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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
|
Because $x^2+3x+1=0$, we have $x^2=-3x-1$ and also $x^2+2x+1=-x$, for $x=a,b$. Hence $$\left(\frac{a}{b+1}\right)^2=\frac{a^2}{(b+1)^2}=\frac{-3a-1}{-b}=\frac{3a+1}{b}$$
By symmetry, the desired expression is $$\frac{3a+1}{b}+\frac{3b+1}{a}=\frac{3a^2+a}{ab}+\frac{3b^2+b}{ab}=\frac{3a^2+a+3b^2+b}{ab}=\frac{3(-3a-1)+a+3(-3b-1)+b}{ab}=\frac{-8(a+b)-6}{ab}$$
Lastly, because $a,b$ are roots of $x^2+3x+1$, we know that $ab=1$ and $a+b=-3$. Plugging this into our final expression gives $$\frac{-8(-3)-6}{1}=18$$
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"timestamp": "2023-03-29T00:00:00",
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|
Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
|
It's probably better to cube $S$:
$$S^3 = (5 + 2\sqrt{13}) + 3\sqrt[3] {5+2 \sqrt {13}}\sqrt[3] {5- 2 \sqrt {13}}\bigg(\sqrt[3] {5+2 \sqrt {13}} + \sqrt[3] {5-2 \sqrt {13}}\bigg) + (5 - 2\sqrt{13})$$
$$= 10 + 3\sqrt[3] {5^2-(2 \sqrt {13})^2}S$$
$$= 10 -9S$$
So $S$ satisfies the equation $S^3 + 9S -10 = 0$. The polynomial $S^3 + 9S -10$ factorizes as $(S-1)(S^2 + S - 10) = 0$, so you expect that $S = 1$.
But you still have to make sure $S$ is not one of the roots of $S^2 + S - 10 = 0$, given by ${\displaystyle -{1 \over 2} \pm {\sqrt{41} \over 2}}$. For this, note that the absolute value of each of these roots is greater than ${\sqrt{41} \over 2} - {1 \over 2} > {\sqrt{36} \over 2} - {1 \over 2} > 2$. On the other hand $\sqrt[3] {5+2 \sqrt {13}}$ is less than $\sqrt[3] {5+2 \sqrt {16}} = \sqrt[3]{13} < 2$. The other term $\sqrt[3] {5-2 \sqrt {13}}$ is negative and of smaller absolute value, so we conclude that $0 < S < 2$, and the only possibility is that $S = 1$.
|
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|
Prove that $\frac{1}{2\pi}\frac{xdy-ydx}{x^2+y^2}$ is closed I would like to prove that $\alpha = \frac{1}{2\pi} \frac{xdy-ydx}{x^2+y^2}$ is a closed differential form on $\mathbb{R}^2-\{0\}$ . However when I apply the external derivative to this expression (and ignore the $\frac{1}{2\pi}\cdot\frac{1}{x^2+y^2}$ factor ), I get:
\begin{equation}
d \alpha = \frac{\partial x}{\partial x}dx\wedge dy +
\frac{\partial x}{\partial y}dy\wedge dy -
\frac{\partial y}{\partial x}dx\wedge dx -
\frac{\partial y}{\partial y}dy\wedge dx
\end{equation}
\begin{equation}
d \alpha = dx\wedge dy - dy\wedge dx
\end{equation}
\begin{equation}
d \alpha = 2 dx\wedge dy
\end{equation}
Which is not closed on $\mathbb{R}^2-\{0\}$. Where is my mistake ?
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Well, let us write $\alpha=f\cdot \omega $ with $f(x,y)=\frac{1}{x^2+y^2}$ and $\omega=xdy-ydx$. Then
\begin{align}
d\alpha=df\wedge \omega+fd\omega&=-\frac{1}{(x^2+y^2)^2}\left(2xdx+2ydy\right)\wedge\omega+
\frac{1}{x^2+y^2}2dx\wedge dy=\\
&=-\frac{1}{(x^2+y^2)^2}\left(2x^2 dx\wedge dy-2y^2dy\wedge dx\right)+
\frac{1}{x^2+y^2}2dx\wedge dy=\\
&=-\frac{1}{(x^2+y^2)^2}\cdot 2\left(x^2+y^2\right) dx\wedge dy+
\frac{1}{x^2+y^2}2dx\wedge dy=
\\
&=0.
\end{align}
|
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|
Telescoping series of form $\sum (an+1) ..(an+k)$ If we are given a series say
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k)$$ we can find a telescoping series by noting that
$(n+1)(n+2)..(n+k)(n+k+1) - n(n+1)..(n+k) = (n+k+1-n) (n+1)...(n+k)$
and hence able to write
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k) = \frac{1}{k+1}\sum_{n=1}^N (n+1)...(n+k+1) - n(n+1)...(n+k)$$
However, for series like $$\sum_{n=1}^N (2n+1)(2n+2) ... (2n+k)$$ or for a general multiple of $n$, such as $$\sum_{n=1}^N (an+1)(an+2) ... (an+k)$$ the same trick to add a term before and after would not work.
Are there any ways to express series like this as a telescoping sum, or failing that, any useful (relatively) closed form identities? If so, a pointer to a reference would be much appreciated!
|
The original identity comes from finding an antidifference for the function $n^{\underline{k}}$. Unfortunately, no such antidifference exists for $(2n)^{\underline{k}}$, due to lack of a "chain rule" for antidifferences. However here's an approach that you might like:
$$a_k=\sum_{n=1}^N(2n+1)(2n+2)\cdots(2n+k)$$
$$b_k=\sum_{n=1}^N (2n)(2n+1)\cdots(2n+k-1)$$
Multiplying the summands of $a_k$ by $(2n+k+1-2n)=k+1$, we get $a_k=\frac{1}{k+1}(a_{k+1}-b_{k+1})$. Multiplying the summands of $b_k$ by $(2n+k-2n+1)=k+1$, we get $b_k=\frac{1}{k+1}(b_{k+1}-\sum_{n=0}^{N-1}(2n+1)(2n+2)\cdots(2n+k+1))=\frac{1}{k+1}(b_{k+1}-a_{k+1}+s_k)$, where $s_k=(2N+1)(2N+2)\cdots(2N+k+1)-(k+1)!$.
Adding, we get $$a_k+b_k=\frac{s_k}{k+1}$$
Replacing $k$ by $k+1$ and rearranging, we get $$-b_{k+1}=a_{k+1}-\frac{s_{k+1}}{k+2}$$
We plug into our above formula for $a_k$ to get $a_k=\frac{1}{k+1}(2a_{k+1}-\frac{s_{k+1}}{k+2})$. Now we may solve for $a_{k+1}$ to get
$$a_{k+1}=\frac{k+1}{2}a_k+\frac{s_{k+1}}{2k+4}$$
which admittedly is no closed form, but it is a first-order recurrence.
|
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Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ I am trying to prove the following inequality
For all positive numbers $a$, $b$ and $c$ we have
$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$
I can probably solve this by reducing it to Schur's inequality.
However, is there any other method?
|
We need to prove that
$$\sum_{cyc}\left(\frac{a^3}{b^2-bc+c^2}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^2+bc-b^2-c^2)}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a+2b-c)-(c-a)(a+2c-b))}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a(a+2b-c}{b^2-bc+c^2}-\frac{b(b+2a-c)}{a^2-ac+c^2}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2((a+b)^3-(a^2+3ab+b^2)c+2(a+b)c^2-c^3)(a^2-ab+b^2)\geq0.$$
Now, since by AM-GM $$(a+b)c^2-(a^2+3ab+b^2)c+(a+b)^3\geq2(a+b)^2c-(a^2+3ab+b^2)c>0,$$
it remains to prove that
$$\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq0.$$
Let $a\geq b\geq c$.
Hence, $$b^2\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq$$
$$\geq b^2(a-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$
$$\geq a^2(b-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$
$$\geq a^2(b-c)^2b^2(a-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b-a)(b^2-bc+c^2)=$$
$$=(a-b)^2(b-c)^2a^2b^2(a+b-c)\geq0.$$
Done!
|
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|
Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\vdots$$
I don't see it's going anywhere. Help appreciated!
|
Let $p=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$
Define :$$x_1=\sqrt{1+2\sqrt{2}}-1$$
$$x_2=\sqrt{1+2\sqrt{2+\sqrt{3}}}-\sqrt{1+2\sqrt{2}}$$
$$.$$
$$.$$
$$x_{n-1}=\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n}}}}-\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n-1}}}}$$
From the summation , we can see that $$p-1=x_1+x_2+....+x_{n-1}$$
$$x_1 \approx 0.956636$$
$$x_2 \approx 0.566284$$
$$x_3 \approx 0.290212$$
$$x_4 \approx 0.141296$$
$$x_5 \approx 0.067556$$
not kosher, but ratio approximation will be $0.4755$
,then $p=\frac{(\sqrt{1+2\sqrt{2}}-1)+1}{0.4755} \approx 3.114$
|
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|
Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$
Question 2. Factor $x^4+(x+y)^4+y^4$
Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\
&=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\
&=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\
&=&(x^2+y^2)^2+(x+y)^4-2x^2y^2
\end{eqnarray}$$
I am stuck in question number 2, I dont know what is next after that line.
|
\begin{equation}
\begin{split}
\ & x^4+y^4+(x+y)^4\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2+2xy)^2\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2)^2+4xy(x^2+y^2)+4x^2y^2\\
\ =& 2((x^2+y^2)^2+x^2y^2+2xy(x^2+y^2))\\
\ =& 2(x^2+y^2+xy)^2
\end{split}
\end{equation}
|
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$a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then find the value of $a+b+c+d$.
I could only get $2$ equations that
$a=2c-b$ and $c=2a-d$.
|
$\bf{My\; Solution::}$ Given $a\;,b$ are the roots of the equation $x^2-2cx-5d=0$ So
$\displaystyle a+b=2c............................(1)\;\;\;\;\;\; ab = -5d.....................(2)$
similarly $c\;,d$ are the roots of the equation $x^2-2ax-5b=0$ So
$\displaystyle c+d=2a............................(3)\;\;\;\;\;\; cd=-5b.....................(4)$
So $a+b+c+d = 2(a+c)............(5)$
Now $\displaystyle \frac{a+b}{c+d}=\frac{2c}{2a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow (a^2-c^2)=(cd-ab)=-5(b-d)$
So $\displaystyle (a+c)\cdot (a-c)=-5(b-d)=-5\left\{(2c-a)-(2a-c)\right\}=-15\left\{c-a\right\}=15(a-c)$
So $\displaystyle (a+c)\cdot (a-c)-15(a-c)=0\Rightarrow (a-c) = 0$ or $(a+c) = 15$
Now $a\neq c\;,$ bcz $a,b,c,d$ are distinct real no.
So $a+c-15=0\Rightarrow a+c = 15$. put into eqn....$(5)$
We get $a+b+c+d = 2(a+c) = 2\cdot 15 = 30\Rightarrow \boxed{\boxed{a+b+c+d = 30}}$
|
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|
Calculating $\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$
Find $$\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$$ where $\lfloor\cdot\rfloor$ is greatest integer function
I am a high school teacher. One of my students came up to ask this limit.
For $\lfloor\frac{\sin x}{x}\rfloor$, I have used $\sin x > x$ using increasing decreasing functions.
I tried to prove $x^2 > \sin x \tan x$ using increasing /decreasing
function but I am not getting it.
|
Let $[.]$ denote GIF.
If $x \to 0$, then $\sin x=x-x^3/6+O(x^5)$ and $\tan x =x+x^3/3+O(x^4)$
then $$f(x)=\left(\frac{x^2}{\sin x \tan x}\right)=\left(\frac{x^2}{x^2(1-x^2/6+...)(1+x^2/3+...)}\right)=\left(\frac{1}{1+x^2/6-x^4/18}\right)$$
Using $(1+z)^k =1+kz+O(z^2)$, we can write
$$1-x^2/6 \le f(x) \le 1-x^2/6+x^4/18 \implies [1-x^2/6] \le [f(x)] \le [1-x^2/6+x^4/18].$$
Hence $$\lim_{x\to 0} [f(x)]=0.$$
|
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|
Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\left(\frac{x+2-4}{x^2-4}\right)=\lim_{x\to2}\left(\frac{x-2}{(x+2)(x-2)}\right)=\lim_{x\to2}\left(\frac{1}{x+2}\right)=\frac14$$
But she gave me the next answer:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&=0
\end{align}$$
I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing?
|
The key here is that you can only break up limits over addition/subtraction/etc when you know that those limits exist.
So, you cannot write
$$
\lim_{x\rightarrow2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)=\lim_{x\rightarrow2}\frac{1}{x-2}-\lim_{x\rightarrow2}\frac{4}{x^2-4},
$$
because these limits are both non-existent. (They each have one-sided limits of $\pm\infty$.)
|
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|
Area enclosed by curves $\; y=x^2;\;\;y^2=2x-x^2$ Find the area enclosed by the curves: $$ y=x^2;\quad y^2=2x-x^2$$
I know how to set-up the problem. I am having difficulty figuring out the integral however. I first found the points of intersection by setting the two equations equal to one another. I got the lower limit to be $0$ and the upper limit to be $1$. Since the second equation is higher than the first, I set up the integral so that the first is being subtracted from the second. Hence, the integral looked like $$\int_{0}^{1} \left(\sqrt{2x-x^2}-x^2\right)\,dx $$ I can't figure out how to integrate that so I can not go further with the problem.
|
You're set up is spot on.
Indeed, we're finding the area bounded below the curve $y = \sqrt{2x - x^2}\,$ and above the curve $y = x^2,\;$ between $x = 0$ and $x = 1$:
So indeed, we need to integrate:
$$\begin{align} I & = \int_0^1 \left(\sqrt{2x - x^2} - x^2\right)\,dx \\ \\
& = \int_0^1 \sqrt{1 - 1 + 2x - x^2} \,dx - \int_0^1 x^2\,dx\\ \\
& = \int_0^1 \sqrt{1 - (x - 1)^2} \,dx - \int_0^1 x^2\,dx
\end{align}$$
Now, for the first integral, we use the trigonometric substitution $(x - 1) = \sin\theta$, so that $dx = \cos\theta\,d\theta$.
Finding the new bounds of integration for the first integral (so we can save ourselves the task of "back substitution" at the very end):
At $x = 0, \sin\theta = -1 \implies \theta = -\pi/2.\;$ At $x = 1, \sin\theta = 0 \implies \theta = 0$.
This gives you, after substituting for the first integral:
$$I = \int_{-\pi/2}^{0} \sqrt{1 - \sin^2 \theta}\cos\theta\,d\theta - \int_0^1 x^2\,dx $$
We will use the identities $$\begin{align}\;1 - \sin^2\theta & = \cos^2 \theta\tag{1} \\ \cos^2 \theta & = \dfrac {1 + \cos (2\theta)}{2}\tag{2}\end{align}$$
$$I =\int_{-\pi/2}^0 \left(\sqrt{\cos^2\theta}\right)\cos\theta\,d\theta = \int_{-\pi/2}^0 \cos^2\theta\,d\theta - \int_0^1 x^2\,dx \tag{1}$$
$$I = \dfrac 12\int_{-\pi/2}^0 \left(1 + \cos(2\theta)\right) \,d\theta - \int_0^1 x^2 \,dx\tag{2}$$
Integration should now be relatively straightforward:
$$I = \left[\dfrac \theta2 + \dfrac 14\sin(2\theta)\right]\Big|_{-\pi/2}^0 \;- \;\dfrac{x^3}3\Big|_0^1\quad = \quad \frac\pi4-\dfrac 13$$
|
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tough algebric problem? I wanted to know how can i prove that if
$xy+yz+zx=1$, then
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}
= \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}}$$
I did let $x=\tan A$, $y=\tan B$, $z=\tan C$
given $xy+yz+zx =1$ we have $\tan A \tan B+ \tan B \tan C+\tan C \tan A=1$
$\tan C(\tan A+\tan B)=1-\tan A \tan B$, or $\tan(A+B)=\tan(\pi/2 -C)$ we have $A+B+C=\pi/2$.
what to do now?
Any help appreciated.
thanks.
|
The accepted answer by `lab bhattacharjee' is perfectly fine. For what it's worth, there's a nice geometric view that might enrich this nearly-4-year-old post a tiny bit.
