2}$. Additionally, the infinite series diverges using the $n$-th term test ($\lim_{n \to \infty} \frac{1}{4} = \frac{1}{4} \neq 0$)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/689117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $x+y+z=xyz$, find $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ I found this question in a maths worksheet of trigonometry (kinda odd, right?), but I dont know how to figure it out.
If $\displaystyle x+y+z=xyz$, find $\displaystyle\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$
First I thought of taking x,y and z as $\displaystyle \tan A, \tan B,$ and $\tan C,$ making $A+B+C=\pi$, but couldnt solve ahead. Any solution not involving trigonometry would do as well.
I also think that this question does not even relate to trigo.....or does it?
|
Case I: At least one of $x^2$, $y^2$ and $z^2$ is equal to $1/3$. This case is possible, e.g., $x=\frac{\sqrt{3}}{3}=-y$ and $z=0$. In this case, the formula $\displaystyle \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ does not make any sense.
Case II: None of $x^2$, $y^2$ and $z^2$ is equal to $1/3$ and $xyz=0$. For example, we say $x=-y$ and $z=0$, which implies that
\begin{eqnarray*}
\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}+\frac{-3x+x^3}{1-3x^2}+0=0.
\end{eqnarray*}
Case III: None of $x^2$, $y^2$ and $z^2$ is equal to $1/3$ and $xyz\neq0$. Take $x=\tan A$, $y=\tan B$ and $z=\tan C$ for some $A,B,C\in(-\pi/2,\pi/2)$, since $x+y+z=xyz$, then $xy,xz,yz\neq 1$, which implies that $A+B$, $A+C$ and $B+C$ can not be $\pm\pi/2$.
Claim I: $A+B+C\neq\pm\pi/2$.
If not, then $\tan(A+B)=\cot C$, that is, $\displaystyle\frac{x+y}{1-xy}=\frac{1}{z}$, that is, $(x+y)z=1-xy$. Hence $(x+y)z^2=z-xyz=z-(x+y+z)=-(x+y)$, that is, $x+y=0$. Since $x+y+z=xyz$ and $z\neq0$, we get $xy=1$, contradiction.
In summary, we can define $\tan(3A)$, $\tan(3B)$, $\tan(3C)$ and $\tan(A+B+C)$. Since $\displaystyle \tan(3\theta)=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$, then we have
\begin{eqnarray*}
\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\tan(3A)+\tan(3B)+\tan(3C).
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\tan(A+B+C)&=&\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C}\\
&=&\frac{x+y+z-xyz}{1-xy-xz-yz}=0.
\end{eqnarray*}
Then there exists some $m\in\{-1,0,1\}$ such that $A+B+C=m\pi$. Hence $\tan(A+B)+\tan C=0$, which implies that $\tan^2(A+B)\neq1/3$. Hence we can define $\tan(3A+3B)$. So we have
\begin{eqnarray*}
0&=&\frac{\tan(3A)+\tan(3B)+\tan(3C)-\tan(3A)\tan(3B)\tan(3C)}{1-\tan(3A)\tan(3B)-\tan(3A)\tan(3C)-\tan(3B)\tan(3C)}.
\end{eqnarray*}
Then we get
\begin{eqnarray*}
\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\tan(3A)+\tan(3B)+\tan(3C)=\tan(3A)\tan(3B)\tan(3C).
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Two dice are rolled, what is the probability that the minimum value of the two dice is $3$? Let $D_1$ and $D_2$ represent the values obtained from dice $1$ and dice $2$. We are looking for $P(\min(D_1, D_2) = 3)$
I.e.
$$
\begin{align}
& P([D_1 = 3, D_2 = 3, 4, 5, 6] \cup [D_2 = 3, D_1 = 3, 4, 5, 6])
= 2P([D_1 = 3, D_2 = 3, 4, 5, 6]) \\[8pt]
& = 2P([D_1 = 3 , D_2 = 3] \cup [D_1 = 3 , D_2 = 4] \cup [D_1 = 3 , D_2 = 5] \cup [D_1 = 3 , D_2 = 6]) \\[8pt]
& = 2P(D_1 = 3, D_2 = 3)^4 \\[8pt]
& = 2\left(\frac{1}{6}\right)^4 \\[8pt]
& = \frac{1}{648}
\end{align}
$$
Does that seem correct? It looks too extreme...
|
Be careful: the probability of the union is not equal to the product of the probabilities (that would be the intersection, and it's only true if the events are independent). Nor is the probability of the union equal to the sum of the probabilities (unless the events are disjoint / "mutually exclusive").
\begin{align*}
P(\min(D_1, D_2 = 3))
&= P([D_1 = 3, D_2 = 3, 4, 5, 6] \cup [D_2 = 3, D_1 = 3, 4, 5, 6]) \\
&= P(D_1 = 3, D_2 = 3) \\
&\quad + P(D_1 = 3, D_2 = 4,5,6) \\
&\quad + P(D_1 = 4,5,6, D_2 = 3) \\
\end{align*}
This last step is valid because the different events are disjoint. So then we have
\begin{align*}
&= P(D_1 = 3, D_2 = 3) + 2P(D_1 = 3, D_2 = 4,5,6) \\
&= P(D_1 = 3) P(D_2 = 3) + 2 P(D_1 = 3) P(D_2 = 4,5,6)
\textbf{ (because of independence)} \\
&= \frac16 \frac16 + 2 \frac16 \frac36 \\
&= \frac{7}{36}
\end{align*}
Of course, a direct count would have been easier: there are seven ways to achieve a minimum of 3 out of 36 possible rolls.
|
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|
If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$
I could not approach the problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.
|
Let
$$
T_1=\sum_{cyc} \frac{a}{b-c}
$$
By hypothesis we have $T_1=0$, but if we put
$T_2=T_1(ab+ac+bc-a^2-b^2-c^2)$, we have
$$
T_2=\sum_{cyc} \frac{a(ab+ac+bc-a^2-b^2-c^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2c}{b-c}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ac^2}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2b}{c-b}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ab^2}{c-b}
$$
$$
T_2=
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}
=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
+\sum_{cyc} \frac{-a^2b}{b-c}
+\sum_{cyc} \frac{-a^2c}{b-c}
$$
and hence
$$
T_2=\sum_{cyc} \frac{a(b-a)(c-a)}{b-c}=
-\sum_{cyc} \frac{a(a-b)(c-a)}{b-c}
$$
So if we put $T_3=\frac{T_2}{(a-b)(b-c)(c-a)}$, we see that
$$
T_3=-\sum_{cyc} \frac{a}{(b-c)^2}
$$
and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Trigonometric Substitution for $\int\sqrt{9-x^2}\,\mathrm dx$ Question: Use the substitution $x=3\sin(t)$ to evaluate the integral of $\int\sqrt{9-x^2}\,\mathrm dx$.
I started by making a right triangle and solving for $\sin(t)$ and $\cos(t)$.
*
*$\sin(t)=\frac{x}{3}$ and $\cos(t)=\frac{\sqrt{9-x^2}}{3}$
Then, I solved for the values $\mathrm dx$ and $\sqrt{9-x^2}$.
*$\sqrt{9-x^2}=3\cos(t)$
*$\mathrm dx=3\,\cos(t)\,\mathrm dt$
Then, I got the integral of $3\,\cos^2(t)\,\mathrm dt$.
*$3\,(\frac{1}{2}\cos(t)\sin(t)+ \frac{1}{2}t)$
*$\frac{3}{2}\cos(t)\sin(t)+\frac{3}{2}t$
Then, I substituted in the values I found for $\sin(t)$, $\cos(t)$, etc.
*$\frac{3}{2}(\frac{\sqrt{9-x^2}}{3})\times\frac{x}{3}+\frac{3}{2}\arcsin(\frac{x}{3})$
That was the wrong answer and I do not know why. Where did I go wrong?
Thank you for your help!
|
$$
\int \sqrt{9-x^2} dx
$$
let $x=3\sin\phi$,$dx=3\cos\phi d\phi$, thus we have
$$
3\int d\phi\cos\phi\sqrt{9(1-sin^2\phi)}=9\int d\phi \cos^2\phi =9\int d\phi \frac{1}{2}(1+cos(2\phi))=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C
$$
Thus we see that
$$
\int \sqrt{9-x^2} dx=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C
$$
where $\phi=\sin^{-1}(x/3)$. Simplifying we obtain
$$
\int \sqrt{9-x^2} dx=\frac{1}{2}\big( \sqrt{9-x^2}+9\sin^{-1}\big(\frac{x}{3}\big) \big)
$$
|
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|
Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider these relations as equivalent ${}\bmod n$ and add them.)
$$1^m+2^m+\cdots+(n-2)^m+(n-1)^m=((n-1)^m+(n-2)^m+\cdots+2^m+1^m)\bmod n$$
$$(n-1)^m+(n-2)^m+\cdots+2^m+1^m=(1^m+2^m+\cdots+(n-2)^m+(n-1)^m)\bmod n$$
By adding them we get:
$$(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)=((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))\bmod n$$
That means that:
$$n|[(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)]-[((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))]$$
Or not??
And that is equal to $$n\mid 0$$
Is this correct? How can I continue??
|
This is not true. Let $m=2$ and $n=3$. Then $3$ does not divide $1^2+2^2$.
|
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|
How to compute a linear transformation which carries the circle $|z|=2$ into $|z+1|=1$?
Find the linear transformation which carries the circle $|z|=2$ into $|z+1|=1$, the point $-2$ into the origin, and the origin into $i$.