The identity to be proven is, given $\alpha + \beta + \gamma = \frac{\pi}2$, $$ \frac12 \big( \sin 2\alpha + \sin 2\beta + \sin 2\gamma \big) = 2 \cos\alpha \cdot \cos\beta \cdot\cos\gamma$$
Consider an acute triangle $\triangle MNP$ as shown in the figure, with its circumcenter denoted as point $O$. Without loss of generality, take the circumcircle as the unit circle. That is,
$$\text{diameter}~~ \overline{PP'} = 2 \qquad \text{radius}~~ \overline{OM} = \overline{ON} = \overline{OP} = 1$$
By definition, all angles are acute so that any one of them can be split into two parts, say, $\angle PMN = \alpha + \beta < \frac{\pi}2$, with a remaining complement $\gamma > 0$, as in $$\alpha + \beta + \gamma = \frac{\pi}2$$
The desired identity comes from calculating the area $\mathcal{A} \equiv |\triangle MNP|$ in two different ways:
$$\begin{align}
&\text{RHS}: & \mathcal{A} &= \frac12 \cdot \overline{MP} \cdot \overline{MN} \cdot \sin(\angle PMN ) \\
&\text{LHS}: & \mathcal{A} &= |\triangle OMP| + |\triangle ONM| +|\triangle OPN|
\end{align}$$
For the RHS, we have $$\overline{MP} = 2\cos\alpha \qquad \overline{MN} = 2\cos\beta$$ since point $O$ is the circumcenter (e.g. $\overline{MO} = 1$ and $\triangle MOP$ is an isosceles, $\overline{MP}$ is bisected by the height to point $O$). That is,
$$\begin{align}
&\text{RHS}: & \mathcal{A} &= \frac12 \cdot \overline{MP} \cdot \overline{MN} \cdot \sin(\angle PMN ) \\
&& &= \frac12 \cdot 2\cos\alpha \cdot 2\cos\beta \cdot \sin(\alpha + \beta) \\
&& &= 2\cos\alpha \cdot \cos\beta \cdot \sin(\frac{\pi}2 - \gamma) \\
&& &= 2\cos\alpha \cdot \cos\beta \cdot\cos\gamma
\end{align}$$
Now, for the LHS we again apply the Law of Sines to each of the small triangles. For example,
$$\begin{align}
\triangle{NOP} &= \frac12 \cdot \overline{ON} \cdot \overline{OP} \cdot \sin(\angle NOP) \\
&= \frac12 \cdot 1 \cdot 1 \cdot \sin(\pi - \angle NOP') \\
&= \frac12 \sin(\pi - 2\gamma)
\end{align}$$
since $\angle NOP'$ is the central angle for the inscribed angle $\gamma$ (and $\triangle PMN$ is acute so $O$ is "inside"). Therefore, we have
$$\triangle{NOP} = \frac12 \sin 2\gamma \qquad \text{and similarly} \qquad \triangle{POM} = \frac12 \sin 2\alpha \quad \triangle{MON} = \frac12 \sin 2\beta$$
which sum to the LHS of the desired identity. Note how this proof (splitting a triangle into 3 inner ones) naturally exhibits the cyclic nature (3-fold symmetry) of the identity. $~~~~~$Q.E.D.
When the triangle is obtuse, the proof is similar: one of $\alpha, \beta, \gamma$ is negative, which corresponding 'inner' triangle becomes `outer', and its effective area is also negative). For the case where two of $\alpha, \beta, \gamma$ are negative, unfortunately the notion of a triangle breaks down (unless one considers projective geometry, I guess).
P.S.
I definitely have seen this triangle-splitting analysis before at various places, with respect to any general point and not just the circumcenter. The corresponding algebraic expression involves more than 3 angles.
|
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|
Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
|
Combinatorial argument:
Take $n$ pairs of people, that is $2n$ people total, and then you can pick:
*
*one person from each pair in $2^n$ ways,
*any $n$ individuals in $\binom{2n}{n}$ ways,
*any subset of those people in $2^{2n}$ ways.
I hope this helps ;-)
|
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|
At least one member of a pythagorean triple is even
I am required to prove that if $a$, $b$, and $c$ are integers such that $a^2 + b^2 = c^2$, then at least one of $a$ and $b$ is even. A hint has been provided to use contradiction.
I reasoned as follows, but drew a blank in no time:
Let us instead assume that both $a$ and $b$ are odd. This means that $a^2$ and $b^2$ are odd, which means that $c^2$ is even. Thus, $c$ is even. But this is not a contradiction -- it's the sum of two odd numbers, after all.
|
Let us try to write this using congruences, which is a very useful and compact notation.
We know that a square can only have remainder $0$ or $1$ modulo $4$. I.e., for any integer $x$ one of these two congruences must be true: $x^2\equiv 0 \pmod 4$ or $x^2\equiv1 \pmod4$. The first case happens if $x$ is even, the second case if $x$ is odd. (Since $(2k+1)^2=4(k^2+k)+1$ and $(2k)^2=4k^2$.)
So we have two possibilities for the remainder of $a^2$ and two possibilities for the remainder of $b^2$ modulo $4$.
$$
a^2 \equiv 0 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 0+0=0 \pmod4\\
a^2 \equiv 1 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 1+0=1 \pmod4\\
a^2 \equiv 0 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 0+1=0 \pmod4\\
a^2 \equiv 1 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 1+1=2 \pmod4
$$
We see that in the last case we would have $c^2\equiv 2\pmod4$, which is not possible.
So only the first three cases can really occur. In the other words, either all of the numbers $a$, $b$, $c$ are even, or exactly one of the is even and the remaining two numbers are odd.
|
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|
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
|
$$\cos \alpha + \sin \alpha = \sqrt{2} (\frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}) = \sqrt{2}(\sin \frac{\pi}{4} \cos\alpha + \cos \frac{\pi}{4} \sin \alpha)
= \sqrt{2}\sin(\frac{\pi}{4} + \alpha) = .2 $$
Taking the inverse sin of each side yields
$$ \alpha = 2.2143 + 2 \pi n_1 \mid n_1 \in \mathbb{Z} \,\,\,\text{or}\,\,\, \alpha = 2\pi n_2 - .643501 \mid n_2 \in \mathbb{Z}$$
|
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|
How to factor $8xy^3+8x^2-8x^3y-8y^2$ How can I factor $8xy^3+8x^2-8x^3y-8y^2$ or the different form $2x(4y^3+4x)-2y(4x^3+4y)$
Is there any general methods that work?
A possible solution should be $8(x^2-y^2)(1-xy)$ But please do not start from here as in the general case I will not know the answer...
Thanks!
Alexander
|
First take out $8$ as factor
$8xy^3+8x^2-8x^3y-8y^2=8(xy^3-y^2+x^2-x^3y)$
Now,
$xy^3-y^2+x^2-x^3y=y^2(xy-1)-x^2(xy-1)=(xy-1)(y^2-x^2)=(xy-1)(y+x)(y-x)$
|
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|
Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
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We have $$\frac{a^2+b^2}{2ab}=\frac31$$
Applying componendo and dividendo, $$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{3+1}{3-1}$$
$$\implies \left(\frac{a+b}{a-b}\right)^2=2$$
|
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|
If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?
If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?
I tried to add $-16a^2$ to both sides and doing some algebra but it didnt help much.
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Here's how to find the factorization, it is very similar to completing the square, we have $$\begin{align*}a^4 + 64 &= a^4 + 16a^2 + 64 - 16a^2\\ &= (a^2 + 8)^2 - 16a^2 \\ &= (a^2 - 4a + 8)(a^2 + 4a + 8)\end{align*}$$
See also http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity
|
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|
show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
|
Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness.
Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$
So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\infty} \frac{3 \sin x - \sin (3x)}{4x^3}\,{\rm d}x. $$Let $0 < r < R$. Consider $\gamma_1$ the line segment joining $r$ to $R$, $\gamma_2$ the circular arc oriented counterclockwise joining $R$ to $-R$, $\gamma_3$ the line segment joining $-R$ to $-r$ and $\gamma_4$ the circular arc oriented clockwise joining $-r$ to $r$. Consider $\gamma = \gamma_1 \ast \gamma_2 \ast \gamma_3 \ast \gamma_4$ the concatenation. Sketch:
Consider the function: $$ f(z) = \frac{3e^{iz}-e^{3iz}}{4z^3}. $$The only singularity is $z = 0$ (triple pole). Expanding in Laurent: $$ f(z) = \frac{1}{4z^3}\left(3\sum_{n \geq 0}\frac{i^nz^n}{n!} - \sum_{n \geq 0}\frac{3^ni^nz^n}{n!}\right) = \sum_{n \geq 0} \left(\frac{(3-3^n)i^n}{4\cdot n!}\right)z^{n-3}. $$ Since $f$ is holomorphic in $\gamma$ and inside it, by Cauchy-Goursat we get: $$ \oint_{\gamma} f(z)\,{\rm d}z = \int_{\gamma_1} f(z)\,{\rm d}z + \int_{\gamma_2} f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z + \int_{\gamma_4}f(z)\,{\rm d}z = 0.$$ Let's analyze everything sistematically.
Parametrizing $\gamma_1(x) = x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_1}f(z)\,{\rm d}z = \int_r^R \frac{3e^{ix}-e^{3ix}}{4x^3}\,{\rm d}x. $$For $\gamma_2$, we have that its lenght is $\pi R$, $|e^{iz}| = e^{{\rm Re}(iz)} = e^{-{\rm Im}(z)} < 1$, and similarly $|e^{3iz}| < 1$, since for all $z$ in $\gamma_2$ we have ${\rm Im}(z) > 0$. Hence: $$ \left|\int_{\gamma_2} f(z)\,{\rm d}z\right| \leq \frac{\pi R(3+4)}{4 R^3} = \frac{\pi}{R^2} \stackrel{R\, \to \,+\infty}{\longrightarrow} 0. $$
Parametrizing $\gamma_3^-(x) = -x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_3}f(z)\,{\rm d}z =- \int_r^R \frac{3e^{-ix}-e^{-3ix}}{4x^3}\,{\rm d}x, $$once the signs in $(-x^3) = -x^3$ and ${\rm d}z = -{\rm d}x$ cancel each other. Notice here that: $$ \int_{\gamma_1}f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z = 2i\int_r^R \left(\frac{\sin x}{x}\right)^3\,{\rm d}x.$$
For $\gamma_4$, we have: $$\begin{align} \int_{\gamma_4}f(z)\,{\rm d}z &= \int_{\gamma_4} \sum_{n \geq 0}\left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \\ &= \int_{\gamma_4}\frac{1}{2z^3}\,{\rm d}z + \int_{\gamma_4}\frac{3}{4z}\,{\rm d}z + \int_{\gamma_4} \sum_{n \geq 3} \left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \end{align}$$
We have: $$ \int_{\gamma_4} \frac{1}{2z^3}\,{\rm d}z = -\frac{1}{4z^2}\Bigg|_{-r}^{r} = 0, \quad \int_{\gamma_4} \frac{3}{4z}\,{\rm d}z = \frac{3}{4}\int_{\gamma_4} -i \frac{{\rm d}z}{-iz} = \frac{3}{4}\int_0^\pi -i\,{\rm d}t = -\frac{3\pi i}{4}, $$ and: $$ \left|\int_{\gamma_4} \sum_{n \geq 3}\frac{(3-3^n)i^n}{4\cdot n!}z^{n-3}\right| \leq \pi r \sum_{n \geq 3}\frac{3^n-3}{4\cdot n!}r^{n-3} = \sum_{n \geq 3}\frac{\pi(3^n-3)}{4 \cdot n!}r^{n-2} \stackrel{r \to 0}{\longrightarrow} 0. $$
Making first $r \to 0$, and then $R \to +\infty$, in $\oint_\gamma f = 0$, we get: $$ 2i\int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x - \frac{3\pi i}{4} = 0 \implies \int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \frac{3\pi}{8}, $$as desired.
|
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|
Providing a closed formula for a linear recursive sequence I am studying for an exam in linear algebra and I have some trouble solving the following:
Let $(a_n)$ be a linear recursive sequence in $\mathbb{Z}_5$ with
\begin{align}
a_0 = 2, a_1 = 1, a_2 = 0 \text{ and } a_{n+3} = 2a_{n+2} + a_{n+1} + 3a_n \text{ for } n \in \mathbb{N}_0
\end{align}
Provide a closed formula for $a_n$.
My findings so far:
$
\begin{pmatrix} a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ \end{pmatrix}
= \begin{pmatrix} 2 & 1 & 3 \\1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_{n} \\ \end{pmatrix}
$
$
\begin{pmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \\ \end{pmatrix}
= A^n \cdot \begin{pmatrix} a_{0} \\ a_{1} \\ a_{2} \\ \end{pmatrix} = A^n \cdot \begin{pmatrix} 2 \\ 1 \\ 0 \\ \end{pmatrix}
$
At some point there must hold the following:
\begin{align}
T^{-1}AT = D = \begin{pmatrix} c_1 & 0 & 0 \\0 & c_2 & 0 \\ 0 & 0 & c_3 \\ \end{pmatrix}
\end{align}
such that D is a diagonal matrix containing all Eigenwerte $c_i$, $i \in {\{1,2,3} \}$ of $A$.
How shall I continue to find the matrix $A$?
|
$$\begin{pmatrix}a_{n+3}\\a_{n+2}\\a_{n+1}\end{pmatrix}=\begin{pmatrix}2&1&3\\1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}a_{n+2}\\a_{n+1}\\a_n\end{pmatrix}=\ldots=\begin{pmatrix}2&1&3\\1&0&0\\0&1&0\end{pmatrix}^n\begin{pmatrix}2\\1\\0\end{pmatrix}$$
Now:
$$\det(tI-A)=\begin{vmatrix}t-2&-1&-3\\
-1&t&0\\
0&-1&t\end{vmatrix}=t^2(x-2)-3-t=(t^2-1)(t+3)\in\left(\Bbb Z/5\Bbb Z\right)[x]$$
Thus, the eigenvalues of the matrix are $\,-1,1,-3 = 1,2,4\pmod 5\;$:
$$\begin{align*}\lambda=1:& \implies \begin{cases}4x-y-3z=0\\{}\\4x+y=0\\{}\\4y+z=0\end{cases}\implies z=y=x\implies& \begin{pmatrix}1\\1\\1\end{pmatrix}\\{}\\
\lambda=2:&\implies\begin{cases}-y-3z=0\\{}\\-x+2y=0\\{}\\-y+2z=0\end{cases}\implies z\;,\;y=2z\;,\;x=4z\implies&\begin{pmatrix}4\\2\\1\end{pmatrix}\\{}\\
\lambda=4:&\implies\begin{cases}2x-y-3z=0\\{}\\-x+4y=0\\{}\\-y+4z=0\end{cases}\implies z\;,\;y=4z\;,\;x=3z\implies&\begin{pmatrix}3\\4\\1\end{pmatrix}\end{align*}$$
The above are corresponding eigenvectors to the corresponding eigenvalues, so now form the matrix:
$$P=\begin{pmatrix}1&4&3\\1&2&4\\1&1&1\end{pmatrix}$$
Find the matrix's inverse and then you'l get
$$P^{-1}AP=\begin{pmatrix}1&0&0\\
0&2&0\\0&0&4\end{pmatrix}=:D\implies P^{-1}A^nP=D^n $$
and this way you'll be able to solve your problem since exponentiating a diagonal matrix is a piece of cake... and don't forget to carry on arithmetic modulo $\,5\,$ all along!
|
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Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ Problem :
Find the sum of :
$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$
My approach :
Here the $n$'th term is given by :
$$t_n = \sin^{-1}\left[\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right]$$
From now how to proceed further please suggest thanks....
|
This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.
Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:
*
*$\frac12 |BC||BD| \sin\theta = \frac12 \sqrt{n+1} \sqrt{n} \sin\theta$
*$\frac12 |CD||AB| = \frac12 ( \sqrt{n} - \sqrt{n-1} )$
Equate them gives us:
$$\sin \theta = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}$$
On the other hand,
$$\begin{align}
\theta &= \measuredangle CBD = \measuredangle CBA - \measuredangle DBA\\
&= \sin^{-1}\frac{|AC|}{|BC|} - \sin^{-1}\frac{|AD|}{|BD|}
= \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}
\end{align}$$
We obtain
$$\sin^{-1}\left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right)
= \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}$$
and we have turned the original series into a telescoping one.
The partial sum of the first $n$ terms of original series becomes the
angle $\measuredangle ABC$.
When $n \to \infty$, the line $BC$ becomes vertical and this geometrically justify why
the limit of the series is $\frac{\pi}{2}$.
|
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|
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
|
In the very first line, you assumed that $a=|a|=\sqrt{a^2}.$ But this is not true in general. In particular, it is false for negative $a.$ And in fact, $$4-\frac92<0,$$ so in this case we do not have that $$4-\frac92=\sqrt{\left(4-\frac92\right)^2}=\left|4-\frac92\right|$$ as assumed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.
$$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$
I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:
Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$
and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$
Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)<x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:
For $n=1$, $a_1(x)=\sqrt{1+x}<1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n>1$.
Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:
Claim: $l(x)=x+1\quad \forall i\ge 0$
Proof: Fix $x$. Let $l(x)<x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon>0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)<\epsilon$$, and if that is true then $$a_n(x)>x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) < \epsilon$$ Now, \begin{align}
x+1-a_n(x) = & x+1-\sqrt{1+xa_{n-1}(x+1)} \\
\ =& \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\
\ =& x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\
\ <& \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\
\ <& \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\
\ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\
\ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\
\ <& \frac{x(x+1)}{x+n-1}
\end{align}
Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}<\epsilon
\Rightarrow x< \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.
Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved.
|
$x>-1\iff \underline{x+1}=\sqrt{(x+1)^2}=\sqrt{1+2x+x^2}=\sqrt{1+x\cdot(\underline{\underline{x+2}})}$
$\begin{align}x>-2\iff \underline{\underline{x+2}}=\sqrt{(x+2)^2}=\sqrt{[(x+1)+1]^2}&=\sqrt{1+2(x+1)+(x+1)^2}=\\&=\sqrt{1+(x+1)(\underline{\underline{\underline{x+3}}})}\end{align}$
$\to x+1=\sqrt{1+x\sqrt{1+(x+1)(\underline{\underline{x+3}})}}\quad-\quad$ Can you see where this is going ? :-)
|
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|
Show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?