In order to find the linear transformation will I use the following formula?
$$(z_1−z_3)(z_2−z)(z_2−z_3)(z_1−z)=(w_1−w_3)(w_2−w)(w_2−w_3)(w_1−w)$$
Or am I wrong?
Similar example but not the same:
Find a linear transformation mapping the circle $|z|=1$ onto the circle $|w-5|=3$ and taking the point $z=i$ to $w=2$.
Given, $|w-5|=3$ so $|(w-5)/3)|=1$. So now, we can use the transformation $z=(w-5)/3$ but $z=i$ is not mapped to $w=2$. So we use the transformation $iz=(w-5)/3$. So, $z=i \implies w=3i*i+5=2$.
How to start ?
|
Not sure if this is correct so please let me know
Let $C_1 = |z| = 2$, and $C_2 = |z + 1| = 1$. Since the origin is in $C_1$ but not in $C_2$, we need to make the circle go inside out. Thus we get $T = \frac{az + b}{cz + d}$.
We know $T = z*(\frac{z + b}{z + d})$ $z*$ be the reflection point of the origin. We let $R$ be the radius of $C_2$, and $\alpha$ be the center of $C_2$.
Now $\infty$ is the inflection point of $0$. Therefore $Z^* = \frac{R}{\bar{z} - \bar{\alpha}} + \alpha = \frac{1}{-i + 1} -1 = \frac{1 + i - 1}{-i + 1} = \frac{i}{1 - i} = \frac{-1}{1 - i} \times \frac{1 + i}{1 + i} = \frac{-1 +i}{1 + 1} = \frac{-1 + i}{2}$.
So $T = \frac{-1 + i}{2} \times \frac{z + b}{z + d}$. So since $T(-2) = 0$ we get $b = 2$. It remains to solve for $d$
. We get $T(0) = i = \frac{-2 + 2i }{2d}$ therefore $d = \frac{-1 + i}{i} = i + 1$.
So $T = \frac{(-1 + i)z + 2}{2z + (i + 1)}$
|
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|
How to solve $6^{2x}-10\cdot 6^x=-21$ using logarithms?
What do I do with $\large 6^{2x}-10\cdot 6^x=-21$?
Since $6$ and $-60$ are not of the same base (nor can they be written as exponents of the same base cleanly) I am having trouble solving for $x$.
|
Let $6^x=y$
$$6^{2x}-10\cdot 6^x=-21$$
$$y^2-10y=-21$$
$$y^2-10y+21=0$$
Factor.
$$(y-3)(y-7)=0$$
$$y=3, \ 7$$
Replace $y$ with $6^x$
$$6^x = 3, \ 7$$
I will solve for both equations separately. Let's start with $6^x=3$
$$6^x=3$$
$$\ln(6^x)=\ln(3)$$
$$x\ln(6)=\ln(3)$$
$$x=\dfrac{\ln(3)}{\ln(6)}$$
Now for $6^x=7$
$$6^x=7$$
$$\ln(6^x)=\ln(7)$$
$$x\ln(6)=\ln(7)$$
$$x=\dfrac{\ln(7)}{\ln(6)}$$
The solutions are:
$$\displaystyle \boxed{x=\dfrac{\ln(3)}{\ln(6)}, \ \dfrac{\ln(7)}{\ln(6)}}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometric reverse substitution integral question. Evaluate the following integral.
$$ \int \frac{1}{(x^2-4x+13)^2}\ dx $$
My try :
$$ \int \frac{1}{(x^2-4x+13)^2}\ dx $$
$$ \int \frac{1}{[(x-2)^2+9]^2}\ dx $$
$$ (x-2)^2-9 = x^2-a^2\ ,\ a=3,\ x=3\sec\theta\ x^2-a^2=9\tan^2\theta\ dx=3\sec\theta \tan\theta\ d\theta $$
$$ \int \frac{1}{(9\tan^2\theta)^2}\ 3\sec\theta\ \tan\theta\ d\theta $$
And I couldn't continue.
|
Put $x^2-4x+3=(x-2)^2+3^2$ and then $\tan \theta=\frac{x-2}{3}\Rightarrow x=2+3\tan \theta$, and $dx=3\sec^\theta d\theta$; $\cos \theta = \frac{3}{\sqrt{(x-2)^2+3^2}}$:
$$\int \frac{dx}{[(x-2)^2+3^2]^2}=\int \frac{3\sec^2\theta d\theta}{3^4 \sec^4\theta} =\frac{1}{3^3}\int \frac{d\theta}{\sec^2\theta}=\frac{1}{27}\int \cos^2\theta d\theta$$
$$= \frac{1}{27}\int \frac{1+\cos 2\theta}{2} d\theta=\frac{1}{54}\int d\theta+\frac{1}{54}\int \cos 2\theta=\frac{1}{54}\theta+\frac{1}{54}\frac{1}{2}\sin 2\theta+c=$$
$$... $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int \cos^4(x)dx$ We have: $\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.$
Find $\int \cos^4x\ dx$ by using the formula twice
What I have so far is:
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}\int \cos^{2} x\
dx$
Now we use the formula for $\int cos^{2} x\ dx$:
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int \cos^{0} x\
dx$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int 1 \
dx$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}[ x ]$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{x}{2}$
Now plug this in to the $\int \cos^4 x\ dx$ equation above
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}[\frac{1}{2} \cos x \sin x + \frac{x}{2}]$
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{8}[\cos x \sin x + {x}]$
This is where I get stuck. I'm aware I could have used an identity for $\cos^2x$ but the question needs me to use the formula twice.
|
You are right.
Look at the alternate forms in WolframAlpha
|
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|
Find the integral $\int \frac{1}{x^2 \cdot \tan(x)} \ dx$ This problem seems pretty tricky. I need to find the integral of
$$\int \dfrac{1}{x^2 \cdot \tan(x)} \ dx$$
Any help would be greatly appreciated!
|
I would use the decomposition of $\frac{1}{\tan x}$ into simple fractions:
$$\frac{1}{\tan x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}$$ So
$$\frac{1}{x^2\tan x}=\frac{1}{x^3}+2\sum_{n=1}^{\infty}\frac{1}{x(x^2-n^2\pi^2)}$$ and
$$\int\frac{dx}{x^2\tan x}=-\frac{1}{2x^2}+\sum_{n=1}^{\infty}\frac{\ln\frac{\left |x^2-n^2\pi^2 \right |}{x^2}}{n^2\pi^2}+\text{const}$$
This is of course a matter of taste, whether the form is closed or not.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function - technical issue. Which sequence is generated by $\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}}$?
We know that:
$$\frac{1}{{{{(1 - x)}^3}}} = \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}} $$
So we have:
$$(5x - 3{x^2}) \cdot \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}} $$
Now it's easy to see that for $x^k$, $a_k$ can be defined by:
$$5\left( {\begin{array}{*{20}{c}}
{k + 1} \\
2 \\
\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}
k \\
2 \\
\end{array}} \right)$$
Because we "used" $x^{k-1}$ and $x^{k-2}$ there may be a problem for $k=0,1$ and I've been told I need to check it directly. Can you help me with that?
|
First, note that
$$\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}} = \frac{3}{(x-1)}+\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}$$
And,
$$\frac{3}{(x-1)} = \sum \limits_{k=0}^{\infty}(-3)x^k$$
$$\frac{1}{(x-1)^2}= \sum \limits_{k=0}^{\infty}(k+1)x^k$$
$$\frac{-2}{(x-1)^3}= \sum_{k=0}^{\infty} x^k(k+1)(k+2)$$
Add them up and you get
$$\sum \limits_{k=1}^{\infty} k(k+4)x^{k}$$
|
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"url": "https://math.stackexchange.com/questions/703482",
"timestamp": "2023-03-29T00:00:00",
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|
How are the Taylor Series derived? I know the Taylor Series are infinite sums that represent some functions like $\sin(x)$. But it has always made me wonder how they were derived? How is something like $$\sin(x)=\sum\limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}\cdot(-1)^n = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\pm\dots$$ derived, and how are they used? Thanks in advance for your answer.
|
This simple derivation of the Taylor series of a function is taken from lecture 37 of the MIT course Single Variable Calculus. More specifically, this derivation is taken from these lecture notes. A similar derivation can also be found here.