I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?
|
Hint: Suppose gcd(a,b)=1 and let $d=gcd(a+b,a^2+b^2) \implies d|(a+b) $ and $ d|(a^2+b^2)$. Let $dr=a+b$ and $ds=a^2+b^2$ where $r,s \in\mathbb{Z}$. We see that by squaring $dr=a+b$ we get $d^2r^2=a^2+2ab+b^2$. Then $d^2r^2-d=a^2+2ab+b^2-a^2-b^2=2ab$. Thus $d(dr^2-1)=2ab\implies d|2ab$ From this we break the proof into two cases where $d$ is odd and $d$ is even.
Case:1.(We want to somehow show d=1). Suppose d is odd then $gcd(d,2)=1$. It follows $d|ab \implies d|a$ or $d|b$. If $d|a$ and since $d|(a+b)$. We see that since $d|(a+b)-d|a=d|b$ and since we assumed $gcd(a,b)=1$ it follows d=1. If $d|b$ then we see that $d|(a+b)-d|b=d|a$. Similarly $d=1$.
Case 2: Suppose d is even and in this case somehow show $d=2$.
|
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|
Geometry problem on circles from a competition Triangle $\triangle ABC$ is an equilateral triangle whose side is $16$. A circle meets the sides of the triangle at $6$ points:
*
*it intersects $AC$ at $G$ and $F$ and $|AG|=2$, $|GF|=13$, $|FC|=1$.
*it intersects $AB$ at $H$ and $J$ and $|AH|=3$, $|HJ|=7$, $|JB|=6$.
*it intersects $BC$ at $D$ and $E$.
Find $|DE|$ using plane geometry concepts.
|
Suppose the triangle is drawn with $A$ as the topmost point, $B$ as the left base point and $C$ as the right base point.
The equation for the circle is $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0, y_0)$ is the centre coordinates and $r$ is the radius. Assume a coordinate system with the origin at $B$. Now the coordinates for the points $G$, $H$ and $J$ are found to be $(9,7 \sqrt{3})$, $(13/2, 13 \sqrt{3}/2)$ and $(3, 7 \sqrt{3})$ respectively. Putting these numbers in the equation for the circle gives you the following equation system:
$$\left\{\begin{array}{l}
(7 \sqrt{3} - y_0)^2 + (9 - x_0)^2 = r^2 \\
(13 \sqrt{3} / 2 - y_0)^2 + (13/2 - x_0)^2 = r^2 \\
(3 \sqrt{3} - y_0)^2 + (3 - x_0)^2 = r^2
\end{array}\right.$$
The solution (with positive $r$) is $x_0 = 10$, $y_0 = 3 \sqrt{3}$ and $r=7$.
Note that circle is completely defined by three points only. By plugging in the coordinates of the point $F$ into the circle equation one can verify that the above solution is correct (assuming the problem makers have done their job).
Now, $|DE|$ is simply the difference between the two solutions of the resulting circle equation when $y=0$, i.e.
$$|DE| = 2 \sqrt{r^2-y_0^2} = 2 \sqrt{22}$$
|
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|
Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what should I do to factorize this polynomial ?
(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)
|
A way to do it is to write $(x+2)(x+3) = x^2 + 6x + 6 - x$, $(x+1)(x+6) = x^2 + 6x + 6 + x$, so
$$
(x+2)(x+3)(x+1)(x+6) = (x^2 + 6x + 6)^2 -x^2
$$
which gives that
$$
(x+2)(x+3)(x+1)(x+6) - 3x^2 = (x^2 + 6x + 6)^2 -4x^2 = (x^2+ 4x + 6)(x^2 +8x + 6).
$$
|
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|
Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3}$
|
This is a little bit not fun, but will give your algebra a good workout. First note that
$$(x+h)^{4/3}-x^{4/3}=\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{2/3}-x^{2/3}\right)
.\tag{1}$$
and the right-hand side of (1) can be written as
$$\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right)
.\tag{2}$$
We have used the identity $a^2-b^2=(a+b)(a-b)$ a couple of times.
So we want to find
$$\lim_{h\to 0}\frac{\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right)}{h}.\tag{3}$$
The first two terms in the numerator of (3) behave very nicely as $h\to 0$. We need to worry only about
$$\frac{(x+h)^{1/3}-x^1/3}{h}.\tag{4}$$
Now we use the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$. Letting $a=(x+h)^{1/3}$ and $b=x^{1/3}$ we multiply top and bottom of (4) by $a^2+ab+b^2$. We get
$$\frac{(x+h)-x}{h\left((x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3}\right)}.$$
Now everything is under control, you can put the pieces together.
Remark: There are many other ways to do the algebra. For example instead of the initial factoring that we did, you could immediately multiply top and bottom by
$$(x+h)^{8/3}+(x+h)^{4/3}x^{4/3}+x^{8/3}.$$
Then on top we end up with $(x+h)^4-x^4$. We can expand this as usual and get a simpler-looking path to the answer.
|
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|
How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
|
My solution:
I. $\sin x\sin2x\sin 3x<0=>$
$ 1 < 1 - \sin x\sin2x\sin 3x = \cos x\cos 2x\cos 3x\leq 1$ False!
II. $\sin x\sin2x\sin 3x = 0 => \cos x\cos 2x\cos 3x\ = 1 =>... x=n\pi, n$ integer
III. $\sin x\sin2x\sin 3x>0=>$
$ 1 = |\sin x\sin2x\sin 3x + \cos x\cos 2x\cos 3x|\leq$
$ |\sin x\sin2x\sin3x| + |\cos x\cos 2x\cos 3x|=$
$ |\sin x\sin2x||\sin3x| + |\cos x\cos 2x||\cos 3x|\leq$
$ \leq|\sin x\sin 2x| + |\cos x\cos 2x| =\pm\sin x\sin2x \pm \cos x\cos 2x$= $\pm\cos (x\pm2x)\leq 1$.
a) $\cos x =\pm1 =>\sin x=0$ False!
b) $\cos 3x =\pm1 =>\sin 3x=0$ False!
|
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|
Help with this inequality I am given four numbers $a,b,c,d$, such that $c>a>b,c>d>b$ and $0 \le a,b,c,d\le 1$ Can the following two inequalities hold strictly
$ad\le bc$ and $(1-a)(1-d)\le (1-b)(1-c)$.
|
My interpretation of your question is that you want to find a set of 4 real numbers such that $0 \leq a, b, c, d \leq 1$, $c > d > b$, $c > a > b$, $ad < bc$, $(1-a)(1-d) < (1-b)(1-c)$
If so, then no set of numbers exist.
Consider $[\frac{1}{c^2} - \frac{1}{c} ] (bc) + [\frac{1}{c}] ( 1-b)(1-c) + [ 1 - \frac{1}{c}] \\= \frac{1}{c^2} (bc) - \frac{1}{c} b - \frac{1}{c} c + 1 = 0 $
Note that the coefficients of $(bc)$ and $(1-b)(1-c)$ are positive. Hence, applying the inequalities, we get that
$ 0 > [\frac{1}{c^2} - \frac{1}{c} ] (ad) + [\frac{1}{c}] (1-a)(1-d) + [ 1 - \frac{1}{c}] \\= \frac{1}{c^2}(ad) - \frac{1}{c} a - \frac{1}{c} d + 1 \\= ( \frac{a}{c} -1) ( \frac{d}{c} -1 ) $
However, the terms on the RHS are both negative, which is a contradiction.
|
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|
Can someone explain this trigonometric limit? I have
$$\lim \limits_{x\to 0} \frac {\tan(2x)}{\sin(x)}$$ and in my case the result is $\frac{2}{1}$ =2 not whether it is right.
This is my procedure.
$$\lim \limits_{x\to 0} \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0} {\dfrac {\sin(2x)}{(\cos(2x))(\sin(x))}}=\dfrac{2x\frac {\sin(2x)}{2x}}{\cos(2x)\frac{x\sin(x)}{x}}$$
I separate the limit.
$$\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2$$
|
That does give you the correct answer, but it takes a bit more work than necessary.
An alternative method is to use the double-angle formula for $\sin$. That is:
$$\sin(2x) = 2\cos(x)\sin(x)$$
Thus:
$$\begin{align}\require{cancel}
\lim_{x\to0} \frac{\tan(2x)}{\sin(x)} &= \lim_{x\to0} \frac{\sin(2x)}{\cos(2x)\sin(x)} \\
&= \lim_{x\to0} \frac{2\cos(x)\cancel{\sin(x)}}{\cos(2x)\cancel{\sin(x)}} \\
&= \lim_{x\to0} \frac{2\color{blue}{\cos(x)}}{\color{red}{\cos(2x)}} \\
&= \frac{2\cdot \color{blue}{1}}{\color{red}{1}} \\
&= \boxed2
\end{align}$$
|
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|
Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
|
HINT:
$$ \lim_{x\to -\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} =\lim_{x\to +\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \lim_{x\to +\infty}\frac{\sqrt{(\frac{1}{x^2}+4)}}{\frac{8}{x^2}+1}$$
|
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|
Solve the equation for x, y and z: $\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$ I am having some trouble with this problem,
Solve for $x,y,$ and $z$.
$$\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$$
Here is my work so far,
$$x - y +z = x+y+z-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy}$$
$$2y-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy} = 0 $$
$$2(y-\sqrt{xy} + \sqrt{xz} - \sqrt{zy}) = 0 $$
$$y-\sqrt{xy} + \sqrt{xz} - \sqrt{zy} = 0$$
|
Your last equation can be written as
$$ \sqrt{zx} = \sqrt{y} \left[ \sqrt{x} - \sqrt{y} + \sqrt{z} \right] = \sqrt{y} \sqrt{ x-y+z} $$
Squaring both sides gives us
$$ zx = xy - y^2 + zy $$
Which simplifies to $$(y-x)(y-z) = 0 $$
Hence, we require $x=y$ or $y=z$. It is clear that in either case, the equation is satisfied.
|
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|
Solving recurrence equation using generating function method I've been trying to solve the following equation intuitively (I only know the method if there are minuses in the equation - $a_{n-1}, a_{n-2}...$).
$$a_{n+2}=4a_{n+1}-4a_{n}$$
$$a_{0}=3$$
$$a_{1}=8$$
$$
\begin{align}
A(x)&=\sum\limits_{n>=0}a_{n}x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1}-\frac{1}{4}a_{n+2})x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1})x^{n}-\frac{1}{4}\sum\limits_{n>=0}(a_{n+2})x^{n} \\
&= \sum\limits_{n>=1}(a_{n})x^{n+1}-\frac{1}{4}\sum\limits_{n>=2}(a_{n})x^{n+2} \\
&= \frac{1}{x}\sum\limits_{n>=1}(a_{n})x^{n}-\frac{1}{4x^{2}}\sum\limits_{n>=2}(a_{n})x^{n} \\
&= \frac{1}{x}[\sum\limits_{n>=0}(a_{n})x^{n} - 3]-\frac{1}{4x^{2}}[\sum\limits_{n>=0}(a_{n})x^{n} - 3 - 8x] \\
&= \frac{1}{x}[A(x) - 3]-\frac{1}{4x^{2}}[A(x) - 3 - 8x]
\end{align}
$$
So I get
$$A(x)=\frac{-4x+3}{4x^{2}-4x+1}$$
Is this correct? I'm asking because the answer to this question according to the source I got it from is $x^{2}-4x+4$ as the denominator...
|
Just another approach for verification.
Assume
$$
a_{n+2}=4a_{n+1}-4a_{n}
$$
Then
$$
\begin{align}
f(x)&=\sum_{k=0}^\infty a_kx^k\\
xf(x)&=\sum_{k=1}^\infty a_{k-1}x^k\\
x^2f(x)&=\sum_{k=2}^\infty a_{k-2}x^k\\
\end{align}
$$
Then we get
$$
\begin{align}
f(x)(1-4x+4x^2)&=a_0+a_1x-4a_0x+\sum_{k=2}^\infty(a_k-4a_{k-1}+4a_{k-2})x^k\\
&=a_0+(a_1-4a_0)x\\[18pt]
f(x)&=\frac{a_0+(a_1-4a_0)x}{1-4x+4x^2}\\
&=\frac{3-4x}{1-4x+4x^2}
\end{align}
$$
|
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|
question about applying a comparison test to a sequence In lecture, we were asked:
Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$ converge?
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$
In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosing a bound for $b_n$. The strategy is to choose an expression larger in the numerator and smaller in the denominator for $b_n$ If I understand correctly, using $2n^5$ in the denominator for $b_n$ doesn't work, because, as n gets large, we have
$$ \frac{2n^3}{n^5} \not < \frac{3n^3}{2n^5}$$
$$ 2 \not < \frac{3}{2} $$
If this is the case, couldn't we choose $b_n = \frac{10n^3}{2n^5}$?
Plugging the first inequality into Wolfram Alpha was not enlightening.
From the lecture:
To apply the comparison test, we need to bound the numerator from above and the denominator from below.
For the numerator of $b_n$, $2n^3$ won't work because we have $3n$ in the numerator of $a_n$. $3n^3$ is large enough.
$2n^5$ in the denominator of $b_n$ won't work.
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{2n^5} =b_n$
Using $n^4$ in the denominator won't work because this gives us a harmonic series for $b_n$, which is divergent.
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{n^4}=b_n $
Using asymptotic analysis, we get $a(x)= \frac{2}{x^2} + \mathcal{O}(\frac{1}{x^4})$, which gives us an integral with p=2, which converges.
|
The easiest way to deal with this situation is to use the Limit Comparison Test, since then you don't have to worry about choosing appropriate constants. However, if you want to use the Comparison Test directly, trying a comparison with $\sum_{n=1}^\infty \frac{a}{n^2}$ and solving $\frac{2n^3+3n-8}{n^5-5n^3-n^2+2}\le \frac{a}{n^2}$ by cross-multiplying will show that any choice of $a$ with $a>2$ will work, since $2n^5+3n^3-8n^2\le an^5-5an^3-an^2+2a$ for n sufficiently large if $a>2$.
|
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|
Proving $\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $ [Corrected question]
I'm struggling at proving the following combinatorical identity:
$$\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $$
I would like to see a combinatorical (logical) solution, or an algebraic solution.
|
Here is what might be the worst possible solution to this problem. We begin like ccorn does by noting
$$\begin{aligned}
\sum_{k=0}^n k\,\binom{n}{k}^2 &= \sum_{k=0}^n k\,\binom{n}{k}^2\,x^{k-1}\bigg|_{x=1} =
\left(\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n \binom{n}{k}^2\,x^k\right)\bigg|_{x=1}
\end{aligned}.$$
We examine
$$\sum_{k=0}^n \binom{n}{k}^2 x^k.$$
Let $y=x^{1/2}$. The above expression can be rewritten as
$$\int_0^{1 } \left(\sum_{k=0}^n \binom{n}{k}y^k e^{2\pi ikt} \right) \left(\sum_{k=0}^n \binom{n}{k}y^k e^{-2\pi ikt} \right) \ dt.$$
Using the binomial theorem, this becomes
$$\int_0^{1 } \left(1+y e^{2\pi it}\right)^n \left(1+y e^{-2\pi it} \right)^n \ dt= \int_0^1 (1 + y(e^{2\pi i t }+e^{-2\pi i t}) + x)^n \ dt $$
$$= \int_0^1 (1 + 2y\cos(2\pi t)+x)^n \ dt.$$
We want the derivative of this evaluated at $1$. We can pass the derivative with respect to $x$ inside the integral sign, so our desired expression is
$$ n\int_0^1 \left(\frac{\cos(2\pi t)}{\sqrt x} +1\right) (2\sqrt x \cos(2\pi t) + x +1)^{n-1} \ dt \bigg|_{x=1}=$$
$$ n\int_0^1 \left({\cos(2\pi t)} +1\right) (2 \cos(2\pi t) + 2)^{n-1} \ dt .$$
$$=2^{n-1}n \int_0^1 \left({\cos(2\pi t)} +1\right)^n \ dt .$$
This becomes $$2^{2n-1}n \int_0^1 (\cos(\pi t) )^{2n} \ dt .$$
Using integration by parts, we have the reduction formula
$$\int_0^1 (\cos(\pi t) )^{2n} \ dt = \frac{2n-1}{2n} \int_0^1 (\cos(\pi t) )^{2n-2} \ dt.$$
This gives the product
$$2^{2n-1}\cdot n \cdot \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1 }{ 2}\cdot 1 =2^n \cdot n \cdot \frac{2n-1}{n}\cdot \frac{2n-3}{n-1} \cdots \frac{1 }{ 1}\cdot 1 $$
$$= n \frac{(2n-1)!}{(n-1)!n!}$$
Now note that $$\binom{2n-1}{n-1}=\frac{(2n-1)!}{(n-1)!n!},$$
so we recover
$$n\binom{2n-1}{n-1}$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\infty} \frac{1}{2^n} = 2$$
I don't know how can I prove it exactly, but we have:
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... \\
\sum_{n=0}^{\infty} \frac{1}{2^n} = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+...$$
Because
$$1 \le 1 \\ \frac{1}{3} < \frac{1}{2} \\ \frac{1}{7} < \frac{1}{4} \\ \frac{1}{15} < \frac{1}{8} \\ \mbox{and so on}$$
we have
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... < 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+... = \sum_{n=0}^{\infty} \frac{1}{2^n}$$
Could you tell me, does it work? Maybe exist more formal proof?
|
This does work. Since your series is positive, the comparison test you've imployed ($2^{n-1}\leq 2^n-1$, hence $\frac{1}{2^n-1}\leq\frac{1}{2^{n-1}}$) implies that the sequence of partial sums of our series is increasing and bounded from above by the sequence $\sum_{n=1}^{N}\frac{1}{2^{n-1}}\to 2$, hence convergent and also bounded by $2$. The strict inequality can be seen starting from the second element, as you've said.
|
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|
Expected value sum of dots We throw $n$-times the die. Let $E_n$ be expected value sum of dots (got in all throws).