Given an infinitely differentiable function $f(x)$, we want to represent it as the power series
$$
f(x) = a_0 + a_1(x - c) + a_2(x-c)^2 + a_3(x-c)^3 + a_4(x-c)^4 + a_5(x-c)^5 + \cdots
$$
The goal is to determine the coefficients $a_0,a_1,a_2,\dots$. The general strategy for doing this is substituting $x=c$ into $f(x)$ and its derivatives and picking out the coefficients. For example,
$$
f(c) = a_0
$$
and so we know that $a_0$ has the value $f(c)$. Next, we can differentiate $f(x)$ to get
$$
f'(x) = a_1 + 2 \cdot a_2(x-c) + 3 \cdot a_3(x-c)^2 + 4 \cdot a_4(x-c)^3 + 5 \cdot a_5(x-c)^4 + \cdots
$$
Note that
$$
f'(c) = a_1
$$
We can then differentiate $f'(x)$ to get
$$
f''(x) = 2 \cdot a_2 + 3 \cdot 2 \cdot a_3(x-c) + 4 \cdot 3 \cdot a_4(x-c)^2 + 5 \cdot 4 \cdot a_5(x-c)^3 + \cdots
$$
Note that
$$
f''(c) = 2 \cdot a_2
$$
or
$$
a_2 = \frac{f''(c)}{2}
$$
We can repeat this process several times to get
\begin{align}
f'''(x) &= 3 \cdot 2 \cdot a_3 + 4 \cdot 3 \cdot 2 \cdot a_4(x-c) + 5 \cdot 4 \cdot 3 \cdot a_5(x-c)^2 + \cdots \\
f''''(x) &= 4 \cdot 3 \cdot 2 \cdot a_4 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot a_5(x-c) + \cdots \\
\end{align}
and we can substitute $x = c$ into these derivatives to get
\begin{align}
f'''(c) &= 3 \cdot 2 \cdot a_3 \\
f''''(c) &= 4 \cdot 3 \cdot 2 \cdot a_4
\end{align}
re-arranging, we get
\begin{align}
a_3 &= \frac{f'''(c)}{3 \cdot 2} \\
a_4 &= \frac{f''''(c)}{4 \cdot 3 \cdot 2}
\end{align}
Therefore, the general pattern is
\begin{align}
a_0 &= f(c) = \frac{f(c)}{0!} \\
a_1 &= f'(c) = \frac{f'(c)}{1!} \\
a_2 &= \frac{f''(c)}{2} = \frac{f''(c)}{2!} \\
a_3 &= \frac{f'''(c)}{3 \cdot 2} = \frac{f'''(c)}{3!} \\
a_4 &= \frac{f''''(c)}{4 \cdot 3 \cdot 2} = \frac{f''''(c)}{4!} \\
&\vdots \\
a_n &= \frac{f^{(n)}(c)}{n!}
\end{align}
So, the power series representation of $f(x)$ is
\begin{align}
f(x) &= \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \cdots \\
&= \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $1^2+2^2+\cdots+n^2 = {n+1\choose2}+2{n+1\choose3}$ Prove that:
$$
1^2+2^2+\cdots+n^2 = {n+1\choose2}+2{n+1\choose3}
$$
Now, if I simplify the right hand combinatorial expression, it reduces to $\frac{n(n+1)(2n+1)}{6}$ which is well known and can be derived by the method of common differences.
This though is in the exercise sheet related to combinatorics and specifically the method of proof by double counting. I can't figure out how to do that. Specifically this looks like "choose 2 objects from n+1 and then choose 3 objects from n+1 twice." But that's that, I can't find a link between that and the sum of squares. Any help?
|
The identity is an application of Worpitzky's identity involving Eulerian numbers.
Worpitzky's theorem states:
$$x^n=\sum_{k=0}^{n}A(n,k) \binom{x+k}{n}$$
where Eulerian number $A(n, k)$ is defined to be the number of permutations of the numbers 1 to n in which exactly k elements are greater than the previous element (permutations with k "ascents"). (Worpitzky's identity is not hard to prove using induction btw)
By Worpitzky's identity, for integer $k\geq 1, k^2=\sum_{i=0}^{2} A(2,i) \binom{n+i}{2} = A(2,0)\binom{k}{2}
+A(2,1)\binom{k+1}{2}=\binom{k}{2}+\binom{k+1}{2}$.
So the sum of the squares of the first $n$ positive integers is
\begin{equation}
\begin{aligned}
\sum_{i=1}^{n} k^2 &=\sum_{i=1}^{n} \binom{k}{2} + \sum_{i=1}^{n} \binom{k+1}{2}\\
&= \binom{n+1}{3}+\binom{n+2}{3}\\
&=\frac{1}{6}n(n+1)(2n+1)
\end{aligned}
\end{equation}
This equality follows from the following identity for the rising sum of binomial coefficients:
$$\sum_{j=0}^{n} \binom{j}{m} = \binom{n+1}{m+1}$$
Similarly, we can find the sum of first n cubic numbers. For integer $k\geq 1$, Worpitzky's identity says that $k^3=A(3,0)\binom{k}{3}+A(3,1)\binom{k+1}{3}+A(3,2)\binom{k+2}{3}=\binom{k}{3}+4\binom{k+1}{3}+\binom{k+2}{3}$
Same as we just did in the last case, the sum of the cubes of the first $n$ positive integers is
\begin{equation}
\begin{aligned}
\sum_{i=1}^{n} k^3 &=\sum_{i=1}^{n} \binom{k}{3} + 4\sum_{i=1}^{n} \binom{k+1}{3}+\sum_{i=1}^{n} \binom{k+2}{3}\\
&= \binom{n+1}{4}+4\binom{n+2}{4}+\binom{n+3}{4}\\
&=\frac{1}{4}(n^4+2n^3+n^2)
\end{aligned}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$.
Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$.
Here $S_n$ denotes $x_1+x_2+\ldots+x_n$. Also $n\in\mathbb N_{>0}$.
It's easy to see that $x_1=2$.
I've substituted $x_n$ with $S_n-S_{n-1}$, got a quadratic equation in terms of $\sqrt{S_n}$ and found out that when $n\ge3$, we have $$\sqrt{S_n}-\sqrt{S_{n-1}}=\sqrt{2}$$
But it seems to be of no use. Any hints would be appreciated.
I could hardly find a tag this question would fit well, so you could suggest some tags as well. Thanks.
|
Noting that $S_n = x_n + S_{n-1}$, you have
$$
x_n^2 + 4x_n + 4 = (x_n + 2)^2 = 8S_n = 8x_n + 8S_{n-1},
$$
or
$$
x_n^2 - 4x_n + (4 - 8S_{n-1}) = 0.
$$
The solutions are
$$
x_n = \frac{1}{2}\left(4 \pm \sqrt{16 - 4(4-8S_{n-1})}\right)=2\pm\sqrt{8S_{n-1}};
$$
only the positive one is relevant, so you have
$$
x_n=2+2\sqrt{2\sum_{i=1}^{n-1}x_i}.
$$
You find that $x_1=2$, $x_2=2+2\sqrt{2\cdot 2}=6$, $x_3=2+2\sqrt{2\cdot(2+6)}=10$, $x_4=2+2\sqrt{2\cdot(2+6+10)}=14$, etc. Pretty clearly $$x_i=4i-2.$$ The proof is by induction, using the fact that $\sum_{i=1}^{n-1}(4i-2)=2(n-1)^2$: assuming that $x_i=4i-2$ for all $i < n$,
$$
x_n=2+2\sqrt{2\cdot2(n-1)^2}=2+4(n-1)=4n-2
$$
as well.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quadratic Partial Fraction Decomposition I am trying to find the inverse laplace transform of $(s^2+4) \over (s-2)(s+2)$.
The solution is $ {2\over(s-2)} - {2\over(s+2)} + 1 $.
But I can't figure out how to break it up so I can find the solution algebraically.
i.e $ (s^2+4)/((s-2)(s+2)) = As+B/(s-2)+C/(s+2) $ etc. What terms should I use?
|
$$\begin{align}\frac{s^2+4}{s^2-4}&=\frac{s^2-4+8}{s^2-4}\\&= 1+\frac{8}{s^2-4}\\&=1+2\frac{4}{(s+2)(s-2)}\end{align}$$
Obviously $4=(s+2)-(s-2)$
substituting, we have
$$1+2\frac{(s+2)-(s-2)}{(s+2)(s-2)}$$
$$1+2\frac{(s+2)}{(s+2)(s-2)}-2\frac{(s-2)}{(s+2)(s-2)}$$
Cancelling,
$$\require{cancel}{1+2\frac{\cancel{(s+2)}}{\cancel{(s+2)}(s-2)}-2\frac{\cancel{(s-2)}}{(s+2)\cancel{(s-2)}}}$$
And thus, we have
$$1+\frac{2}{s-2}-\frac{2}{s+2}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try:
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$ minimum I can't find it,thank you
|
Given:
$$ab+\frac 1{a^2}+\frac 1{b^2}=ab+\frac 1{a^2}+\frac 1{b^2}+\frac {2}{ab}-\frac {2}{ab}$$
$$\Rightarrow ab+\frac 1{a^2}+\frac 1{b^2}=ab+(\frac 1a-\frac 1b)^2+\frac 2{ab}$$
Let us analyse the RHS:
The minumum value of $(\frac 1a-\frac 1b)^2$ is 0 when a=b and the minumum value of $ab+\frac 2{ab}$ is $2\sqrt 2$(A.M.-G.M inequality). Hence the minimum value of the required expression is $2\sqrt 2$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that the range of linear transformation is a linear subspace I need some help figuring out how to prove this question.
True or false, the set $S = \left \{ A\mathbf{y}: \mathbf{y} \in \mathbb{R}^4\right \}$ is a subspace of $\mathbb{R}^3$ where A is a fixed $3\times4$ real matrix.
Well I will need to show that the zero vector is in the set S. Then show the closure axioms hold. Or I can show for $A\mathbf{u}, A\mathbf{v}$ and some scalar $c$ in our field, $A\mathbf{u}+cA\mathbf{v} \in S$
What I have so far:
For $\mathbf{y} = \begin{pmatrix}
0\\
0\\
0\\
0 \end{pmatrix} \in \mathbb{R}^4$. It is clear that, $A\mathbf{y} = \begin{pmatrix}
0\\
0\\
0
\end{pmatrix} \in S $. (Hopefully this is right so far)
Now, we need to prove the closure axioms.
Suppose $A\mathbf{u}, A\mathbf{v} \in S$ where $\mathbf{u} = \begin{pmatrix}
u_{1}\\
u_{2}\\
u_{3}\\
u_{4} \end{pmatrix}, \mathbf{v} = \begin{pmatrix}
v_{1}\\
v_{2}\\
v_{3}\\
v_{4} \end{pmatrix} \in \mathbb{R}^4 $
Here is where I'm stuck, I know I am suppose to show $A\mathbf{u} + cA\mathbf{v} \in S$
Any clues or hints would be appreciated. Thanks. Please try to only post partial solutions or hints to get me going.
|
I think I figured it on, now I just need some confirmation so continuing from what I have above, we have $A\mathbf{u}+cA\mathbf{v} = A\begin{pmatrix}
u_{1}\\
u_{2}\\
u_{3}\\
u_{4} \end{pmatrix} + cA\begin{pmatrix}
v_{1}\\
v_{2}\\
v_{3}\\
v_{4} \end{pmatrix} $
$
= A\begin{pmatrix}u_{1}\\
u_{2}\\
u_{3}\\
u_{4} \end{pmatrix} + A\begin{pmatrix}
cv_{1}\\
cv_{2}\\
cv_{3}\\
cv_{4} \end{pmatrix} $
$
= A\begin{pmatrix}
u_{1}+cv_{1}\\
u_{2}+cv_{2}\\
u_{3}+cv_{3}\\
u_{4}+cv_{4} \end{pmatrix} \in S$ as $\begin{pmatrix}
u_{1}+cv_{1}\\
u_{2}+cv_{2}\\
u_{3}+cv_{3}\\
u_{4}+cv_{4} \end{pmatrix} \in \mathbb{R}^4$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum\limits_{r=0}^{2n} r \binom{2n}{r}^2 = 2n \binom{4n-1}{2n}$ I expanded
$(1+x)^{2n}$ = $\sum\limits_{r=0}^{2n} \binom{2n}{r} x^r $
Differentiating both sides, we get
$2n(1+x)^{2n-1}$ = $0$ + $\binom{2n}{1}$ + $2\binom{2n}{2}x$ + $3\binom{2n}{3}x^2$ .....
Put $x=1$.
As you see, I'm not able to get the 'squares'.Any idea on how to go about it?
|
$$
\begin{align}
\sum_{r=0}^{2n}r\binom{2n}{r}^2
&=\sum_{r=1}^{2n}r\binom{2n}{r}\binom{2n}{2n-r}\tag{1}\\
&=\sum_{r=1}^{2n}2n\binom{2n-1}{r-1}\binom{2n}{2n-r}\tag{2}\\
&=2n\binom{4n-1}{2n-1}\tag{3}\\
&=2n\binom{4n-1}{2n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}$
$(3)$: Vandermonde's Identity
$(4)$: $\binom{n}{k}=\binom{n}{n-k}$
Here is a proof of the Vandermonde Identity step using the binomial identity.
The binomial theorem says
$$
\begin{align}
(1+x)^{2n-1}(1+x)^{2n}
&=\sum_{j=0}^{2n-1}\binom{2n-1}{j}x^j\sum_{k=0}^{2n}\binom{2n}{k}x^k\\
&=\sum_{s=0}^{4n-1}\color{#00A000}{\sum_{r=0}^s\binom{2n-1}{r}\binom{2n}{s-r}x^s}
\end{align}
$$
but it also says
$$
(1+x)^{4n-1}=\sum_{s=0}^{4n-1}\color{#00A000}{\binom{4n-1}{s}x^s}
$$
Equating the coefficients of $x^s$, we get
$$
\sum_{r=0}^s\binom{2n-1}{r}\binom{2n}{s-r}=\binom{4n-1}{s}
$$
Setting $s=2n-1$ and substituting $r\mapsto r-1$ gives
$$
\sum_{r=1}^{2n}\binom{2n-1}{r-1}\binom{2n}{2n-r}=\binom{4n-1}{2n-1}
$$
|
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|
Prove if $n^3$ is odd, then $n^2 +1$ is even I'm studying for finals and reviewing this question on my midterm. My question is stated above and I can't quite figure out the proof.
On my midterm I used proof by contraposition by stating:
If $n^2 +1$ is odd then $n^3$ is even.
I let $n^2+1 = (2m+1)^2 + 1$
$= (4m^2 + 4m + 1) + 1$
$= 2(2m^2 + 2m + .5) + 1$
Let $2m^2 + 2m + .5 = k$
$n^2 + 1 => 2k + 1$
Therefore proving that $n^2 + 1$ was odd making $n^3$ even.
I know my logic was messed up somewhere..some guidance would be nice.
|
If $n$ is even, then $n=2k$ and $n^3=8k$, and we have that $n^3$ is even.So, if $n^3$ is
odd then $n$ is odd, otherwise if $n$ is even, by the previous statement we would have $n^3$ even given a contradiction. So, $n$ is odd, and then $n=2t+1$ so $n^2+1=(2t+1)^2+1=
4t^2+2t+1+1=2(2t^2+t+1)$ then, $n^2+1$ is even.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
limit $\lim_{n \to \infty} \frac{(2n+1)(2n+2) n^n}{(n+1)^{n+2}}$? Wolfram alpha says that this limit (arising from a ratio test to determine the radius of convergence of a series) should be $4/e$. How does it get this result?
|
Notice that
$$\frac{2n+1}{n+1} = 2 - \frac{1}{n+1}.$$
Then
\begin{eqnarray*}
\lim_{n \to \infty} \dfrac{(2n+1)(2n+2) n^n}{(n+1)(n+1)^{n+1}} &=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2(n+1)n^n}{(n+1)^{n+1}} \\
&=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2n^n/n^n}{(n+1)^{n}/n^n} \\
&=&\lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2}{(1+\frac1n)^n} \\
&=& \frac{4}{e}
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of Unique Sequences with Circular Shifts I'm trying to determine the number of binary sequences containing exactly $a$ $1$'s and $b$ $0$'s if two sequences are considered identical if they can be reached from the each other by a series of circular shifts (moving the first bit to the end). For instance, for $a=b=4$, $10110010$ and $11001010$ would be considered the same.
At first, I assumed that there would simply be $\frac{(a+b-1)!}{a!b!}$ unique sequences because there are $\frac{(a+b)!}{a!b!}$ different sequences without circular shifts and each unique sequences with can reach $a+b$ different sequences with them; but some sequences, such as $10101010$ can become themselves after a series of circular shifts (in this case $2$, $4$, or $6$). Thus, there are actually more unique sequences. Thank you in advance for any assistance you can provide on this topic.
Edit: If the general solution is too difficult, I'm especially interested in situations where $a+b = p$ or $a = b = p$ for some prime $p$.
|
The total number of sequences (without circular shifting) with precisely $a$ $1$'s and $b$ $0$'s is $\tbinom{a+b}{a}=\tbinom{a+b}{b}$. If either $a=0$ or $b=0$ then there is of course only one such sequence.
For any sequence of length $a+b$, the number of cyclic shifts $d$ required to return the original sequence is a divisor of $a+b$. In particular, if $a+b=p$ is prime we must have $d=1$ or $d=p$. If $d=1$ for some sequence then either $a=0$ or $b=0$. Hence if $a\neq0$ and $b\neq0$ then every sequence has precisely $a+b=p$ distinct cyclic shifts, including itself. The number of distinct sequences is therefore precisely
$$\frac{1}{a+b}\binom{a+b}{a}=\frac{1}{p}\cdot\binom{p}{a}=\frac{(p-1)!}{a!\cdot b!}.$$
If $a+b=2p$ then for any sequence the number of cyclic shifts $d$ required to return to the original sequence is either $1$, $2$, $p$ or $2p$. Again if it is $1$, then either $a=0$ or $b=0$, and then there is only one such sequence. So suppose $a\neq0$ and $b\neq0$ for now.
There is only one sequence returned to its original after $2$ cyclic shifts; this is $1010\ldots1010$ up to shifting. For this we must have $a=b=p$.
A sequence that returns to its original after $p$ cyclic shifts is a sequence of length $p=(a+b)/2$ repeated twice. For the prime case we have already counted
$$\frac{1}{(a+b)/2}\binom{(a+b)/2}{a/2}=\frac{1}{p}\cdot\binom{p}{a/2}=\frac{(p-1)!}{(a/2)!\cdot(b/2)!},$$
such sequences, and its a nice exercise to check that distinct sequences of length $p$ yield distinct sequences of length $2p$ when repeating them twice. So number of such sequences is precisely
$$\frac{(p-1)!}{(a/2)!\cdot(b/2)!}.$$
For this we must have both $a$ and $b$ even.
This allows us to count the number of distinct sequences if $a+b=2p$. If $a$ and $b$ are not both even, and $a\neq p$ and $b\neq p$, then ever sequence has precisely $a+b=2p$ distinct cyclic shifts, so just like in the prime case the number is
$$\frac{1}{a+b}\binom{a+b}{a}=\frac{1}{2p}\cdot\binom{2p}{a}=\frac{(2p-1)!}{a!\cdot b!}.$$
If $a$ and $b$ are both even (and nonzero) but $a\neq p$ and $b\neq p$, then we count seperately the sequences that are returned after $p$ cyclic shifts, and those that are not. The number is then
$$\frac{1}{a+b}\left(\binom{a+b}{a}-\binom{(a+b)/2}{a/2}\right)+\frac{1}{(a+b)/2}\binom{(a+b)/2}{a/2}=\frac{1}{2p}\left(\binom{2p}{a}+\binom{p}{a/2}\right).$$
If $a=b=p$ but $a$ and $b$ are not both even, then we count seperately the sequences that are returned after $2$ cyclic shifts, and those that are not. The number is then:
$$\frac{1}{a+b}\left(\binom{a+b}{a}-2\right)+1=\frac{1}{2p}\left(\binom{2p}{p}-2\right)+1.$$
The only remaining case is $a=b=2$, which is easily counted by 'brute force'.