Compute
*
*$E_1$
*$E_2$
*$E_3$
*$E_4$
So i know how can I do it, for example in 1. I have:
*
*$E_1 = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = \frac{7}{2}$.
Similar:
$E_2 = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36}+ 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36} = 7$
It is easy but for $E_3$ and $E_4$ it will be compute very long. I suppose that exist easier way to do this task. I will grateful for your help.
|
Let $X_n$ denote the number of dots thrown on the $n$th try. Then $$E_n = E[\sum_1^n X_i] = \sum_1^nE[X_i] = n E_1$$
do you see how to finish?
|
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|
$4^\text{th}$ power of a $2\times 2$ matrix $$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I tried to use some trigonometric expressions but it gets very complicated and couldn't solve it.
|
\begin{eqnarray*}
A
& = &
\left(%
\begin{array}{rr}
\cos\left(x\right) & -\sin\left(x\right)
\\
\sin\left(x\right) & \cos\left(x\right)
\end{array}
\right)
=
\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}
\\
A^{2}
& = &
\cos^{2}\left(x\right) - \sin^{2}\left(x\right)
-
2{\rm i}\sin\left(x\right)\cos\left(x\right)\,\sigma_{y}
=
\cos\left(2x\right) - {\rm i}\sin\left(2x\right)\,\sigma_{y}
\\&&\mbox{Then}
\\
A^{4}
& = &
\cos\left(4x\right) - {\rm i}\sin\left(4x\right)\,\sigma_{y}
=
\left(%
\begin{array}{rr}
\cos\left(4x\right) & -\sin\left(4x\right)
\\
\sin\left(4x\right) & \cos\left(4x\right)
\end{array}
\right)
\quad\Longrightarrow\quad
{\rm det}\,\left(A^{4}\right) = 1
\\[5mm]&&\mbox{}
\end{eqnarray*}
$$
ad + bc = \cos^{2}\left(4x\right) - \sin^{2}\left(4x\right) = \cos\left(8x\right)
$$
$\sigma_{y}$ is a Pauli Matrix: http://en.wikipedia.org/wiki/Pauli_matrices
Indeed, ${\rm det}\,\left(A^{4}\right) =\left({\rm det}\,A\right)^{4} = 1^{4} = 1$.
Also,
$$
A'
=
-\sin\left(x\right) - {\rm i}\cos\left(x\right)\,\sigma_{y}
=
-{\rm i}\left\lbrack\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}\right\rbrack
=
-{\rm i}A
\ \Longrightarrow\ A = {\rm e}^{-{\rm i}\,x\,\sigma_{y}}
$$
since $A'' + A = 0$ with $\left.A\right\vert_{x\ =\ 0} = 1$
and $\left.A'\right\vert_{x\ =\ 0} = -{\rm i}\,\sigma_{y}$. That means
$$
A^{n} = {\rm e}^{-{\rm i}\,n\,x\,\sigma_{y}}
=\left(%
\begin{array}{rr}
\cos\left(nx\right) & -\sin\left(nx\right)
\\
\sin\left(nx\right) & \cos\left(nx\right)
\end{array}
\right)\,,
\quad
A\ \mbox{is a Rotation Matrix}
$$
|
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|
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
|
From this and this, $\sin7x=7t-56t^3+112t^5-64t^7$ where $t=\sin x$
Now, if $\sin7x=0, 7x=n\pi, x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$
Clearly, $\sin\frac{r\pi}7$ are the roots of $7-56t^2+112t^4-64t^6=0$ where $r=1,2,3,4,5,6$
As $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7,$
$\sin^2\frac{r\pi}7$ are the roots of $7-56s+112s^2-64s^3=0$ where $r=1,2,4$
Putting $y=\frac1s,$ $\displaystyle7-\frac{56}y+\frac{112}{y^2}-\frac{64}{y^3}=0$
$\displaystyle\implies 7y^3-56y^2+112y-64=0$
Now, using Vieta's Formula, $\displaystyle \frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\frac{56}7=8$
|
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|
Initial value problem with $x^2u''-2xu'+2u=24/x^2$ I have some trouble finding the coefficients and powers of $x$ in the following problem.
Solve $x^2u''-2xu'+2u=\frac{24}{x^2}$ with $u(1)=2$ and $u'(1)=-5$.
I first set $$Lu_H=x^2u''-2xu'+2u=0$$ to find the homogeneous solution. After differentiating and substituting into the homogeneous equation I obtain $$Lu_H=\sum_{k=0}^{\infty}\left[x^k\left(A_{k+2}k(k-1)x^2-2A_{k+1}kx+2A_k\right)\right].$$ (This seems to be the problem as I get one coefficient correct later on but another one incorrect).
For the particular solution, I try $$u_P=\frac{C}{x^2}$$ for some constant $C$, which yields $C=2$. (By the way, this part seems to be correct as there is a term in the answers which matches up).
We now have $$u=u_H+u_P=\sum_{k=0}^{\infty}\left[x^k\left(A_{k+2}k(k-1)x^2-2A_{k+1}kx+2A_k\right)\right] +\frac{2}{x^2}.$$
Differentiating and substituting the initial values implies that $$\sum_{k=0}^{\infty}\left[A_{k+2}k(k-1)-2A_{k+1}k+2A_k \right]=0 \qquad (\ast)$$ and $$\sum_{k=0}^{\infty}A_{k+1}\left[k(k-1)(k-2)-2k(k+1)+2k \right]=-1.$$
I seem to have done something wrong somewhere, however, as the answers give $u=x-x^2+2x^{-2}$. The first thing is the equation $(\ast)$, which gives me the wrong value for the coefficient. The second thing I don't understand is where the terms in $x$ and $x^2$ come from in the answers.
I can see where my coefficients should be going and that substituting this back in will solve the original equation, but if someone could explain why we know that these are the terms that go with the coefficients that would be just dandy.
Many thanks.
|
Easier way to see the solution:
$$\frac{d}{dx} \frac{u}{x} = \frac{u'}{x}-\frac{u}{x^2}$$
$$\frac{d^2}{dx^2} \frac{u}{x} = \frac{u''}{x}-2 \frac{u'}{x^2} + 2 \frac{u}{x^3}$$
So...divide both sides of the ODE by $x^3$ and get
$$\frac{u''}{x}-2 \frac{u'}{x^2} + 2 \frac{u}{x^3} = \frac{d^2}{dx^2} \frac{u}{x} = \frac{24}{x^5}$$
Integrate twice w.r.t $x$:
$$\frac{d}{dx} \frac{u}{x} = -\frac{6}{x^4} + C_1$$
$$\frac{u}{x} = \frac{2}{x^3}+C_1 x+C_2 \implies u(x) = \frac{2}{x^2}+C_1 x^2+C_2 x$$
Apply initial conditions:
$$u(1)=2 \implies C_1+C_2=0$$
$$u'(1) = -5 \implies 2 C_1+C_2 = -1 \implies C_1=-1 \quad C_2=1$$
Therefore
$$u(x) = \frac{2}{x^2}-x^2+x$$
|
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|
Prove that $ \sigma = \left( \begin{array}{ccccccccc} 1&2&3&4&5 \\ 2&3&4&1&5 \end{array} \right) \neq s_1 \circ s_2 \circ \ldots \circ s_k $ Prove that we can't write permutation $ \sigma = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 1 & 5 \end{array} \right) $ as $s_1 \circ s_2 \circ \ldots \circ s_k $ where $s_1, s_2 , \ldots, s_k \in \left\{ \sigma_1= \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 4 & 5 \end{array} \right), \sigma_2 = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 4 & 5 & 3 \end{array} \right) \right\} $.
We have
$ \sigma = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 1 & 5 \end{array} \right) = (1234)
\\ \sigma_1 = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 4 & 5 \end{array} \right) = (123)
\\ \sigma_2 = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 4 & 5 & 3 \end{array} \right) = (345) $
$ \mathrm{ord}( \sigma ) = 4 \\ \mathrm{ord} (\sigma_1) = 3 \\ \mathrm{ord}( \sigma_2) = 3$
Then I was thinking about $\mathrm{ord}(s_1 \circ \ldots \circ s_k)$ and maybe I should prove that $\mathrm{ord}(s_1 \circ \ldots \circ s_k) \neq 4 = \mathrm{ord}(\sigma)$ ?
But I have no idea how to prove it. Furthermore, I'm not sure, is it possible? Maybe is another way to do this task? Thanks in advance for any help.
|
Hint/Spoiler: Have you checked the parities of all the permutations here?
|
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|
Wallis Product (Long infinite Product) I'm almost finished proving Wallis product in a given question. But the last step is:
What am I doing wrong? Am I on the right track? Hints?
Show that $$\prod_{k=1}^{m} \left(\frac{(2k-1)(2k+1)}{(2k)^2}\right) = \frac{(2m+1)((2m!))^2}{2^{4m}(m!)^4}$$
Currently I have tried expanding a few terms as so:
$$\frac{1 * 3}{2^2} * \frac{3 * 5}{4^2} * \frac{5 * 7}{6^2}*.........*\frac{(2k-7)(2k-5)}{(2(k-3))^2} *\frac{(2k-5)(2k-3)}{(2(k-2))^2}*\frac{(2k-3)(2k-1)}{(2(k-1))^2}*\frac{(2k-1)(2k+1)}{(2k)^2}$$
I note that the denominator is $$4*m = 2^{2m}$$
For the numerator I get $(2m+1)*[(2m-1)!]^2$
Which is not really close to what I want...
I want to try this manipulation
$$\frac{1 * 3}{2^2} *\frac{3^2}{3^2} * \frac{3 * 5}{4^2} * \frac{5^2}{5^2} \frac{5 * 7}{6^2}*.........*\frac{(2k-7)(2k-5)}{(2(k-3))^2} *\frac{(2k-5)^2}{(2k-5)^2}*\frac{(2k-5)(2k-3)}{(2(k-2))^2}*\frac{(2k-3)^2}{(2k-3)^2}*\frac{(2k-3)(2k-1)}{(2(k-1))^2}*\frac{(2k-1)^2}{(2k-1)^2}*\frac{(2k-1)(2k+1)}{(2k)^2}$$
Note: Excuse my errors: It's 3 AM here, I am brain dead.
|
You are quite close. If we multiply out, then on top we get $2m+1$ times the product of the squares of the odd numbers up to $2m-1$. So we get
$$(2m+1) \left(1\cdot 3\cdot 5\cdots(2m-1)\right)^2.$$
"Improve" $1\cdot 3\cdot 5\cdot (2m-1)$ to $(2m)!$ by filling in the even numbers, that is, by multiplying by $2\cdot 4\cdot 6\cdots (2m)$. We must also divide by this number, which is $2^m m!$. With this manipulation, the original top becomes
$$(2m+1)\frac{((2m)!)^2}{2^{2m}(m!)^2}.\tag{1}$$
The manipulation of the original bottom is simpler: the product $\prod (2k)^2$ is just $$2^{2k}(m!)^2.\tag{2}$$
Finally, we divide (1) by (2).
|
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|
How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
|
If you multiply a number $n$ by any prime number, then its number of divisors is multiplied by $\frac{m+1}m$ where $p^{m-1}$ is the highest power of $p$ dividing$~n$. This is because the multiplicity of $p$ in divisors of $n$ can be any one of the $m$ values $0,1,\ldots,m-1$, but that multiplicity can be chosen among the $m+1$ values $0,1,\ldots,m$ for divisors of $pn$ (for prime numbers other than $p$, the same set of multiplicities is available in both cases).
So in the problem here let $k$ be the number of divisors of $n$ without factors $2$ or $3$, and $a-1,b-1$ the multiplicities of the primes $2$ and $3$ in$~n$, respectively. So $k,a,b$ are positive integers and $n$ has $kab$ divisors in all. We are given that $28=k(a+1)b$ and $30=ka(b+1)$, which implies that $\frac{28}{30}=\frac{14}{15}=\frac{(a+1)b}{a(b+1)}$. The factor $7$ in the numerator cannot come from $b$ (given that $b+1$ divides $~30$ and that $\frac{a+1}a\neq1$), so it must come from the factor $a+1$. Since $a$ divides $30$, the only possibility is $a=6$, which indeed gives a solution namely $(k,a,b)=(1,6,4)$. So we find $n=k2^{a-1}3^{b-1}=2^53^3$, and $6n=2^63^4$ has $k(a+1)(b+1)=35$ divisors.
|
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|
Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$
If $x \sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x^2)\frac{dy}{dx}+1=0.$
The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.
|
the points $(0,0)$ and $(-1,-1)$ no longer lie on the graph.
Btw on actual differentiation it yield $\frac{dy}{dx} = \frac{y^2 - 2x(1+y)}{x^2 - 2y(1+x)}$.
|
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|
Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way? I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p - b^p=p\cdot(ab)^{\alpha_1}(a+b)^{\alpha_2}(a^2+ab+b^2)^{\alpha_3}P_{\omega}(a,b)$ for any prime number p where $P_{\omega}(a,b)$ is an irreducible homogenous polynomial of degree $\omega$ and $\alpha_1 + \alpha_2 + \alpha_3 + \omega = p-1$. I wonder if it's true in general.
Here are some examples:
$(a+b)^2 - a^2 - b^2 = 2ab$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$
$(a+b)^5 - a^5 - b^5 = 5ab(a+b)(a^2+ab+b^2)$
$(a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2+ab+b^2)^2$
$(a+b)^{11}- a^{11} - b^{11} = 11 ab(a + b)(a^2 + a b + b^2)(a^6 + 3 a^5 b + 7 a^4 b^2 + 9 a^3 b^3 + 7 a^2 b^4 + 3 a b^5 + b^6)$
I've proved the following statements:
For any prime number p:
$p|(a+b)^p - a^p-b^p$
$ab|(a+b)^p - a^p-b^p$
For any odd prime number:
$(a+b) | (a+b)^p - a^p-b^p$ because $(a+b)^p - a^p-b^p$ vanishes when $a=-b$ if p is odd.
For any prime number $p \geq 5$:
$ (a^2 + ab + b^2) | (a+b)^p - a^p - b^p$
Proof: Any prime number greater than 3 is of the form $p=6k+1$ or $p=6k+5$. On the other hand, if we set $\Large \omega= e^\frac{2\pi i}{3}$ we have: $1 + \omega + \omega^2 = 0$, therefore
$(1+\omega)^3 = 1 + 3\omega + 3\omega^2 + \omega^3 = 2 + 3(\omega+\omega^2) = -1$
$(1+\omega^2)^3 = 1 + 3\omega^2 + 3\omega^4 + \omega^3 = 2 + 3(\omega^2+\omega) = -1$
We have $a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)$ in $\mathbb{Z}[\omega]$, I'll show that both $a-b\omega$ and $a-b\omega^2$ divide $(a+b)^p - a^p - b^p$ for every $p>3$:
if $p=6k+1$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+1} = b^{6k+1} (1+\omega)^{6k} (1+\omega) = b^{6k+1}(1+\omega)$
$(a+b)^p - a^p - b^p = b^{6k+1}(1+\omega) - (b\omega)^{6k+1} - b^{6k+1} = b^{6k+1}(1+\omega) - b^{6k+1}(\omega+1) = 0$
If $p=6k+5$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+5} = b^{6k+5} (1+\omega)^{6k} (1+\omega)^5 = (-1)b^{6k+1}(1+\omega)^2$
$(a+b)^p - a^p - b^p = (-1)b^{6k+5}(1+\omega)^2 - (b\omega)^{6k+5} - b^{6k+5} = (-1)b^{6k+5}(1+\omega)^2 - b^{6k+5}(\omega^2+1) = p^{6k+5}(-1-2\omega-\omega^2-\omega^2-1)=0$
The same could be shown for $a=b\omega^2$.
Therefore both $(a-b\omega)$ and $(a-b\omega^2)$ are factors of $(a+b)^p-a^p-b^p$ but I'm not sure if that is sufficient to conclude $a^2+ab+b^2=(a-b\omega)(a-b\omega^2) | (a+b)^p - a^p - b^p$
So, I guess I have proved that $p$, $ab$, $a+b$ and $a^2+ab+b^2$ all divide $(a+b)^p-a^p-b^p$, but I have no idea how to show that $P_{\omega}(a,b)$ must be irreducible. I have verified this conjecture up to $p=97$ and it's true I think. Any ideas on how to prove that?
|
It is true in general. These are often called Cauchy polynomials — see, for example, Fermat's Last Theorem for Amateurs, Chapter VII. The power of $a^2+ab+b^2$ is either $1$ or $2$ as $p \equiv -1$ or $+1$ modulo $6$, respectively. There are many proofs in the literature (Cayley, Glaisher, etc.).
|
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|
Divide polynomials with exponents and simplify The expression is
$$\frac{p^2q^2}{m^2-n^2} \cdot \frac{m^2+2mn+n^2}{3p^2+2pq-q^2}.$$
How could I divide and simplify this?
|
We can write the product of two rational fractions as one fraction, and use commutativity of multiplication to rearrange factors, as needed:
$$\frac{p^2q^2}{m^2-n^2} \cdot \frac{m^2+2mn+n^2}{3p^2+2pq-q^2} = \frac{p^2q^2(m^2 + 2mn+n^2)}{(3p^2 + 2pq - q^2)(m^2 - n^2)}$$
Here, you have quadratics in each of the numerator and denominator that can be nicely factored.
$$\frac{p^2q^2(\overbrace{m^2 + 2mn+n^2}^{(m+n)^2})}{(3p^2 + 2pq - q^2)(m^2 - n^2)} = \frac{p^2q^2\color{blue}{\bf (m + n)}(m+n)}{(p+q)(3p-q)\color{blue}{\bf (m+n)}(m-n)}$$
We can cancel the common factor appearing in each of the numerator and denominator, which gives us:$$\frac{p^2q^2(m+n)}{(p+q)(3p-q)(m-n)}$$
This is about as simplified as it gets.
|
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|
Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
|
Putting $\frac1x=h$
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
$$=\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}$$
Method $1:$
Using Binomial Expansion (1, 2) , $(1+x)^n=1+nx+O(x^2)$ and as $\lim_{h\to0}\frac{\sin h}h=1\implies \sin h $ is $O(h)$
$$\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}=\lim_{h\to0}\frac{\frac13h\sin h+O(h^4)}{h^2}=\frac13\lim_{h\to0}\frac{\sin h}h$$
Method $2:$
As $a^3-b^3=(a-b)(a^2-ab+b^2),$
$$\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}$$
$$=\lim_{h\to0}\frac{(1+h\sin h)-1}{h^2\{(1+h\sin h)^{\frac23}+(1+h\sin h)^{\frac13}+1\}}$$
$$=\lim_{h\to0}\frac{\sin h}h\cdot\frac1{\lim_{h\to0}\{(1+h\sin h)^{\frac23}+(1+h\sin h)^{\frac13}+1\}}\text{ as }h\ne0 \text{ as }h\to0$$
$$=1\cdot\frac1{\{(1+0)^{\frac23}+(1+0)^{\frac13}+1\}}$$
|
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|
Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y}{dx^2}ny^{n-1} = 0.$$
Therefore $$y'' = \displaystyle \frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}(y')^2}{ny^{n-1}}.$$
Substituting the first derivative:
$$y'' = \displaystyle\frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}\left(\dfrac{-x^{n-1}}{y^{n-1}}\right)^2}{ny^{n-1}}.$$
I've been trying tons of different steps and can't establish which way to eliminate the $y$ terms from the numerator. Could someone offer a hint on how to proceed. Thanks.
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$y'=-x^{n-1}y^{1-n}$.
$y''=-(n-1)x^{n-2}y^{1-n}-x^{n-1}(1-n)y^{-n}y'$
$y''=-(n-1)x^{n-2}y^{1-n}+x^{n-1}(1-n)y^{-n}x^{n-1}y^{1-n}$
$y''=-(n-1)x^{n-2}y^ny^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$
$y''=-(n-1)x^{n-2}(1-x^n)y^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$
Should be easy from there.
|
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Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? $(x+3)^2 \ge 0$
Am I missing the point in anyway?
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Since $(x+3)^2\geq 0$ for all real values of $x$, $x^2+6x+10=x^2+6x+9+1=(x+3)^2+1\geq 1$, and the inequality follows.
|
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|
Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
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$$\frac{x}{3x-5}\leq \frac{2}{x-1}$$
Subtracting the right-hand side from each side, then finding the common denominator and subtracting resulting numerators gives us:$$\begin{align}
\dfrac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\leq 0 & \iff \dfrac{x^2 - 7x + 10}{(x-1)(3x-5)} \leq 0 \\ \\
& \iff \dfrac{(x-2)(x-5)}{(x-1)(3x-5)} \leq 0\quad x \neq 1, \;x \neq \frac 53\\ \\
\end{align}$$
This inequality will be satisfied when exactly one or exactly three of the factors is/are negative or when $x = 2$ or $x = 5$, regardless of the signs of the other factors.
A nice way to approach this sort of problem is through the use of a sign-chart, ordering the roots (or points at which one factor switches from positive to negative):
$x \quad = \qquad 1 \qquad \frac 53 \qquad 2\;\qquad 5$
$\qquad\quad+\qquad - \qquad+\;\;\,0\;\; -\;\;0\quad +$
This gives us $$\dfrac{(x-2)(x-5)}{(x-1)(3x-5)} \leq 0 \;\;\iff\;\; 1 \lt x <\dfrac 53\;\;\text{OR}\;\;2 \leq x \leq 5$$
|
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|
Simpler solution to a geometry problem In a set of geometry problems, I got this one:
If in a triangle $ABC$ with segments $AB=8$, $BC=4$, and $3A+2B=180^{\circ}$, calculate the side $AC$
My solution was
Let $A=2\alpha$,$B=90^{\circ}-3\alpha$, where $\alpha<30$, then the second condition is always met.
So
$$tan(2\alpha)=\frac{cos(3\alpha)}{2-sen(3\alpha)}$$
$$\frac{2sen(\alpha)cos(\alpha)}{cos(2\alpha)}=\frac{cos(\alpha)(2cos(2\alpha)-1)}{2-sen(\alpha)(2cos(2\alpha)+1)}$$
$$\frac{2sen(\alpha)}{cos(2\alpha)}=\frac{2cos(2\alpha)-1}{2-sen(\alpha)(2cos(2\alpha)+1)}$$
$$4sen(\alpha)-2sen^2(2cos(2\alpha)+1)=2cos^2(2\alpha)-cos(2\alpha)$$
$$4sen(\alpha)+(cos(2\alpha)-1)(2cos(2\alpha)+1)=2cos^2(2\alpha)-cos(2\alpha)$$
$$4sen(\alpha)+2cos^2(2\alpha)-cos(\alpha)-1=2cos^2(2\alpha)-cos(2\alpha)$$
$$sen(\alpha)=\frac{1}{4}$$
We now construct the altitude to $BC$
And since $sin(\alpha)=\frac{1}{4}$ we set $AC=4k$, $CE=k$, $AE=\sqrt{15}k$
Then it follows from pythagoras that
$$(k+4)^2+15k^2=64$$
$$16k^2+8k-48=0$$
$$2k^2+k-6=0$$
$$(2k-3)(k+2)=0$$
Since $k$ is positive, $k=\frac{3}{2}\iff AC=4k=6$
But the $2,3,4$(from the sides $4,6,8$) pattern makes me think there is an easier way, so it makes me think I missed something obvious. Any hints are ideas are greatly appreciated.
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We might also "drop a perpendicular" from vertex $ \ B \ $ to the extended line $ \ AC \ \ , \ $ the length of which we will call $ \ y \ \ . \ $ Using your notation for angle measure, we have $ \ y \ = \ 8 \sin (2 \alpha) \ = \ 4 \sin (90º - \alpha) \ = \ 4 \cos \alpha \ \ . $ From this, we find
$$ 8 \ · \ 2 · \sin \alpha · \cos \alpha \ = \ 4 · \cos \alpha \ \ \Rightarrow \ \ \sin \alpha \ = \ \frac14 \ \ , \ $$ since $ \ \cos \alpha \ \neq \ 0 \ \ . $ (This is just a variant of imranfat's method, since the Law of Sines can be derived using altitudes of the triangle.)
The Law of Cosines then gives us
$$ x^2 \ \ = \ \ 8^2 \ + \ 4^2 \ - \ 2·8·4·\cos(90º - 3 \alpha) \ \ = \ \ 80 \ - \ 64 \sin (3 \alpha) \ \ , \ $$
with $ \ x \ $ being the length of side $ \ AC \ \ . $
From the "triple-angle" identity for sine, we obtain $$ \sin (3 \alpha) \ \ = \ \ 3·\sin \alpha \ - \ 4·\sin^3 \alpha \ \ = \ \ 3·\frac14 \ - \ 4·\left(\frac14 \right)^3 \ \ = \ \ \frac{11}{16} \ \ . \ $$
Thus, $$ x^2 \ \ = \ \ 80 \ - \ 64·\frac{11}{16} \ \ = \ \ 80 \ - \ 44 \ \ = \ \ 36 \ \ \Rightarrow \ \ x \ = \ 6 \ \ . $$
[We can check the consistency of our calculations using our constructed right triangle: since $ \ \cos \alpha \ = \ \frac{\sqrt{15}}{4} \ \ , \ $ we have $ \ y \ = \ 4·\cos \alpha \ = \ \sqrt{15} \ \ , \ \ z \ = \ 4·\cos(90º - \alpha) \ = \ 4·\sin \alpha \ = \ 1 \ \ , \ $ and so
$$ y^2 \ + \ (x + z)^2 \ \ = \ \ (\sqrt{15})^2 \ + \ (6 + 1)^2 \ \ = \ \ 15 \ + \ 7^2 \ \ = \ \ 8^2 \ \ . \ ] $$
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Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$
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Without loss of generality
let $0\le a\le b\le c\le 1$,then
\begin{align*}
&\dfrac{a}{c+b+1}+\dfrac{b}{a+c+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&\le\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(a+b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(a+1)(b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(1-a^2)(1-b^2)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{1-c}{a+b+1}=1
\end{align*}
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Inequality With Recurrent Relation $a_{1}=1$, $~$ $a_{n+1}-a_{n}=\sqrt{\dfrac{a_{n+1}^{2}-1}{2}}+\sqrt{\dfrac{a_{n}^{2}-1}{2}}$ , $~$ $a_{n+1}>a_{n}$
Prove that $~$ $\displaystyle\sum_{k=1}^{\infty}\frac{1}{a_{n}}<e$
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Note that sequence $\{a_n, ~ n\in \mathbb{N}\}$ is non-decreasing (because RHS can't be negative, if $a_n\in \mathbb{R}$).
First, let's consider increasing sequence $\{a_n, ~ n\in \mathbb{N}\}$:
$$
a_{n+1}>a_n.\tag{1}
$$
1.
Let's show that there is other (easy) recurrent relation
$$
a_n = 6a_{n-1}-a_{n-2}, \qquad (n\geqslant 3),\tag{2}
$$
where $a_1=1,a_2=3$.
So, we'll have integer sequence
$$
1,~~ 3,~~ 17,~~ 99,~~ 577,~~ 3363,~~ 19601,~~ \ldots\tag{3}
$$
It is easy to see that $a_2$ can be $1$ either $3$ (as solution of quadratic equation).
We'll consider case $a_2=3$ here, because case $a_2=1$ doesn't hold ineq. $(1)$.
Denote $b_n = \sqrt{\dfrac{a_n^2-1}{2}}$ . Then
$$
a_{n+1}-a_n = b_{n+1}+b_n,\qquad (n\in \mathbb{N}),\tag{4}
$$
and also $a_{n+1}-a_n = \sqrt{\dfrac{a_{n+1}^2-1}{2}}+b_n$.
Now we'll find $a_{n+1}$ as expression of $a_n, b_n$.
Denote $\color{purple}{x} = a_{n+1}$.
$$\color{purple}{x}-a_n = \sqrt{\dfrac{\color{purple}{x}^2-1}{2}} + b_n;$$
$$\color{purple}{x}-(a_n+b_n) = \sqrt{\dfrac{\color{purple}{x}^2-1}{2}}; ~~~~~ (\color{purple}{x}\geqslant a_n+b_n);\tag{5}$$
$$2\Bigl(\color{purple}{x}^2-2(a_n+b_n)\color{purple}{x} +(a_n+b_n)^2\Bigr)= \color{purple}{x}^2-1.$$
$$\color{purple}{x}^2-4(a_n+b_n)\color{purple}{x} +2(a_n+b_n)^2+1= 0.$$
Solving this quadratic equation, we find $2$ solutions:
$a_{n+1} = \color{purple}{x} = a_n$; (but this value doesn't hold ineq. $(1)$, and holds $(5)$ when $b_n=0$ only);
$a_{n+1} = \color{purple}{x} =3a_n+4b_n$.
We will focus on the last expression:
$$a_{n+1} = 3a_n + 4b_n, \qquad (n\in \mathbb{N}).\tag{6}$$
$(4),(6)\implies$
$$b_{n+1} = a_{n+1}-(a_n+b_n) = 2a_n+3b_n, \qquad (n\in \mathbb{N}).\tag{7}$$
$(6),(7)\implies$
$$a_{n+1} = 3(3a_{n-1} + 4b_{n-1}) +4(2a_{n-1}+3b_{n-1}) = 17a_{n-1} + 24b_{n-1}, \qquad (n\geqslant 2);$$
$$a_{n+1} = (18a_{n-1} + 24b_{n-1}) -a_{n-1} = 6(3a_{n-1} + 4b_{n-1}) -a_{n-1} = 6a_n-a_{n-1},\qquad (n\geqslant 2).$$
Statement $(2)$ is proved.
2.
Since $a_{n-2}<a_{n-1}$ (see $(1)$), then
$$
a_n > 5a_{n-1}, \qquad (n\geqslant 3),
$$
other words
$$
a_n > 3 \cdot 5^{n-2}, \qquad (n\geqslant 3).\tag{8}
$$
3.
Then
$$
\sum_{n=1}^\infty \dfrac{1}{a_n} < 1+\dfrac{1}{3} + \sum_{n=3}^\infty \dfrac{1}{3\cdot 5^{n-2}} = 1+\dfrac{1}{3} + \dfrac{1}{12} = \dfrac{17}{12} <2<e.\tag{9}
$$
Bound $\dfrac{17}{12} = 1+\dfrac{5}{12}$ is more sharp estimation than $e$.
I missed the case, when can be $a_{n+1} = a_n$ (non-decreasing sequence).
It can be when $b_{n+1}=b_n=0$ (see $(4)$), other words $a_{n+1}=a_n=1$.
So, if we will consider $a_2=a_1$, then, as Don Antonio noted in comments,
it generates constant sequence $1,1,1,1,\ldots$.
For this sequence we will have
$\sum\limits_{n=1}^\infty \dfrac{1}{a_n} = \sum\limits_{n=1}^\infty 1~~$ (divergent series).
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How prove this $x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$ show that
$$x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$$ have nice methods? Thank you
my try:
$$\Longleftrightarrow \cos{x}\cdot x^3<(\sin{x})^3$$
let
$$f(x)=\cos{x}\cdot x^3-(\sin{x})^3$$
$$\Longrightarrow f'(x)=-\sin{x}\cdot x^3+3\cos{x}\cdot x^2-3(\sin{x})^2\cos{x}$$
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We have that
$$\sin^3x-x^3 \cos x=\frac14\left(3\sin x-\sin(3x)\right)-x^3 \cos x\ge0 \iff 3\sin x-\sin(3x)-4x^3\cos x \ge 0$$
and since by Taylor's series
*
*$\sin x\ge x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7$
*$\sin (3 x)\le 3x-\frac 9 2 x^3+\frac{81}{40}x^5-\frac{243}{560}x^7+\frac{243}{4480}x^9$
*$\cos x\le 1-\frac12 x^2+\frac1{24}x^4$
then
$$ 3\sin x-\sin(3x)-4x^3\cos x\ge $$
$$\ge 3x-\frac12x^3+\frac1{40}x^5-\frac1{1680}x^7-3x+\frac 9 2 x^3-\frac{81}{40}x^5+\frac{243}{560}x^7-\frac{243}{4480}x^9-4x^3+2x^5-\frac1{6}x^8=$$
$$=-\frac1{1680}x^7+\frac{243}{560}x^7-\frac{243}{4480}x^9-\frac1{6}x^8=$$
$$=\frac{13}{30}x^7-\frac16 x^8-\frac{243}{4480}x^9\ge 0 \iff 729x^2+2240x-5824\le 0$$
which leads to $0\le x\le a\approx 1.68$.
|
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compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$. I programmed and found that $\frac{\ln 2}{4}+{\frac{\pi}{4}}/2$ is precise to the eighth digit.
|
$$
\begin{align}
\log(2)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\\
\frac12\log(2)&=\sum_{n=0}^\infty\hphantom{-\frac1{4n+2}}\frac1{4n+2}\hphantom{\;+\frac1{4n+3}}-\frac1{4n+4}\\
\frac\pi4&=\sum_{n=0}^\infty\frac1{4n+1}\hphantom{-\frac1{4n+2}}-\frac1{4n+3}\\
\frac12\left(\log(2)-\frac12\log(2)+\frac\pi4\right)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}
\end{align}
$$
|
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Inequalities with radicals I have a question regarding the following inequality; It may sound petty but yet;
The following inequality $$ \sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$$ has the solution $\frac {-5}{4}\le x \lt 1 $
During the process of solving it we need to check the correctness of the solution to $2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$ which is: $\frac {-5}{4}\le x \le \frac {7}{11} $ and is created due to squaring . Should I take a value of it and place it in $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$ or in $2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$ to check it's validity? To which inequality should it be applied? Or are they both equivalent in regard to this? The thing is squaring gives us more solutions than there are initially, so as I suppose what is true for the second inequality may not be true for the first..