To generalise these results, if you are familiar with some group theory you could try applying Burnside's lemma. You may also want to look at the problem of counting necklaces, which is related.
|
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|
Function extremal - calculus of variations Find a curve passing through (1,2) and (2,4) that is an extremal of the function:
$J(x,y')=\int_1^2 xy'(x)+(y'(x))^2dx$
I don't know what methods to use at all.
|
This is a very simple problem. You are trying to find the function $y$ which gives a minimum of the following integral:
$$
J = \int\left(xy' + y'^2\right)dx = \int\mathcal{L}(y, y', x)dx
$$
Where $y$ is a function of $x$. The very first thing to notice is that $\mathcal{L}$ has no dependence on $y$! Therefore $\frac{\partial \mathcal{L}}{\partial y} = 0$. This leads to the following Euler-Lagrange equation:
$$
\frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial y'}\right) = 0
$$
If the full derivative with respect to $x$ is $0$, then the this partial must be a constant. Let me write this out in two steps:
\begin{align}
\frac{\partial \mathcal{L}}{\partial y'} =& \alpha \\
\frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial y'}\right) =& \frac{d}{dx}\left(\alpha\right) = 0
\end{align}
Now simply set the partial with respect to $y'$ to a constant:
$$
\frac{\partial \mathcal{L}}{\partial y'} = x + 2y' = \alpha
$$
This is a separable equation, just integrate both sides:
$$
2y' = \alpha - x \\
2y = \alpha x - \frac{x^2}{2} + \beta \\
y = \alpha x - \frac{x^2}{4} + \beta \\
y = -\frac{1}{4}\left(x^2 - 2\alpha x\right) + \beta \\
y = -\frac{1}{4}\left(x - \alpha\right)^2 + \beta - \alpha^2 \\
y = -\frac{1}{4}\left(x - \alpha\right)^2 + \beta
$$
Notice that $\alpha$ and $\beta$ changed a lot in the last few steps. Rememeber these are constants, so I can make them whatever I like. Just as one example, in the last step I had $\beta - \alpha^2$ then changed to just $\beta$. This is a constant minus another constant which gives some other constant. That is, the new $\beta$ is whatever the old one was minus $\alpha^2$.
Now find $\alpha$ and $\beta$ for your problem:
$$
y(1) = 2 = -\frac{1}{4}(1 - \alpha)^2 + \beta \\
y(2) = 4 = -\frac{1}{4}(2 - \alpha)^2 + \beta \\
$$
Subtract the equations to find $\alpha$:
$$
2 = -\frac{1}{4}\left((2 - \alpha)^2 - (1 - \alpha)^2\right) \\
-8 = 4 + \alpha^2 - 4\alpha - (1 + \alpha^2 - 2\alpha) \\
-8 = 3 - 2\alpha \rightarrow \alpha = \frac{11}{2}
$$
Now use either equation to find $\beta$:
$$
2 = -\frac{1}{4}\left(1 - \frac{11}{2}\right)^2 + \beta \\
\beta = 2 + \frac{1}{4}\cdot\frac{9^2}{2^2} = \frac{113}{16}
$$
Giving:
$$
y(x) = -\frac{1}{4}\left(x - \frac{11}{2}\right)^2 + \frac{113}{16} = \frac{113 - (2x - 11)^2}{16}\\
y' = -\frac{2x - 11}{4} \rightarrow y'^2 = \frac{(2x - 11)^2}{16}
$$
Finally, if you want to find the value of the integral:
\begin{align}
J =& \int\limits_1^2\left(-x\frac{2x - 11}{4} + \frac{(2x - 11)^2}{16}\right)dx = \frac{1}{16}\int\limits_1^2\left(44x - 8x^2 + (2x - 11)^2\right)dx \\
J =& \frac{1}{16}\left.\left(22x^2 - \frac{8}{3}x^3 + \frac{1}{2\cdot 3}(2x - 11)^3\right)\right|_1^2 \\
J=& \frac{1}{16}\left(22\cdot3 - \frac{8}{3}\cdot7 + \frac{1}{6}\cdot386\right) = \frac{1}{16}\left(\frac{198 - 56 + 193}{3}\right) = \frac{335}{48}
\end{align}
|
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|
How to show that $7\mid a^2+b^2$ implies $7\mid a$ and $7\mid b$? For my proof I distinguished the two possible cases which derive from $7 \mid a^2+b^2$:
Case one: $7\mid a^2$ and $7 \mid b^2$
Case two (which (I think) is not possible): $7$ does not divide $a^2$ and $7$ does not divide $b^2$, but their sum.
I've shown that case $1$ implies $7\mid a$ and $7\mid b$,
so I just have show that case 2 isn't possible
- I'll be happy for any input.
MfG,
Karl
|
a = 7k + m, and b = 7p + n ==> a^2 + b^2 = m^ + n^2 ( mod 7 ).
Case 1: m = 1, n = 1 => m^2 + n^2 = 2. ( mod 7 )
Case 2: m = 1, n = 2 => m^2 + n^2 = 5.
Case 3: m = 1, n = 3 => m^2 + n^2 = 3.
Case 4: m = 1, n = 4 => m^2 + n^2 = 3.
Case 5: m = 1, n = 6 => m^2 + n^2 = 2.
....
All cases yield non-zero mod 7. So it can happen only when both a and b are 0 mod 7.
|
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|
Estimate the error of interpolating ,,, Estimate the error of interpolating (${lnx}$) . at ${x=3}$ with an interpolation polynomial with base points ${x=1 , x=2 , x=4 , x=6 }$ .
|
Well, the polynomial $p$ that you need has
$$
p(1) = ln(1) = 0\\
p(2) = ln(2)\\
p(4) = ln(4) = 2 ln(2) \\
p(6) = ln(6)
$$
Writing
$$
p(x) =
a_1 \frac{(x-2)(x-4)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_2 \frac{(x-1)(x-4)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_3 \frac{(x-1)(x-2)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_4 \frac{(x-1)(x-2)(x-4)}{(x-1)(x-2)(x-4)(x-6)},
$$
you can see, by plugging in $x = 1, 2, 4,$ and $6$ that $a_1 = 0; a_2 = ln(2); a_3 = ln(4)$, and $a_4 = ln(6)$. So your polynomial is
$$
p(x) =
ln(2) \frac{(x-1)(x-4)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
ln(4) \frac{(x-1)(x-2)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
ln(6) \frac{(x-1)(x-2)(x-4)}{(x-1)(x-2)(x-4)(x-6)}.
$$
Plug in $x = 3$, and take the difference from $ln(3)$ and you've got your answer.
|
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|
How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$? I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of the first $9$ positive odd numbers is $1,3,5,7,9,11,13,15,17 = 81 = 9^2$, but where do I go from here.
|
We have, (x+1)^2 - x^2 = 2x + 1
Using this, we can say,
1^2 = 0^2 + 1 = 0 +1 = 1
2^2 = 1^2 + 3 = 1 + 3 = 4
3^2 = 2^2 + 5 = 1 + 3 + 5 = 9 and so on...
This explains why the sum is n^2.
|
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|
One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.
|
I don't think your formula exhaust all solutions of the Diophantine equation
$$2(a^2+b^2+c^2+d^2) = (a+b+c+d)^2\tag{*1}$$
Your equation can be rewritten as
$$(a+b+c+d)^2 = (-a+b-c+d)^2 + (-a+b+c-d)^2 + (-a-b+c+d)^2$$
This is the equation for a Pythagorean quadruple. The set of all Pythagorean quadruples can be parametrized
by 5 integers $\alpha,\beta,\gamma,\delta$ and $\lambda$:
$$\begin{cases}
\hphantom{-}a + b + c + d &= \lambda (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)\\
-a + b - c + d &= \lambda(\alpha^2+\beta^2 - \gamma^2 - \delta^2)\\
-a + b + c - d &= 2\lambda(\alpha\delta + \beta\gamma)\\
-a - b + c + d &= 2\lambda(\beta\delta - \alpha\gamma)
\end{cases}
$$
Using this, we see all solutions of $(*1)$ must have the form
$$
\begin{cases}
a &= \frac{\lambda}{2}
\left(\gamma^2 + \delta^2 + (\alpha-\beta)\gamma - (\alpha+\beta)\delta\right)\\
b &= \frac{\lambda}{2}
\left(\alpha^2 + \beta^2 + (\alpha+\beta)\gamma + (\alpha-\beta)\delta\right)\\
c &= \frac{\lambda}{2}
\left(\gamma^2 +\delta^2 - (\alpha-\beta)\gamma + (\alpha+\beta)\delta\right)\\
d &= \frac{\lambda}{2}
\left(\alpha^2+\beta^2 - (\alpha+\beta)\gamma - (\alpha-\beta)\delta\right)
\end{cases}\tag{*2}
$$
Furthermore, if one substitute any integers $\lambda, \alpha,\beta,\gamma,\delta$
into $(*2)$ and if the resulting $a, b, c, d$ are integers, then it will be a solution
of $(*1)$. It is not hard to check this happens when and only when
$$\lambda(\alpha+\beta+\gamma+\delta)\quad\text{ is an even number }\tag{*3}$$
Conclusion
- all solutions of $(*1)$ can be parametrized by $(*2)$ subject to the constraint $(*3)$.