Any advice?
|
As a follow-up to my comment above, I'd like to walk through a solution of this inequality.
$$ \sqrt{-6x+10} + \sqrt{-x + 2} > \sqrt{4x+5}$$
*
*First determine the restrictions on $x$ implied by the radicals.
$$
-6x + 10 \geq 0 \;\Rightarrow\; x \leq 5/3.
$$
$$
-x + 2 \geq 0 \;\Rightarrow\; x \leq 2.
$$
$$
4x+5 \geq 0 \;\Rightarrow\; x \geq -5/4.
$$
On these restrictions alone, we know that $-5/4 \leq x \leq 5/3$.
2 . Use algebra to solve the inequality under the assumption that $-5/4 \leq x \leq 5/3$ (so that all radicands are guaranteed nonnegative).
$$
\begin{eqnarray*}
(\sqrt{-6x+10} + \sqrt{-x + 2})^2 &>& 4x+5 \\
-7x+12 + 2\sqrt{(-6x+10)(-x+2)} &>& 4x + 5\\
2\sqrt{(-6x+10)(-x+2)} &>& 11x - 7\\
\end{eqnarray*}
$$
Now at this point there are two cases to consider. If $11x - 7 < 0$, then the inequality is automatically satisfied. Therefore, we have a partial solution of $-5/4 \leq x \leq 7/11$. On the other hand, if $11x - 7 \geq 0$, then there is still work to do. Assume now that $11x - 7 \geq 0$ and square again:
$$
\begin{eqnarray*}
4(-6x+10)(-x+2) &>& (11x - 7)^2\\
24x^2 -88x + 80 &>& 121x^2 - 154x + 49\\
0 &>& 97x^2 - 66x - 31\\
0 &>& (x-1)(97x+31)
\end{eqnarray*}
$$
The roots of the quadratic are $x=1$ and $x=-31/97$, and since the parabola is opening upward, the last inequality is satisfied for $-31/97 < x < 1$. Together with the previous restrictions and previous work done, we finally have a complete solution:
$$ -5/4 \leq x < 1 $$
|
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CS problem, turned to mathematics I am trying to solve some of the projecteuler problems using a much of a programmers approach. However, I would like to get more into the math, and therefore would try to do some mathematical reasoning to this specific projecteuler exercise:
Surprisingly there are only three numbers that can be written as the sum of fourth
powers of their digits:
$ 1634 = 1^4 + 6^4 + 3^4 + 4^4 $
$ 8208 = 8^4 + 2^4 + 0^4 + 8^4 $
$ 9474 = 9^4 + 4^4 + 7^4 + 4^4$
As $ 1=1^4$
is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their
digits.
There are various ways of presenting a reasonable solution to this problem, using simple programming skills. However, I would like to be able to do the reasoning on pen and paper and not rely on the computational power. Can anybody tell me the steps of how to solve this problem with mathematical reasoning?
|
If a number $n$ has seven digits, the sum of fifth powers of digits is $\le 7\cdot 9^5<10^6\le n$ and with each additional digit the upper limit for the power-sum grows by at most $9^5$ whereas the lower lmit for $n$ grows by a factor of $10$. We conclude that $n$ has at most six digits.
Moreover, from $d^5\equiv d\pmod{10}$ we see that the sum of all but the last digit must be a multiple of $10$.
We not that $a<b$ implies $a^5+b^5<(a+1)^5+(b-1)^5$, i.e. a sum of fifth powers for a given sum of digits is maximized when the digits are as close togethre as possible (i.e. assume at most tow consecutive values).
From $6\cdot 9^5=354294$, we conclude that $n$
*
*has at most five digits (see below)
*or has leading digits $3$, $\le5$ and then at most four $9$s (contradiction: $4\cdot 9^5+5^5+3^5<350000$)
*or has leading digit $2$ and at most five nines, so $n\le 2^5+6\cdot 9^5=295277$. In fact from $2+4\cdot 9=38$, we conclude that the sum of the five leading digits is at most $30$ and thus $n\le 2^5+4\cdot 7^5+9^5=126309<200000$, so this case is not possible either
*or has leading digit $1$ and at most five nines. Like above the sum of the first five digits is a multiple of $10$, so either $=30$ or $\le 20$. In the latter case we get $n\le1^5+3\cdot 5^5+4^5+9^5=69449<10000$. Remains the case with digit sum $30$. Dropping the leading $1$, the sum $29$ can be obtained with four digits as: $9+9+9+2, 9+9+8+3, 9+9+7+4, 9+9+6+5,9+8+8+4,9+8+7+5,9+7+7+6,8+8+8+5,8+8+7+6, 8+7+7+7$. For each of these ten cases one can compute the sum of fifth powers of these digits and the leading $1$ and add $0^9,\ldots ,9^9$ for the last digit to check if a valid number occurs. - Only one solution is found this way
For the five-digit (or less) case a lot of more case distinctions are necessary, though much can still be gained from the condition that the sum of all but th elast digit is a multiple of ten.
However, just learning from above the limit that $n<200000$ and then making a brute-force search, is probably the recommended way.
|
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|
Taking Radicals Out of Denominator How does one take radicals out of the denominator of
$$\frac{1}{a + b(\sqrt{2} + \sqrt{3})}$$
assuming $a,b \in \mathbb{Q}$ not both equal to $0$. I know the trick is to multiply this expression by $1$ s.t. the radicals are removed, but I can't think of such $c/d = 1$ to try.
|
Try
$$ \frac{ a + b \sqrt{2} - b \sqrt{3} } { a + b \sqrt{2} - b \sqrt{3} } \times \frac{ a - b \sqrt{2} + b \sqrt{3} } { a - b \sqrt{2} + b \sqrt{3} } \times \frac{ a - b \sqrt{2} - b \sqrt{3} } { a - b \sqrt{2} - b \sqrt{3} } .$$
The idea is that you have to multiply by all possible conjugates.
|
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|
How to show that $10^n - 1$ is divisible by $9$ How can I show that $10^n-1, 10^{n-1}-1,...., 10-1$ are all divisible by 9? I was considering using Euclid's algorithm, but I can't find a way to get that to work.
|
$$ a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) $$
Proof of the identity by expansion:
$$\begin{align}
&(a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \\\\
&=a^{n} + a^{n-1}b + a^{n-2}b^2 + \ldots + ab^{n-1}\\\\
&\;\;\;\;\;\;\;\;- ba^{n-1} - b^2a^{n-2}-\ldots -b^{n-1}a - b^n\\\\
&=a^n-b^n
\end{align}$$
|
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|
How to prove $n=7k$ or $7k+1$, if $n=x^2=y^3$. $n$ is both a square and a cube, I need to prove $n=7k$ or $n=7k+1$, for some integer $k$.
I need some guide lines.
|
If $n$ is both a square and a cube, it is a perfect $6$-th power.
Let $z$ be a number. Then $z$ has shape $7k$, or $7k+1$, or $7k+2$, and so on up to $7k+6$.
Then $z^6$ has shape (respectively) $7k$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, and $7k+1$.
To do the calculations, let us take the example $z$ of the shape $7k+2$.
Imagine expanding $(7k+2)^6$ using the Binomial Theorem. The first $6$ terms are obviously divisible by $7$, and the last term is $2^6=64$. This has remainder $1$ on division by $7$. So $(7k+2)^6$ has remainder $1$ on division by $7$, and therefore has shape $7t+1$.
The other calculations are very similar.
Remark: We do not need to work with $6$-th powers. The squares of numbers of shape $7k+a$, where $a$ goes from $0$ to $6$, are, in order, of shape
$$7k \qquad 7k+1 \qquad 7k+4 \qquad 7k+2\qquad 7k+2 \qquad 7k+4\qquad 7k+1.$$
The cubes, are, in order, of the shape
$$7k \qquad 7k+1 \qquad 7k+1 \qquad 7k+6\qquad 7k+1 \qquad 7k+6\qquad 7k+6.$$
Only $7k$ and $7k+1$ occur in both lists.
|
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|
On the roots of $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$ The Problem Prove that $\tan \frac{\pi}{15}$ is a root of the equation $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$, and find the other roots.
Source: Question 17 from the complex numbers chapter from Bostock's Further Pure Mathematics
Thoughts Using the identity $\tan 5\theta =\frac{5\tan\theta-10\tan^3\theta-\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$, I can show that $\tan \frac{\pi}{15}$ is a root of $t^5+5\sqrt3t^4+10t^3-10\sqrt3t^2-5t+\sqrt3=0$, but it isn't clear how to proceed. Alternatively, I can derive and use a formula for $\tan 4\theta$ to obtain a quartic equation as desired, but that will require me to find $\tan \frac{4\pi}{15}$ which I cannot.
I suspect this is not difficult though I can't see how to proceed. Hints would be appreciated.
|
First of all, $\tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
Set $\tan5\theta=\sqrt3=\tan\frac\pi3$
$\implies 5\theta=n\pi+\frac\pi3=\frac{(3n+1)\pi}5$ where $n$ is any integer
So, the values of $\theta$ can be set from $n=0,1,2,3,4$
Observe that for $n=3,\tan\theta=\tan\frac{2\pi}3=\tan\left(\pi-\frac\pi3\right)=-\tan\frac\pi3=-\sqrt3$
So rearrange $\displaystyle\sqrt3=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$ and divide by $\tan\theta -(-\sqrt3)=\tan\theta+\sqrt3$ to get the equation with roots $\displaystyle\tan\frac{(3n+1)\pi}{15}$ where $n=0,1,2,4$
|
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Why does $\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$ equal zero? Limit $$\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$$
equals zero.
Could you help me prove it?
|
Let $y = \dfrac1x$. We then have
$$L = \lim_{x \to 0^+} \left(\sqrt{\dfrac1x+2} - \sqrt{\dfrac1x}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) \times \dfrac{\sqrt{y+2} + \sqrt{y}}{\sqrt{y+2} + \sqrt{y}}$$
This gives us
$$L = \lim_{y \to \infty} \dfrac{y+2 - y}{\sqrt{y+2} + \sqrt{y}} = \lim_{y \to \infty} \dfrac2{\sqrt{y+2} + \sqrt{y}}$$
Can you now finish it off?
|
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|
What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
|
$$x^2+y^2\geq 2xy\implies 2(x^2+y^2)\geq (x+y)^2$$
Hence, $$(x^2+y^2)\geq \frac{6^2}{2}=18$$
|
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|
Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
|
Let's use degree measure (for writing convenience):
we need to prove identity
$$
\tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ = 1.\tag{1}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \cos 12^\circ \cos 48^\circ \cos 54^\circ \cos 72^\circ;\tag{2}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \sin 78^\circ \sin 42^\circ \sin 36^\circ \sin 18^\circ;
$$
$$
(\sin 12^\circ \sin 72^\circ ) \cdot ( \sin 48^\circ \sin 54^\circ) =^?= (\sin 18^\circ \sin 78^\circ ) \cdot ( \sin 36^\circ \sin 42^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ - \cos 102^\circ) =^?=
(\cos 60^\circ - \cos 96^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ + \cos 78^\circ) =^?=
(\cos 60^\circ + \cos 84^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
-\cos 84^\circ \cos 6^\circ + \cos 60^\circ \cos 78^\circ =^?=
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ;\tag{3}
$$
$$
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ=^?=0.\tag{4}
$$
Yes, $(4)$ is true identity, because
$$
2\sin 6^\circ \cos 6^\circ = \sin 12^\circ,
$$
$$
\sin 6^\circ \cos 6^\circ = \frac{1}{2}\sin 12^\circ,
$$
$$
\cos 84^\circ \cos 6^\circ = \cos 60^\circ \cos 78^\circ.\tag{5}
$$
So, we get equation
$$
\sin (x) = 1. \tag{6}
$$
Solution is obvious...
|
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|
Finding basis for the space spanned by some vectors. Find a subset of vectors $\{v_1, v_2, v_3, v_4, v_5\}$ that forms the basis for the space spanned by these vectors: $$v1=\left ( \begin{array}{c} 1\\-2\\0\\3 \end{array}\right), v2= \left ( \begin{array}{c} 2\\-5\\-3\\6\end{array}\right)\, ,\, v3=\left ( \begin{array}{c}1\\-1\\3\\1\end{array} \right)\, ,\, v4=\left(\begin{array}{c} 2\\-1\\4\\-7\end{array} \right )\, ,\, v5=\left (\begin{array}{c}3\\2\\14\\-17\end{array} \right).$$
|
You don't need to guess; just write down the matrix having the vectors as columns:
$$\begin{bmatrix}
1 & 2 & 1 & 2 & 3 \\
-2 & -5 & -1 & -1 & 2 \\
0 & -3 & 3 & 4 & 14 \\
3 & 6 & 1 & -7 & -17
\end{bmatrix}$$
and proceed with Gaussian elimination; first do $R_2+2R_1$ (sum to the second row the first multiplied by $2$) and then $R_4+(-3)R_1$ to get
$$\begin{bmatrix}
1 & 2 & 1 & 2 & 3 \\
0 & -1 & 1 & 3 & 8 \\
0 & -3 & 3 & 4 & 14 \\
0 & 0 & -2 & -13 & -24
\end{bmatrix}$$
I usually do pivot reduction, so multiply the second row by $-1$ and then do $R_3+3R_2$ to get
$$\begin{bmatrix}
1 & 2 & 1 & 2 & 3 \\
0 & 1 & -1 & -3 & -8 \\
0 & 0 & 0 & -5 & -10 \\
0 & 0 & -2 & -13 & -24
\end{bmatrix}$$
Now swap the third and fourth rows; if you also do pivot reduction you get
$$\begin{bmatrix}
1 & 2 & 1 & 2 & 3 \\
0 & 1 & -1 & -3 & -8 \\
0 & 0 & 1 & 13/2 & 12 \\
0 & 0 & 0 & 1 & 2 \\
\end{bmatrix}$$
Since we have pivots in the first four columns, we conclude that $v_1, v_2, v_3, v_4$ span your subspace. But, of course, since the dimension of the subspace is $4$, it is the whole $\mathbb{R}^4$, so any basis of the space would do.
These computations are surely easier than computing the determinant of a $4\times 4$ matrix. Note that if the dimension of the subspace were less than $4$, computing a determinant built with any set of four vectors would lead to nothing, while the elimination always works.
|
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|
Extended euclidean algorithm So I am trying to figure this out.
And for one of the problem the question is x*41= 1 (mod 99)
And the answer lists
x | 41x mod 99
0 99
1 41
-2 17
5 7
-12 3
29 1
And conclude x=29
How did they get this value(can someone explain)? To better put, how do you calculate this?
|
When two numbers $e,\varphi$ are given, $e<\varphi$, $GCD(e,\varphi)=1$, and we need to find $x$, such that
$$
x\cdot e = 1 (\bmod~\varphi),\tag{1}
$$
then denote
$$
r_0 = \varphi, \qquad v_0 = 0;
$$
$$
r_1 = e, \qquad\; v_1 = 1;
$$
then for each $n\geqslant 1$ to build values:
$$
s_n = \left\lfloor \frac{r_{n-1}}{r_n} \right\rfloor;
$$
$$
r_{n+1} = r_{n-1}-r_n s_n;
$$
$$
v_{n+1} = v_{n-1}-v_n s_n;
$$
and repeat it until $r_n=1$ $(r_{n+1}=0)$.
Last value $v_n$ (when $r_n=1$) will figure as solution of equation $(1)$.
It is comfortable to build appropriate table:
$$
\begin{array}{|c|c|r|r|}
\hline
n) & r_n & v_n & s_n & \color{gray}{check: ~~ e \cdot v_n \equiv r_n (\bmod~\varphi)} \\
\hline
0) & r_0 = \varphi = 99 & \mathbf{0} & - & - \\
1) & r_1 = e = 41 &
\mathbf{1} &
{\small\left\lfloor\frac{99}{41}\right\rfloor}= 2 &
\color{gray}{41\cdot 1 \equiv 41 (\bmod~99)} \\
2) & 17 & -2 &
{\small\left\lfloor\frac{41}{17}\right\rfloor}= 2 &
\color{gray}{41\cdot (-2) = -84 \equiv 17 (\bmod~99)} \\
3) & 7 & 5 &
{\small\left\lfloor\frac{17}{7}\right\rfloor}= 2 &
\color{gray}{41\cdot 5 = 205 \equiv 7 (\bmod~99)} \\
4) & 3 & -12 &
{\small\left\lfloor\frac{7}{3}\right\rfloor}= 2 &
\color{gray}{41\cdot (-12) = -492 \equiv 3 (\bmod~99)} \\
5) & 1 & x=\underline{\underline{29}} &
{\small\left\lfloor\frac{3}{1}\right\rfloor}= 2 &
\color{gray}{41\cdot 29 = 1189 \equiv 1 (\bmod~99)} \\
\color{gray}{6)} & \color{gray}{0} & & & \\
\hline
\end{array}
$$
|
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Proving the identity: $\sin3x + \sin x = 2\sin2x\cos x$ I need some help proving this identity:
$$\sin3x + \sin x = 2\sin2x\cos x$$
I don't know where to start. I thought about expanding $\sin 3x$ into $\sin (2x + x)$ but I don't think that does me any good. Any hints would be appreciated.
Thanks!
|
$$\sin 3x+\sin x=2\sin 2x\cos x$$
$$LHS
=\sin(2x+x)+\sin x\\
=\sin2xcosx+\sin x\cos2x+\sin x\\
=2(\sin x \cos x)\cos x+\sin(cos^2x-\sin^2x)+\sin x\\
=2\sin x \cos^2x+\sin x \cos^2x-\sin^3x+\sin x\\
=\cos^2x(2\sin x+\sin x)-\sin^3x+\sin x\\
=(1-\sin^2x)3\sin x-\sin^3x+\sin x\\
=3\sin x-\sin^3x-\sin^3x+\sin x\\
=4\sin x-4\sin^3x\\
=4(\sin x-\sin x(1-\cos^2x)\\
=4(\sin x \cos^2x)\\
=4\sin x \cos^2x$$
RHS
$$=2(2\sin x \cos x)\cos x\\
=4\sin x \cos^2x$$
|
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|
How to prove or disprove $n$ is prime? Let $n$ be an odd number such that
$$2^{\frac{n-1}{2}}\equiv -1 \pmod{n}$$
How do I prove or disprove $n$ must is prime?
This problem is from when I solved another problem. Thank you
|
The basic method for fast exponentiation is to first find the base 2 expansion of $1638$, i.e., notice that $1638=2+2^2+2^5+2^6+2^9+2^{10}$.
On the other hand, $2^2\equiv 4 \pmod{3277}$, $2^{2^2}\equiv 16 \pmod{3277}$, $2^{2^3}\equiv 256 \pmod{3277}$ and $2^{2^4}\equiv -4 \pmod{3277}$, $2^{2^5}\equiv 16 \pmod{3277}$, $2^{2^6}\equiv 256 \pmod{3277}$, $2^{2^7}\equiv -4 \pmod{3277}$, $2^{2^8}\equiv 16 \pmod{3277}$, $2^{2^9}\equiv 256 \pmod{3277}$, $2^{2^{10}}\equiv -4 \pmod{3277}$. Note that to get each term, I'm just squaring the previous term, so this is a pretty quick calculation.
Therefore, $2^{1638}\equiv 2^{2+2^2+2^5+2^6+2^9+2^{10}}\equiv (4)(16)(16)(256)(256)(-4)\equiv -1 \pmod{3277}$.
|
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Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
|
Note that real solutions of $x$ occur for $|x|\lt\sqrt 2$. Then start by dividing by $2$:
$$1=(2-x^2)\sqrt{2-x^2}=(2-x^2)^{3/2}$$
$$\implies 1=2-x^2 \implies x^2=1$$
Which implies that $x=\pm 1$.
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|
Prove the included formula relating cos(nx) and cos(x) I'm struggling with the below problem. Can anyone shed some light on it?
Show that the below formula is a correct relation between $y = \cos n\theta$ and $x = cos \theta$ for all $n$:
$$ x = \frac 12 \sqrt[n]{y + \sqrt{y^2 - 1}} + \frac 12 \sqrt[n]{y - \sqrt{y^2 - 1}} $$
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This may not be exactly the proof you want, but here is a proof (here, we will only use the fact that $i=\sqrt{-1}$ a couple of times, and nothing else about the complex number $i$...I will make note of when we use this).:
We proceed first to prove the identity
$\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n$
Clearly, this is true for $n=1$, so we proceed by induction. Assume this is true up to $n=k$, and then prove it for $k+1$.
$(\cos(\theta) + i\sin(\theta))^{k+1} = (\cos(\theta)+i\sin(\theta))^k (\cos(\theta) + i\sin(\theta))$
Using our induction hypothesis,
$= (\cos(k\theta) + i\sin(k\theta))(\cos(\theta) + i\sin(\theta))$
$= (\cos(k\theta)\cos(\theta) - \sin(k\theta)\sin(\theta)) + i(\cos(k\theta)\sin(\theta) + \sin(k\theta)\cos(\theta))$
$= \cos((k+1)\theta) + i\sin((k+1)\theta)$
We have thus proven it for $n=k+1$, hence it is proven for all positive integers.
For negative integers, note that the formula is true for $n=0$, because in both cases, it is just 1. Then,
$(\cos(\theta) + i\sin(\theta))^{-n} = ((\cos(\theta) + i\sin(\theta))^n)^{-1} = (\cos(n\theta) + i\sin(n\theta))^{-1}$
$ = (\cos(n\theta) - i\sin(n\theta)) = (\cos(-n\theta) + i\sin(-n\theta))$
The last two equalities follow by multiplying and dividing the last thing on the previous line by the complex conjugate. Now, return to the formula we just proved
$\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n$
Using this, we get
$(\cos(\theta) + i\sin(\theta)) = \sqrt[n]{\cos(n\theta) + i\sin(n\theta)}$
$= \sqrt[n]{\cos(n\theta) + \sqrt{-\sin^2(n\theta)}} = \sqrt[n]{\cos(n\theta) + \sqrt{-(1-\cos^2(n\theta))}} = \sqrt[n]{\cos(n\theta) + \sqrt{\cos^2(n\theta) - 1}}$
(In the second to last equality, we used $i=\sqrt{-1}$.) Also, using the identity with $n$ replaced by $-n$, we get
$(\cos(\theta) - i\sin(\theta)) = \sqrt[n]{\cos(n\theta) - i\sin(n\theta)} = \sqrt[n]{\cos(n\theta) - \sqrt{-\sin^2(n\theta)}}$
$= \sqrt[n]{\cos(n\theta) - \sqrt{-(1-\cos^2(n\theta))}} = \sqrt[n]{\cos(n\theta) -\sqrt{\cos^2(n\theta) - 1}}$
(In the second to last equality, we used $i=\sqrt{-1}$.) Add $1/2$ the first to $1/2$ the second to get
$\frac{1}{2}(\cos(\theta) + i\sin(\theta)) + \frac{1}{2}(\cos(\theta) - i\sin(\theta)) = \cos(\theta)$
$ = \frac{1}{2}\sqrt[n]{\cos(n\theta) + \sqrt{\cos^2(n\theta) - 1}} + \frac{1}{2}\sqrt[n]{\cos(n\theta) -\sqrt{\cos^2(n\theta) - 1}}$
Substituting $x = \cos(\theta)$, $y = \cos(n\theta)$ gives
$x = \frac{1}{2}\sqrt[n]{y + \sqrt{y^2-1}} + \frac{1}{2}\sqrt[n]{y-\sqrt{y^2-1}}$
Why the downvote? If I made a mistake, point it out in comments and I will fix it.
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Computing an inverse modulo $25$ Supposed we wish to compute $11^{-1}$ mod $25$. Using the extended Euclid algorithm, we find that $15 \cdot 25 - 34 \cdot 11 =1$. Reducing both sides modulo $25$, we have $-34 \cdot 11 \equiv 1$ mod $25$. So $-34 \equiv 16 $ mod $25$ is the inverse of $11$ mod $25$.
im realy confused about this problem, can someone go over this please and explain how this is done
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Note that $25 | 15 \cdot 25$, so $15 \cdot 25 \equiv 0 \pmod{25}$. On the other hand, we see that
$$1 \equiv -15 \cdot 25 - 34 \cdot 11 \equiv 0 + (-34) \cdot 11 \pmod{25}$$
Finally,
$$-34 \equiv -34 + 50 \equiv 16 \pmod{25}$$
so we can rewrite the above as
$$1 \equiv 16 \cdot 11 \pmod{25}$$
and $16$ is the desired inverse.
Note that by direct computation,
$$16 \cdot 11 = 176 = 7 \cdot 25 + 1$$
Alternatively, we can use the Euclidean algorithm and find that
\begin{align*}
25 &= 2 \cdot 11 + 3 \\
11 &= 3 \cdot 3 + 2 \\
3 &= 1 \cdot 2 + 1
\end{align*}
Thus,
\begin{align*}
1 &= 3 - 1 \cdot 2 \\
&= 3 - 1 \cdot (11 - 3 \cdot 3) = 4 \cdot 3 - 1 \cdot 11 \\
&= 4 \cdot (25 - 2 \cdot 11) - 1 \cdot 11 = 4 \cdot 25 - 9 \cdot 11
\end{align*}
So $1 \equiv -9 \cdot 11 \equiv 16 \cdot 11 \pmod{25}$ as before.
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Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we have
$$\Delta x=\frac{b-a}{n}=\frac{4}{n}$$
and
$$\xi_1=-2+\frac{4}{n};\;\;\xi_2=-2+2\frac{4}{n};\;\;\ldots\;\;;\xi_n=-2+n\frac{4}{n}$$
then
$$\xi_i=-2+\frac{4i}{n}=\frac{4i-2n}{n},$$
so that
$$ \begin{align} \int_{-2}^2 {4-x^2}\;dx=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{16i^2-16ni+4n^2}{n^2}\right)\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \left(\frac{4n^2-16i^2+16ni-4n^2}{n^2} \right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \frac{-16i^2+16ni}{n^2}\cdot\frac{4}{n} \\
=&\lim_{n\to+\infty}\sum_{i=1}^{n} \frac{-64i^2+64ni}{n^3}\\
=&\lim_{n\to+\infty} \frac{64}{n^3}\left(-\sum_{i=1}^{n} i^2+n\sum_{i=1}^{n} i\right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{n(n+1)(2n+1)}{6}+n\cdot \frac{n(n+1)}{2} \right)\\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{\color{#ff0000}{2}n^3+3n^2+n}{6}+\frac{n^3+n^2}{2} \right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(\frac{-\color{#ff0000}{2}n^3-3n^2-n+3n^3+3n^2}{6}\right)\\
=&\lim_{n\to+\infty}\frac{32}{3} \left(\frac{\color{#ff0000}{1\times}n^3-n}{n^3}\right) \\
=&\lim_{n\to+\infty}\frac{64}{\color{#ff0000}{2\times}3}-\frac{32}{3n^2}=\frac{64}{\color{#ff0000}{2\times}3} \color{#ff0000}{=}\frac{32}{3}=\int_{-2}^{2} {4-x^2}\;dx
\end{align}$$
Where is the mistake?
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We have $$\Delta x=\frac{b-a}{n}=\frac{4}{n}$$ and $$\xi_1=-2+\frac{4}{n};\;\;\xi_2=-2+2\frac{4}{n};\;\;...\;\;;\xi_n=-2+n\frac{4}{n}$$ then $$\xi_i=-2+\frac{4i}{n}=\frac{4i-2n}{n}$$ $$\int_{-2}^2 {4-x^2}\;dx=\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\frac{4}{n}\\\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\cdot\frac{4}{n}\\
\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{16i^2-16ni+4n^2}{n^2}\right)\right)\cdot\frac{4}{n}\\
\lim_{n\to+\infty} \sum_{i=1}^n \left(\frac{4n^2-16i^2+16ni-4n^2}{n^2} \right)\cdot\frac{4}{n}\\
\lim_{n\to+\infty} \sum_{i=1}^n \frac{-16i^2+16ni}{n^2}\cdot\frac{4}{n}=\lim_{n\to+\infty}\sum_{i=1}^{n} \frac{-64i^2+64ni}{n^3}\\\lim_{n\to+\infty} \frac{64}{n^3}\left(-\sum_{i=1}^{n} i^2+n\sum_{i=1}^{n} i\right)=\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{n(n+1)(2n+1)}{6}+n\cdot \frac{n(n+1)}{2} \right)=\\
\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{2n^3+3n^2+n}{6}+\frac{n^3+n^2}{2} \right)=\lim_{n\to+\infty} \frac{64}{n^3} \left(\frac{-2n^3-3n^2-n+3n^3+3n^2}{6}\right)\\
\lim_{n\to+\infty}\frac{32}{3} \left(\frac{n^3-n}{n^3}\right)=\lim_{n\to+\infty}\frac{32}{3}-\frac{32}{3n^2}=\frac{32}{3}=\int_{-2}^{2} {4-x^2}\;dx$$
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Probability of rolling 6's on 3 dice - adjusted gambling odds? I'm struggling to work out odds on a game that were working on. It's probably best if I write an example as I'm really not a mathematician!
I'm working on a dice game where the player bets 1 coin and rolls 3 dice. If any of the dice are a 6 we payout based on the following.
If 1 dice is a six we pay x
If 2 are a six we pay y
and if all of them are a 6 we pay z
This seemed simple at first as the odds of rolling 1 six are 1/2, 2 sixes would be 1/14 and 3 sixes would be 1/216 so they would be our odds.
The issue is that in theory would have to roll the dice 216 times to get the jackpot and receive the £216 payout (at £1 per bet). However, if I did do that I would also win 108 times at 2/1, 15 times at 14.4/1 and once at 216/1 (ignoring the luck factor etc).
Therefor for my £216 stake i would win a total of £648.
So what should my payouts be for 1x6, 2x6, and 3x6 (assuming 0% house edge and all that) to be fair to the player?
Thanks for you time.
Mark
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The probabilities you have are wrong. Denote by $X$ the number of sixes rolled. Then the correct probabilities are:
$P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
$P(X = 1) = P$ (first dice six) $ + P$(second dice six) $ + P$(third dice six) $= \frac{1}{6} \frac{5}{6} \frac{5}{6} + \frac{5}{6} \frac{1}{6} \frac{5}{6} + \frac{5}{6} \frac{5}{6} \frac{1}{6} = 3 \cdot \frac{25}{216} = \frac{75}{216} = \frac{25}{72}$.
$P(X = 2) = P$ (first dice not six) $ + P$(second dice not six) $ + P$(third dice not six) $= \frac{5}{6} \frac{1}{6} \frac{1}{6} + \frac{1}{6} \frac{5}{6} \frac{1}{6} + \frac{1}{6} \frac{1}{6} \frac{5}{6} = 3 \cdot \frac{5}{216} = \frac{15}{216} = \frac{5}{72}$.
$P(X = 3) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$
To answer your question, if you want a fair game (expectation zero profit/loss) you can pick $x,y,z$ however you want, as long as they satisfy $75x + 15y + z = 216$. Namely, on average in 216 rolls, you will get one six $75$ times, two sixes $15$ times and three sixes once. Reasonable values could be $x = 2, y = 3.4, z = 15$.
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Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
I think I got it, but is this proof correct?
We can write any integer x in the form: $x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4$, and $x = 6k + 5$.
If $x = 6k$, then $x^3 = 216k^3$. Then $x^3 - x = 216k^3 - 6k = 6(36k^3 - k)$. Thus, $6 | (x^3 - x)$. Thus, $x^3 \equiv x \pmod 6$.
If $x = 6k + 1$. Then $x^3 - x = (216k^3 + 108k^2 + 18k + 1) - (6k + 1) = 6(36k^3 + 18k^2 + 2k)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
If $x = 6k + 2$, then $x^3 - x = (216k^3 + 216k^2 + 72k + 8) - (6k + 2) = 6(36k^3 + 36k^2 + 11k + 1)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$
If $x = 6k + 3$, then $x^3 - x = (216k^3 + 324k^2 + 162k + 27) - (6k + 3) = 6(36k^3 + 54k^2 + 26 + 4)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6 $
If $x = 6k + 4$, then $x^3 - x = (216k^3 + 432k^2 + 288k + 64) - (6k + 4) = 6(36k^3 + 72k + 47k + 10)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
If $x = 6k + 5$, then $x^3 - x = (216k^3 + 540k^2 + 450k + 125) - (6k + 5) = 6(36k^3 + 90k + 74k + 20)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
In all cases, we have shown that $x^3 \equiv x \pmod 6$. QED.
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Yes, that works just fine.
Another way is to observe that if $$x^3\equiv x\pmod 6$$ then also $$x^3-x\equiv 0\pmod 6$$
and so the original statement is equivalent to the assertion that $x^3-x$ is always a multiple of 6.
Since $x^3-x$ factors as $(x-1)\cdot x\cdot (x+1)$ it is a product of three consecutive numbers; one of these must be a multiple of 3 and at least one must be even, so the product is certainly a multiple of 6.
You can also use your method, with much less trouble, by dealing with the general case of $x=6k+i$, where $i\in\{0,1,2,3,4,5\}$, rather than dealing with the six values of $i$ one by one. Then you are asking if $x^3\equiv x\pmod 6$ when $x=6k+i$, and by following exactly the same steps that you did in your solution, but once instead of six times, you get
$$\begin{align}
(6k+i)^3 & \stackrel?{\equiv} 6k+i \pmod 6 \\
216k^3 + 72k^2i + 18ki^2 + i^3 & \stackrel?{\equiv} 6k + i \pmod6\\
6(36k^3 + 12k^2i + 3ki^2) + i^3 & \stackrel?{\equiv} 6(k) + i \pmod6\\
i^3 & \stackrel?{\equiv} i \pmod6\\
\end{align}
$$
and now you only have to check whether the claim $i^3\equiv i\pmod 6$ is true for the six possible values of $i$, namely $\{0,1,2,3,4,5\}$. This eliminates five-sixths of the algebra, and so five-sixths of the opportunities to make a silly mistake.