The formula you have is a special case of above parametrization. It can be reproduced by
following substitutions:
$$\begin{cases}
\lambda &= 1\\
\alpha &= s - p,\\
\beta &= 2(s+p+k),\\
\gamma &= k+p,\\
\delta &= k+s.
\end{cases}$$
|
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|
Solve: $\frac{d^2x}{dt^2}+\Bigl(\frac{d^2y}{dt^2}\Bigr)^2=0$ We have the following coupled Diferential equation:
$\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2$
$\dfrac{d^2y}{dt^2}=\dfrac{y}{x}$
Then find the solution $x$ and $y$ in terms of $t$ .
What we have that
$\dfrac{d^2x}{dt^2}+\left(\dfrac{d^2y}{dt^2}\right)^2=0$
Now how to solve this equation. Please help. Thanks in advance!
|
$\begin{cases}\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2~......(1)\\\dfrac{d^2y}{dt^2}=\dfrac{y}{x}............(2)\end{cases}$
From $(2)$, $x=\dfrac{y}{\dfrac{d^2y}{dt^2}}......(3)$
Put $(3)$ into $(1)$ :
$\dfrac{d^2}{dt^2}\left(\dfrac{y}{\dfrac{d^2y}{dt^2}}\right)=-\dfrac{y^2}{\dfrac{y^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}}$
$\dfrac{d}{dt}\left(\dfrac{\dfrac{dy}{dt}}{\dfrac{d^2y}{dt^2}}-\dfrac{y\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}\right)=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$\dfrac{\dfrac{d^2y}{dt^2}}{\dfrac{d^2y}{dt^2}}-\dfrac{\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}-\dfrac{\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}-\dfrac{y\dfrac{d^4y}{dt^4}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}+\dfrac{2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^3}=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$1-\dfrac{y\dfrac{d^4y}{dt^4}+2\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}+\dfrac{2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^3}=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$\biggl(\dfrac{d^2y}{dt^2}\biggr)^3-y\dfrac{d^2y}{dt^2}\dfrac{d^4y}{dt^4}-2\dfrac{dy}{dt}\dfrac{d^2y}{dt^2}\dfrac{d^3y}{dt^3}+2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^5$
$y\dfrac{d^2y}{dt^2}\dfrac{d^4y}{dt^4}+2\dfrac{dy}{dt}\dfrac{d^2y}{dt^2}\dfrac{d^3y}{dt^3}-2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2-\biggl(\dfrac{d^2y}{dt^2}\biggr)^5-\biggl(\dfrac{d^2y}{dt^2}\biggr)^3=0$
|
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|
prove that $\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}$ if $a+b+c=3$ $a,b,c\in[0,2]$
observation
by triangle inequality $|a|+|b|\ge |a+b| $
$|a-1|+|b-1|+|c-1|\ge |a+b+c-3|$
but $a+b+c=3$
hence
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge \sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2} +0$
hence our question is
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}\ge 3\sqrt{3}$
wlog we try to prove
$\sqrt{4-a^2}\ge\sqrt{3}$
$4-a^2\ge 3 \implies a^2-1\le0 $
$a\in \left [-1,1\right ]$ but $a\in [0,2]$ so $a\in [0,1]$
this would mean the inequality works for $a,b,c\in [0,1]$
$a=b=c=1$
but i need to prove it for $a,b,c\in[0,2]$
for all $ a+b+c=3$
it is evident that $6=a^2+b^2+c^2+2(ab+bc+ac)$
by GM-HM
$\sqrt{ab}\ge \frac{ab}{a+b}$
so
$\sqrt{4-a^2}=\sqrt{(4-a)(4+a)}\ge\dfrac{16-a^2}{8}$
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2} \ge 6-\frac{a^2+b^2+c^2}{8}\ge^{AM-GM} 6-3\dfrac{\sqrt[3]{a^2b^2c^2}}{8}$
|
Here is a simple solution.
First I would like to emphasize that $a,b,c$ must be non-negative for the inequality to hold. Otherwise a counter-example is $(a,b,c)=(-1,2,2)$.
Thus, we need to prove the following inequality for $0\le a,b,c\le 2$:
$$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}.$$
It suffices to show that the following inequality holds for all $a\in [0,2]$:
\begin{equation}
\sqrt{4-a^2} + (\sqrt{3}-1)|a-1| \ge \sqrt{3} + 1 -a \tag{1}
\end{equation}
(Then similarly we get two other inequalities for $b,c$ and taking the sum of these 3 inequalities we get the result.)
Proof for $(1)$:
*
*If $1\le a\le 2$: $(1)$ becomes
\begin{align}
&\sqrt{4-a^2} + (\sqrt{3}-1)(a-1) \ge \sqrt{3} + 1 -a \\
\Longleftrightarrow &\sqrt{4-a^2} \ge \sqrt{3}(2-a)\\
\Longleftrightarrow &4(2-a)(a-1)\ge 0,
\end{align} which is clearly true.
*If $0\le a< 1$: $(1)$ becomes
\begin{align}
&\sqrt{4-a^2} + (\sqrt{3}-1)(1-a) \ge \sqrt{3} + 1 -a \\
\Longleftrightarrow &\sqrt{4-a^2} \ge 2-(2-\sqrt{3})a.
\end{align}
Note that the right hand side of the last inequality is positive, squaring the both sides and taking the difference we get the equivalent inequality $4(2-\sqrt{3})a(1-a)\ge 0$, which is clearly true.
Therefore, $(1)$ is proved and the conclusion follows. Equality holds if and only if $(a,b,c)=(1,1,1)$ or a permutation of $(0,1,2)$.
Remark. According to the proof, we see that the inequality also holds for the weaker condition $a+b+c\le 3$.
|
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|
Confusing rational numbers Question:
If $$x = \frac{4\sqrt{2}}{\sqrt{2}+1}$$ Then find value of, $$\frac{1}{\sqrt{2}}*(\frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x - 2\sqrt{2}})$$
My approach:
I rationalized the value of $x$ to be $8-4\sqrt{2}$, then substituted values to get:
$$\frac{1}{\sqrt{2}}* (\frac{10 - 4\sqrt{2}}{6 - 4\sqrt{2}}+\frac{8-6\sqrt{2}}{8-2\sqrt{2}})$$
and solved until I got:
$$\frac{24-15\sqrt{2}}{8\sqrt{2}-11}$$
But this doesn't seem to please the options.
Can anyone please guide me on how to approach this problem (not all the steps, that would be huge, but the beginning steps or any hints).
Thanks a lot.
|
$$\frac x2=\frac{2\sqrt2}{\sqrt2+1}$$
Applying Componendo & Dividendo, $$\frac{x+2}{x-2}=\frac{2\sqrt2+(\sqrt2+1)}{2\sqrt2-(\sqrt2+1)}=\frac{3\sqrt2+1}{\sqrt2-1}=\frac{(3\sqrt2+1)(\sqrt2+1)}{2-1}$$
$$=7+4\sqrt2$$
Similarly, $$\frac x{2\sqrt2}=\frac2{\sqrt2+1}$$
Apply Componendo & Dividendo again
|
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|
Inequality doubt with taylor expansion Can I prove that $\forall x>0$ $$e^{x/(1+x)} < 1+x$$
Showing that $e^{x/(1+x)} = 1+x-\frac{x^2}{2}+o(x^2)$ and so $-\frac{x^2}{2}+o(x^2)<0$ for all $x>0$? How i can be sure that $o(x^2)$ doesn't interfere in the inequality also for large values of $x$?
|
Or using the Taylor expansion for $e^t$:
$$e^{\frac{x}{x+1}} = 1+ \frac{x}{x+1} + \frac1{2!} \left(\frac{x}{x+1} \right)^2 + \cdots < 1+ \frac{x}{x+1} + \left(\frac{x}{x+1}\right)^2 + \cdots = \frac1{1-\frac{x}{x+1}} = 1+x$$
|
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|
Bounding $\sum_{n=n_1}^\infty x^n (n+1)^2$ I need to upperbound the sum
$$\sum_{n=n_1}^\infty x^n (n+1)^2$$
where $0<x<1$ is a parameter.
I know it can be done starting from
$$\sum_{n=n_1}^\infty x^n (n+1)^2\le \sum_{n=0}^\infty x^n (n+1)(n+2)-\sum_{n=0}^{n_1} x^n (n+1)(n+2)$$ using nasty calculations (integrating twice, then taking the second derivative from the closed form formula of the integral). I was wondering if this can be done in a simpler way.
|
Write
$$
\sum_{n \in S} x^n(n+1)^2
= \sum_{n \in S} \frac{d}{dx} x^{n+1} (n+1)
= \frac{d}{dx} \left( x \sum_{n \in S} x^n (n+1) \right)
= \frac{d}{dx} \left( x \frac{d}{dx} \sum_{n \in S} x^{n+1} \right)
$$
for any index set $S$. In the case that $S = \{n_1,n_1+1,\ldots\}$, this becomes
\begin{align}
\sum_{n = n_1}^\infty x^n(n+1)^2
&= \frac{d}{dx} \left( x \frac{d}{dx} x^{n_1 + 1}\sum_{n =0} x^n \right)
\\&= \frac{d}{dx} \left( x \frac{d}{dx} \frac{x^{n_1 + 1}}{1-x} \right) \\&=
\dfrac{{x}^{n_1}\left( x+{\left( 1+n_1 -n_1x \right) }^{2}\right) }{{\left( 1-x\right) }^{3}}
\end{align}
|
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What converges this series? What exactly converges the series?
$\sum _{k=3}^{\infty \:}\frac{2}{k^2+2k}$
I tried taking out the constant
$=2\sum _{k=3}^{\infty \:}\frac{1}{k^2}$
then $p=2,\:\quad \:p>1\quad \Rightarrow \sum _{k=3}^{\infty \:}\frac{1}{k^2}$
but I really don't know what i'm doing
|
Note that $\dfrac{2}{k^2+2k}=\dfrac{1}{k}-\dfrac{1}{k+2}$. So our sum is
$$\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+ \left(\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{10}\right)+\cdots.$$Remove the parentheses, and note the mass cancellations (telescoping). Almost everything dies, and we are left with $\dfrac{1}{3}+\dfrac{1}{4}$.
Remark: The above argument was informal. It can be replaced by a formal argument in which we take the sum $s_n$ of the first $n$ terms, and show that $\lim_{n\to\infty}s_n=\frac{1}{3}+\frac{1}{4}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding limits of diagonalised matrix I diagonalised
$A=$
$\left[\begin{array}[c]{rr} 0.6 & 0.9\\ 0.4 & 0.1\end{array}\right]$
and got
$SAS^-$$^1$$= V = (-1/13)$
$\left[\begin{array}[c]{rr} -1 & -1\\ -4 & 9\end{array}\right]$
$\left[\begin{array}[c]{rr} 1 & 0\\ 0 & -0.3\end{array}\right]$
$\left[\begin{array}[c]{rr} 9 & 1\\ 4 & -1\end{array}\right]$
now I am supposed to somehow deduce what matrix $V^k$ approaches as $k$ approaches $∞$. How am I supposed to figure that out?
Also, I am supposed to find the limit for $SV^kS^-$$^1$ when $k$ approaches $∞$ and specify what the columns of this matrix portray.
What came to mind is that when all $|λ|<1$ then $A^k$ approaches $0$, and $λ_1=1$ and $λ_2 = -0.3$. I'm not really sure if that's of any use though. I'm somewhat lost and any help is much appreciated!
|
Your solution looks close.
Write the matrix as fractions:
$$A=\left[\begin{array}[c]{rr} \dfrac{6}{10} & \dfrac{9}{10}\\ \dfrac{4}{10} & \dfrac{1}{10}\end{array}\right]$$
I have my eigenvalues and eigenvectors swapped from the order you show yours in.
Diagonalization yields:
$$A = PJP^{-1} =
\left(\begin{array}{cc}
-1 & \frac{9}{4} \\ 1 & 1 \\\end{array} \right)\left(\begin{array}{cc}
-\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right) \left(\begin{array}{cc}
-\frac{4}{13} & \frac{9}{13} \\ \frac{4}{13} & \frac{4}{13} \\ \end{array}
\right)$$
This yields:
$$A^k = PJ^kP^{-1} = \left(
\begin{array}{cc}
\frac{1}{13} 2^{2-k} \left(-\frac{3}{5}\right)^k+\frac{9}{13} & ~\frac{9}{13}-\frac{1}{13} \left(-\frac{1}{10}\right)^k 3^{k+2} \\
\frac{4}{13}-\frac{1}{13} \left(-\frac{3}{5}\right)^k 2^{2-k} & ~\frac{1}{13} 3^{k+2} \left(-\frac{1}{10}\right)^k+\frac{4}{13} \\
\end{array}
\right)$$
Note: Once we have diagonalized the matrix, we have:
$$J^k = \left(\begin{array}{cc}
-\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right)^k = \left(\begin{array}{cc}
\left(-\frac{3}{10}\right)^k & 0 \\ 0 & (1)^k \\\end{array}\right)$$
Now, as far as the limit goes, do you see what happens to the $k$ terms as $k$ approaches infinity? They approach zero, so we are left with:
$$\displaystyle \lim_{k \to \infty} A^k = \left(
\begin{array}{cc}
\frac{9}{13} & ~\frac{9}{13} \\
\frac{4}{13} & ~\frac{4}{13} \\
\end{array}
\right) $$
An easier approach would have been to take the limit of the diagonalized matrix first, and then multiply out, which yields:
$$\displaystyle \lim_{k \to \infty} J^k = \left(
\begin{array}{cc}
0 & 0 \\
0 & 1 \\
\end{array}
\right) $$
|
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|
Given length of two medians and one altitude , find the length of one side. In $\triangle ABC$, altitude $AD = 18$, median $BE = 9\sqrt5$ and median $CF = 15$. Find $BC$.
(Note that I've drawn median AG)
By appolonius theorem ,
$$2(15)^2+ 2x^2=(2y)^2+(2z)^2$$
$$2(9\sqrt5)^2+2y^2=(2x)^2+(2z)^2$$
By herons formula and the normal area ( = 1/2 baseheight),
$$18z=\sqrt(x+y+z)(x+y-z)(x+z-y)(z+y-x)$$
I solved this system using wolfram alpha and it is giving the right answer ($z=10$ so $BC=20$).
But needless to say , it is a very tedious task to solve this system of equations. So a more elegant solution will be apreciated.
|
First, note that your diagram is somewhat inaccurate. The (undrawn) $\overline{EF}$ should be parallel to $\overline{BC}$, and half as long. (This is the "Midpoint Theorem for Triangles".) With that in mind ...
Let $P$ and $Q$ be the feet of perpendiculars from $E$ and $F$ to $\overline{BC}$. Then
$$|\overline{EP}| = |\overline{FQ}| = \frac{1}{2}|\overline{AD}| = 9$$
By the Pythagorean Theorem in $\triangle EPB$ and $\triangle FQC$,
$$\begin{align}
|\overline{EP}|^2 + |\overline{BP}|^2 = |\overline{BE}|^2 \quad &\to \quad |\overline{BP}| = 18 \\
|\overline{FQ}|^2 + |\overline{CQ}|^2 = |\overline{CF}|^2 \quad &\to \quad |\overline{CQ}| = 12
\end{align}$$
Then, because
$$|\overline{BP}| + |\overline{CQ}| = |\overline{BC}| + |\overline{PQ}| = |\overline{BC}| + |\overline{EF}| = |\overline{BC}| + \frac{1}{2}|\overline{BC}| = \frac{3}{2}|\overline{BC}|$$
we have
$$\frac{3}{2}|\overline{BC}| = 18 + 12 = 30 \qquad \to \qquad |\overline{BC}| = 20$$
|
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|
$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} $$
I try by expanding, by subtraction, but no solution /:
|
By Fermat's little theorem, if $n\equiv m\ \ \text{mod}\ (p-1)$, then $a^m \equiv a^n\ \ \text{mod}\ p$.
Now, since any power of two modulo $18$ is one of the six numbers in $S=\{2,4,8,10,14,16\}$, it remains to check that:
$$2^k+3^k+5^k \equiv 0 \ \ \text{mod}\ 19$$
For all $k\in S$.
|
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|
Finding the limit of a function $n^3 / 3^n$ $$\lim_{n→\infty} \frac{n^3}{3^n} =0 $$
The answer is 0 but how would i go about proving that?
|
Let $a_n = \frac{n^3}{3^n} $. Then
$$ \left| \frac{ a_{n+1}}{a_n} \right| = \frac{(n+1)^3}{3^{n+1}} \frac{3^n}{n^3} = \frac{1}{3} \left( \frac{n+1}{n} \right)^3 \to \frac{1}{3} < 1$$
Therefore $a_n \to 0 $
|
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|
Finding volume of convex polyhedron given vertices I am trying to compute the volume of the convex polyhedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,2,0)$, $(0,0,3)$, and $(10,10,10)$. I am supposed to use a triple integral but am struggling with how to set it up.
|
First we can find the equation for the plane that goes through $A = (1,0,0)$, $B = (0,2,0)$, and $C = (0,0,3)$. It is: $x + \dfrac{y}{2} + \dfrac{z}{3} = 1$ or $6x + 3y + 2z - 6 = 0$. Now let $O = (0,0,0)$, and $D = (10, 10, 10)$. We now find the volume $V_1$ of the tetrahedron $OABC$. $V_1 = \frac{1}{6}\cdot OA\cdot OB\cdot OC = \dfrac{1\cdot 2\cdot 3}{6} = 1$. Let $d$ = distance from $O$ to the plane $ABC$, and $S$ = area of triangle $ABC$. Then $3V_1 = S\cdot d$, and $d = |\dfrac{6\cdot 0 + 3\cdot 0 + 2\cdot 0 - 6}{\sqrt{6^2 + 3^2 + 2^2}}| = \dfrac{6}{7} $. So $3\cdot 1 = S\cdot \dfrac{6}{7}$. Thus $S = \dfrac{7}{2}$. Next let $k$ be the distance from $D$ to the plane $ABC$. Then $k = \dfrac{6\cdot 10 + 3\cdot 10 + 2\cdot 10 - 6}{7} = \dfrac{104}{7}$. So let $V_2$ be the volume of the tetrahedron $ABCD$. So $V_2 = \dfrac{S\cdot k}{3} = \dfrac{1}{3}\cdot \dfrac{7}{2}\cdot \dfrac{104}{7} = \dfrac{52}{3}$. So let $V$ be the volume of the polyhedra $OABCD$, then $V = V_1 + V_2 = 1 + \dfrac{52}{3} = \dfrac{55}{3}$.
|
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|
Square root of a Mersenne number is irrational Defining a Mersenne Number like this:
k = $2^n -1$
I have to prove that the square root of a Mersenne number is irrational (has no solution in $\mathbb Q$). I know that it can be proven that the square root of a non-perfect square number is always irrational, but is there a particular proof for a Mersenne Number?
|
In binary, $2^n - 1$ is:
$$
111...1\text{, } n \text{ } 1\text{'s}
$$
We can attempt to construct a number that squared gives this, start with $1\times1 = 1$ (that works). Obviously both numbers must be odd (since $2^n - 1$ is certainly odd). Now let's try to add to this to still get all $1$'s:
\begin{align}
&0011 \\
\times & 0011 \\\hline\\
&0011 \\
+& 0110 \\\hline\\
&1001
\end{align}
We can't possibly put $01$ as a test since this is just $1\times1 = 1$. Does it matter what comes next? What if it should have been:
\begin{align}
&00101 \\
\times & 00101 \\\hline\\
&00101 \\
& 00000 \\
+&10100\\\hline
&11001
\end{align}
Right away, you see that the "second" digit cannot possibly be $0$. This invariably gives $01$ as the last two digits which cannot possibly represent a $2^n - 1$ number. So the $2^\text{nd}$ digit must be $1$, let's try that:
\begin{align}
&000111 \\
\times & 000111 \\\hline\\
&000111 \\
& 001110 \\
+&011100\\\hline
&xxxx01
\end{align}
This should not be surprising because we already showed above that $11\times11 \neq 2^n - 1$. If the next digit cannot be $0$ and it cannot be $1$, then there is no possible next digit and thus $2^n - 1$ must not be a perfect square unless $n = 1$ such that $2^1 - 1 = 1 = 1\times1$.
All of this follows from the fact that $(4x + 1)^2 = 16x^2 + 8x + 1 = 1 \pmod{4}$ and $(4x + 3)^2 = 16x^2 + 24x + 9 = 1 \pmod{4}$ whereas $2^n - 1 = 2^n - 4 + 3 = 4(2^{n - 2} - 1) + 3 = 3 \pmod{4}$ (when $n \geq 2$). Since the modulus isn't equal, the values cannot possibly be equal (unless $x = 0$ and we have $1^2 = 1 = 2^1 - 1$).
(we don't need to check $(4x + 2)^2$ or $(4x + 0)^2$ because these numbers are divisible by $2$ and thus certainly not equal to $2^n - 1$ which is definitely not divisible by $2$)
edit All of the above assumes $n > 0$. You can manually see that if $n=0$ you get $2^0 - 1 =1-1= 0 = 0^2$. For negative $n$, it's trivially true that the square root is irrational since you would get a negative number which would at the very least require an $i$, which is irrational.
|
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|
Separate Into Real and Imaginary Parts Separate the following trigonometric function into Real and Imaginary Parts
$$\tan^{-1}e^{i\theta} $$
or
$$\tan^{-1}(\cos\theta+i\sin\theta)$$
I Have made till here
Assuming $x+iy$ is the final outcome after separation
$$\therefore \tan^{-1}e^{i\theta}=x+iy$$
$$\therefore e^{i\theta}=\tan(x+iy)$$
$$\therefore e^{i\theta}= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy} $$
$$\therefore \cos\theta+i\sin\theta= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy} $$
But How to proceed from here to find the value of $ x $ and $ y $ so as to get $x+iy$
Or is there any other way to get the answer other than the steps that I have used.
|
I have come up with this solution Hope this might be the correct one
Let$$\displaystyle\tan^{-1}e^{i\theta}=x+iy$$
$$e^{i\theta}=\tan(x+iy)$$
$$\cos\theta+i\sin\theta=\tan(x+iy)$$
$$\cos\theta-i\sin\theta=tan(x-iy)$$
Now,$$ \tan2x=tan[(x+iy)+(x-iy)]$$
$$=\frac{\tan(x+iy)+\tan(x-iy)}{1-\tan(x+iy)\tan(x-iy)}$$
$$=\frac{(\cos\theta+i\sin\theta)+(\cos\theta-isin\theta)}{1-(\cos\theta+isin\theta)(\cos\theta-isin\theta)}$$
$$=\frac{2\cos\theta}{1-(\cos^2\theta+isin^2\theta)}=\frac{2\cos\theta}{1-1}=\frac{2cos\theta}{0}=\infty$$
$$\therefore 2x=\frac{\pi}{2}\therefore x=\frac{\pi}{4}$$
Also,
$$ \tan2iy=tan[(x+iy)-(x-iy)]$$
$$=\frac{\tan(x+iy)-\tan(x-iy)}{1+\tan(x+iy)\tan(x-iy)}$$
$$=\frac{(\cos\theta+i\sin\theta)-(\cos\theta-isin\theta)}{1+(\cos\theta+isin\theta)(\cos\theta-isin\theta)}$$
$$=\frac{2i\sin\theta}{1+(\cos^2\theta+isin^2\theta)}=\frac{2i\sin\theta}{2}$$
$$\therefore i\tanh2y=i\sin\theta\therefore \tanh2y=\sin\theta$$
$$\therefore 2y=\tanh^{-1}\sin\theta\therefore y=\frac{1}{2}\tanh^{-1}\sin\theta$$
Real part x=$\frac{\pi}{4}$,
Imaginary part y=$\frac{1}{2}\tanh^{-1}\sin\theta$
|
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|
Show that every nonzero integer has balanced ternary expansion? show that every nonzero integer can be uniquely represented in the form
$e_k3^k + e_{k-1}3^{k-1}+ … + e_13+e_0$ where $e_j= -1, 0, 1$ for $j = 0,1,2,…k$ and $e_k \neq 0$
|
Proof by Strong Induction.
Let $P(n) : n = 3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b^0 \text{ where each b = -1, 0 or 1 }$.
Base Case: P(1) is true. Since $1 = 3^0$.
Assuming $P(1), P(2), \cdots, P(n-1)$ is true to prove it for $n$.
Considering cases:
Case 1: $n = 3m$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0)$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*0$$
Case 2: $n = 3m+1$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0) + 1$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*1$$
Case 3: $n = 3m-1$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0) - 1$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*(-1)$$
So we find that any number can be expressed like a balanced ternary expansion.
|
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|
If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to sub $c$ into both of the equations:
$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality
$c^2+cb+a=c^2+ac+b$
$ => cb+a=ac+b$
$=> b(c-1)=a(c-1)$
which gives me then a=b which is contradictory because the integers are supposed to be distinct.
Update #2
Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?
|
Express $$x^2+bx+a=(x-c)(x-d)$$
$$x^2+ax+b=(x-c)(x-f)$$
Matching the coefficients of the powers of $x$ for both equations, we have
$$c+d=-b$$
$$c+f=-a$$
$$cd=a$$
$$cf=b$$
Solving for $d$ and $f$ in terms of $a$ and $b$ we have
$$d=a,f=b$$
leading to $c$ having only one possible value, which is $1$.
Thus we have $$c+a=-b\Rightarrow a+b+c=0$$
as well as
$$a+b+1=0\Rightarrow a+b = -1$$
Also, $$c(a+c)=a+1=-b$$
Thus, options (1), (2) and (3) are true.
|
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|
Finding all primes $(p,q)$ for perfect squares. Find all prime pairs $(p,q)$ such that $2p-1, 2q-1, 2pq-1$ are all perfect squares.
Source: St.Petersburg Olympiad 2011
I have only found the pair $(5,5)$ so I am thinking that maybe a modulo $5$ approach could work.
|
I'd like to offer another answer based on the unique factorization of $\mathbb{Z}[i]$.
Notice that if we let $a^2=2p-1$, $b^2=2q-1$ and $c^2=2pq-1$, we can write $$p=\left(\frac{a+1}{2}\right)^2+\left(\frac{a-1}{2}\right)^2, q=\left(\frac{b+1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2$$ and $pq=\left(\frac{c+1}{2}\right)^2+\left(\frac{c-1}{2}\right)^2$. This already implies, that $a,b,c$ are odd (otherwise $p,q$ wouldn't be whole numbers) and by Fermat's theorem that $p,q$ are of the form $4k+1$ since they are sums of two squares. (Also this representation is unique).
Now let $\alpha=\frac{a+1}{2}+\frac{a-1}{2}i$, $\beta=\frac{b+1}{2}+\frac{b-1}{2}i$ and $\gamma=\frac{c+1}{2}+\frac{c-1}{2}i$. Then $\alpha$ and $\beta$ are Gaussian primes, since their norm is prime. Also, if we let $\epsilon$ be a unit ($\pm 1$ or $\pm i$) then by unique factorization of Gaussian integers, we can write $\alpha \beta=\epsilon \gamma$.
This equation leads to four cases, with two equations in each case (the real and the imaginary part). Writing out and adding/subtracting to eliminate $c$ gives the following four possibilities:
$$
(a-1)(b-1)-2 = \pm2 \\
(a+1)(b+1)-2 = \pm2
$$
By unique factorization of the integers, this leads to:
$$
a, b = \pm(1, k), \pm(k, 1), \pm(3, 3), \pm(2, 5), \pm(5, 2)
$$
Since $a=\pm1$ or $b=\pm1$ implies that one of $p$ or $q$ equals $1$ and both $a$ and $b$ need to be odd, we are left with $a,b=\pm(3,3)$ which implies $p,q=5,5$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.