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The probability that it will take at least $k$ tosses until the first head A fair coin is tossed until a head is obtained. So for example, the sample space can be modeled as {$H, TH, TTH, TTTH, ...$}. The probability measure is modeled so that $P(H)=\frac 12$, $P(TH)=\frac 14$, $P(TTH)=\frac 18$ etc.
Find the probability that the total number of tosses is at least $k$.
Progress: the probability of getting first head on the $k$th toss is $1/2^k$, so I was trying to use this fact, although not sure how to calculate it as "at least k".
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Let $X$ be the number of moves. Then
$$P(X=k) = \frac{1}{2^k}$$
(a) $$P = 1 - P(X > k) = 1 - P(k \text{ tails}) = 1 - \frac{1}{2^k} = \frac{2^k - 1}{2^k}$$
(b) $$P = \sum_{k=0}^\infty P(X=2k+1) = \sum_{k=0}^\infty \frac{1}{2^{2k+1}} = \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k} = \frac{1}{2} \frac{1}{1- \frac{1}{4}} = \frac{2}{3}$$
(c) $$P = \sum_{k=0}^\infty P(X=3(k+1)) = \ldots = \frac{1}{7}$$
(d) $$P = P_b + P_c - \sum_{k=0}^\infty P(X = 3(2k+1)) = \ldots = \frac 2 3 + \frac 1 7 - \frac 1 8 \cdot \frac{1}{1 - \frac 1 {64}} = \frac{43}{63}$$
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Prove that $6$ divides $n(n + 1)(n + 2)$ I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows:
Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$.
Note: An integer $a$ divides an integer $b$, written $a|b$, if there exists $q ∈ Z$ such that $b = qa$. Alternatively, $a|b$ if dividing $b$ by $a$, $b ÷ a$, results in an integer.
Should I do a proof by induction?
All help/input is appreciated!
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If we let $m=n+1$, we need to prove that $6 \vert m(m^2-1)$ for all $m \geq 2$. Equivalently, we want to show that
$$m^3 \equiv m \bmod6$$
First note that $m \equiv 0,1,2,3,4,5 \pmod 6$. Now observe that following:
$$m \equiv 0 \bmod 6 \implies m^3 \equiv 0^3 \bmod 6 \implies m^3 \equiv 0 \bmod6 $$
$$m \equiv 1 \bmod 6 \implies m^3 \equiv 1^3 \bmod 6 \implies m^3 \equiv 1 \bmod6$$
$$m \equiv 2 \bmod 6 \implies m^3 \equiv 2^3 \bmod 6 \implies m^3 \equiv 8 \bmod6 \implies m^3 \equiv 2 \bmod6$$
$$m \equiv 3 \bmod 6 \implies m^3 \equiv 3^3 \bmod 6 \implies m^3 \equiv 27 \bmod6 \implies m^3 \equiv 3 \bmod6$$
$$m \equiv 4 \bmod 6 \implies m^3 \equiv 4^3 \bmod 6 \implies m^3 \equiv 64 \bmod6 \implies m^3 \equiv 4 \bmod6$$
$$m \equiv 5 \bmod 6 \implies m^3 \equiv 5^3 \bmod 6 \implies m^3 \equiv 125 \bmod6 \implies m^3 \equiv 5 \bmod6$$
Hence, in all cases, we have $m^3 \equiv m \bmod 6$, which means that $$6 \vert (m^3-m) \implies 6 \vert n(n+1)(n+2)$$
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Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$ This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)(\sqrt{x^2+2}+\sqrt{2})}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} = \lim_{x \to 0}\frac{({x^2+2}-2)(\sqrt{x^2+1}+1)}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} =\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{1}{2\sqrt{2}}$$
But Wolframalpha gave me another answer!
So did I do it right?
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If you correctly simplify the very last line in your solution, you'll see you get the same answer: $\frac 2{2\sqrt 2}$
$$\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{2}{2\sqrt{2}} = \frac 1{\sqrt 2}$$
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$|3^a-2^b|\neq p$, from a contest I recently came across an old contest problem: (I did not find the solution anywhere)
Find the least prime number which cannot be written in the form $|3^a-2^b|$ where $a$ and $b$ are nonnegative integers.
At the beginning I thought that the answer was the number $2$ but the ''nonnegative integers'' allows us to take $a=1$ and $b=0$.
Also we have:
$3=2^2-3^0$
$5=3^2-2^2$
$7=2^3-3^0$
$11=3^3-2^4$
$13=2^4-3^1$
$17=3^4-2^6$
and so on...
Can anybody help me with this?
Thanks in advance
P.S. I am familiar with Pillai's theorem http://mathworld.wolfram.com/PillaisTheorem.html
but i would like to see -if possible- a solution much more simple because it was a contest problem.
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One readily finds representations for all primes $<41$.
Assume $41=|3^a-2^b|$ that is $2^b\pm 41=3^a$.
By not finding powers of three among the numbers $2^b\pm41$ with $b\le 2$, we conclude $b\ge 3$.
Then $3^a\equiv \pm41\pmod{8}$, which is equivalent to $a\equiv 0\pmod 2$.
Especially $3^a\equiv1\pmod {8}$ and hence $41=3^a-2^b$.
Likewise, we can check and exclude small values of $a$, hence know that $2^b\equiv -41\pmod{3}$, which is equivalent to $b\equiv 0\pmod 2$.
But then $a$ and $b$ are both even and we obtain a factorization
$41=(3^{a/2}-2^{b/2})(3^{a/2}+2^{b/2})$ and conclude $3^{a/2}+2^{b/2}=41$, $3^{a/2}-2^{b/2}=1$, i.e. $3^{a/2}=\frac{41+1}2$, which is absurd.
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Elementary Algebra Inequality question $$
\frac{1}{x-1} < -\frac{1}{x+2}
$$
(see this page in wolframalpha)
Ok, so I think the main problem is that I don't really know how to do these questions. What I tried to do was move $-1/(x + 2)$ to the LHS and then tried to get a common denominator. I ended up with
$$
\frac{(x + 2) + (x -1)}{(x-1)(x+2)} < 0.
$$
So then I went
$$
\frac{2x + 1}{(x-1)(x+2)}<0
$$
and got $x < -1/2$ and $x \ne 1$ and $x\ne -2$. Therefore, the answer should be $x$ is $(-\infty, -2)$ or $(-2, -1/2)$. But it's not.
|
$\frac{1}{x-1}<-\frac{1}{x+2} \rightarrow \frac{1}{x-1}+\frac{1}{x+2}<0 \rightarrow$
$\frac{(x+2)+(x-1)}{(x+2)(x-1)} <0 \rightarrow \frac{(x+2)+(x-1)}{(x+2)(x-1)}((x+2)(x-1)^2<0 \rightarrow (x+2+x-1)(x+2)(x+1)<0 $
for $x$ different to $-2,1$. (-2 and 1 arent soultions cause they are undefined.)
using the fact that rational functions are continuous wherever they are defined and the intermediate value theorem:
$(x+2+x-1)(x+2)(x-1)=2(x+\frac{1}{2})(x+2)(x+1)<0$
Since the polynomial 2(x+\frac{3}{2})(x+2)(x+1) has three roots:(each with multiplicity 1)
$-\frac{1}{2},-1,-2$ and -1000 is a solution to the inequality we can know that the solution set is $(-\infty,-2)\cup (-1,-\frac{1}{2})$
|
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|
Integral of $\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}$dx Question:
$$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$
What we did:
we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this:
$\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1} $, for which we know how to solve the first and second integral but not the third...
Thanks
|
Alternatively, if we call our integral $I$ and perform a dummy variable substitution, it all falls out rather nicely! $$I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x \ \overset{x \mapsto \frac{\pi}{2} - x}= \ \int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x $$ Adding the two integrals together gives $$ 2I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x = \int_{0}^{\pi/2} \text{d}x = \dfrac{\pi}{2} \implies \ I = \dfrac{\pi}{4}$$
|
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|
How to obtain $y$
The question was written with dark-blue pen. And I tried to solve this question. I obtained $x$ as it is below. But I cannot obtain $y$ Please show me how to do this.
By the way, $\gamma (t)$ may not be clearly readable. So, I wrote again.
$$\gamma (t)=( \cos ^2 (t)-1/2, \sin(t)\cos (t), \sin (t))$$
Thanks for helping.
-sorry for not writing with MathJax. -
|
From your work,
$$x^2+y^2=\frac{1}{4} \Rightarrow y^2=\frac{1}{4}-x^2,$$
and
$$x=\cos^2 t -\frac{1}{2}.$$
Substituting the latter into the former produces
\begin{align*}
y^2 &=\frac{1}{4}-x^2 \\
&=\frac{1}{4}-\left( \cos^2 t - \frac{1}{2} \right)^2 \\
&=\frac{1}{4}-\left( \cos^4 t - \cos^2 t + \frac{1}{4} \right) \\
&=\cos^2 t - \cos^4 t \\
&=\cos^2 t\left( 1-\cos^2 t \right) \\
&=\cos^2 t \sin^2 t \\
\Rightarrow y &= \sin t \cos t,
\end{align*}
as was your intention.
|
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|
Determinant of a General Expression Matrix I don't really know how to do this question:
" Let $A = $
$\begin{pmatrix}1 & 1& 1 & \cdots & 1 \\ 1 & 1-x & 1 & \cdots & 1 \\ 1 & 1& 2-x &\cdots & 1 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1& 1 & 1 &\cdots& n-1-x \end{pmatrix}$"
be an n×n matrix with n≥3. Solve the equation det(A)=0 of x∈R and find the largest root.
|
Subtract the first column from the last, then $$\det A=\det\begin{pmatrix}1 & 1& 1 & \cdots & 1 & 0 \\ 1 & 1-x & 1 & \cdots & 1 & 0 \\ 1 & 1& 2-x &\cdots & 1 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1& 1 & 1 &\cdots& n-2-x & 0\\1& 1 & 1 &\cdots& 1 & n-2-x \end{pmatrix}=$$$$(n-2-x)\det\begin{pmatrix}1 & 1& 1 & \cdots & 1 \\ 1 & 1-x & 1 & \cdots & 1 \\ 1 & 1& 2-x &\cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1& 1 & 1 & \cdots& n-2-x \end{pmatrix}.$$ How can you continue from here?
|
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|
Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show there exists two more solutions to this equation by using number theory vs college algebra. Does it have to do with the relationship that $0\leq y^2 \leq 10$ ? So does this say anything about solutions being irrational?
|
The number of solutions has nothing to do with some of the intersection points being rational, which is a happy coincidence. The way to approach this problem is plain old algebra. Add twice the first equation to the second, and divide through by 3 to get
$$2x y + 3x^2 = 2.$$
Then manipulating the second equation gives
$$9x^4 + 4x^2y^2 = 10x^2$$
$$9x^4 + (2-3x^2)^2 = 10x^2$$
This is a quadratic equation in $x^2$, which has solutions $x^2 = 1, \frac{\sqrt{2}}{3}.$
When $x^2 = 1$, from the second equation we have $y^2 = \frac{1}{4}$, checking these in the first equation reveals that two of the sign choices are spurious, leaving $(-1, \frac{1}{2})$ and $(1, -\frac{1}{2})$ as true solutions.
Similarly, when $x^2 = \frac{\sqrt{2}}{3}$, you get two solutions $(-\frac{\sqrt{2}}{3}, -\sqrt{2})$ and $(\frac{\sqrt{2}}{3}, \sqrt{2})$.
Since your two polynomials have no common polynomial divisor, by Bezout's Theorem, the system
of polynomial equations has at most four solutions, and we have found them all. (Geometrically, your equations describe a nondegenerate hyperbola and ellipse, which can intersect in at most four points.)
|
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|
Complex Number - Find all roots of the equation $$e^{i \frac{{\pi}}{3}}z^5+4e^{i\frac{(2+3){\pi}}{6}}z^3 + z^2 + 4i = 0.$$
By using Euler's formula, I got:
$$e^{i \frac{{\pi}}{3}} = \cos{\frac {\pi}{3}} + i\sin{\frac {\pi}{3}} = (\frac{1}{2} + i{\frac {\sqrt{3}}{2}}) $$
$$4e^{i \frac{(2+3){\pi}}{6}} = 4(\cos{\frac {5\pi}{6}} + i\sin{\frac {5\pi}{6}}) = 4 ({\frac {-\sqrt{3}}{2}} + i \frac{1}{2}) $$
but how do i proceed from here? Any help is much appreciated!
Thank you.
|
Hint: The polynomial can be factored as
$$
e^{i\pi/3} (z^3+a)(z^2+b).
$$
More: Multiply it out:
$$
e^{i\pi/3} (z^3+a)(z^2+b) = e^{i\pi/3}z^5+be^{i\pi/3}z^3 + ae^{i\pi/3}z^2 + abe^{i\pi/3}.
$$
To make this match the original polynomial, $e^{i\pi/3}z^5 + 4e^{i5\pi/6}z^3 + z^2 + 4i$, we need
$$
be^{i\pi/3} = 4e^{i5\pi/6}, \qquad ae^{i\pi/3} = 1, \qquad abe^{i\pi/3} = 4i.
$$
From the first equation we get
$$
b = 4e^{i3\pi/6} = 4i
$$
and from the second we get
$$
a = e^{i5\pi/3}.
$$
Thus
$$
e^{i\pi/3}z^5 + 4e^{i5\pi/6}z^3 + z^2 + 4i = e^{i\pi/3}(z^3 + e^{i\pi/3})(z^2+4i).
$$
Now just solve
$$
z^3 + e^{i\pi/3} = 0
$$
and
$$
z^2+4i = 0
$$
to find the roots of the equation.
|
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|
How to prove $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$? If $m$ and $n$ are any two positive integers then how do we show that the inequality $$\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$$ always holds ?
|
let
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}$$
then we easy to
$$f(1,1)=f(1,2)=f(2,1)=\dfrac{1}{12}\le\dfrac{4}{45}$$
so,when $m,n\ge 2,$,then we let $j=m+n+2\ge 6$
then
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}\le\dfrac{1}{j-1}-\dfrac{4}{j^2}\le\dfrac{4}{45}$$
It is clearly,
because $$f(x)=\dfrac{1}{x-1}-\dfrac{4}{x^2}\Longrightarrow f'(x)=-\dfrac{1}{(x-1)^2}+\dfrac{8}{x^3}=\dfrac{8(x-1)^2-x^3}{x^3(x-1)^2}<0,x\ge 6$$let
where $j\ge 6$.
|
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|
Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.
|
let $a=x^2,b=y^2,c=z^2$
$$\Longleftrightarrow x^2+y^2+z^2=3\Longrightarrow \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3$$
note
$$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2z^2x^2+2y^2z^2$$
use
AM-GM inequality,then we have
$$4\dfrac{x^2}{y}+2x^2y^2+x^4\ge (2\cdot 2+2+1)x^{\frac{2\cdot 2\times 2+2\cdot 2+2\cdot 2}{2\cdot 2+2+1}}=7x^{\frac{16}{7}}$$
and the same
$$4\dfrac{y^2}{z}+2y^2z^2+y^4\ge 7y^{\frac{16}{7}}$$
$$4\dfrac{z^2}{x}+2z^2x^2+z^4\ge 7z^{\frac{16}{7}}$$
so
$$4\sum\dfrac{x^2}{y}+(x^2+y^2+z^2)^2\ge 7\sum x^{\frac{16}{7}}$$
$$\Longleftrightarrow x^{\frac{16}{7}}+y^{\frac{16}{7}}+z^{\frac{16}{7}}\ge x^2+y^2+z^2$$
use AM-GM inequality
we have
$$7x^{\frac{16}{7}}+1=x^{\frac{16}{7}}+x^{\frac{16}{7}}+\cdots+x^{\frac{16}{7}}+1\ge 8\sqrt[8]{x^{\frac{16}{7}\cdot 7}}=8x^2 $$
$$7\sum x^{\frac{16}{7}}+\sum x^2\ge \sum 8x^2$$
so
$$\sum x^{\frac{16}{7}}\ge \sum x^2$$
In general,we have
$x^n+y^n+z^n=3,2p+q>2n,p,q,n\in N^{+}$,then
$$\dfrac{x^p}{y^q}+\dfrac{y^p}{z^q}+\dfrac{z^p}{x^q}\ge 3$$
|
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|
Intersection of two tangents on a parabola proof There are two tangent lines on a parabola $x^2$. The $x$ values of where the tangent lines intersect with the parabola are $a$ and $b$ respectively. The point where the two tangent lines intersect has an $x$ value of $c$.
Prove that $c=(a+b)/2$
I have tried taking the derivative of $x^2$ and using the point slope form to find each tangent line's equation, and then making each tangent equal to each other to find that $c$ is equal to $a+b$, but each time I try it out I can't get anything out of the algebra
|
Let's try to work through that algebra, then.
The derivative of $x^2$ is $2x$, so those tangent lines are given by
$$
y = 2a(x-a) + a^2\\
y = 2b(x-b) + b^2
$$
Thus, at the $x$-coordinate of the lines' intersection, we have
$$
2a(x-a) + a^2 = 2b(x-b) + b^2
$$
Expanding the above expressions, we have
$$
2ax - 2a^2 + a^2 = 2bx - 2b^2 + b^2\implies\\
2ax - a^2 = 2bx - b^2
$$
The tricky part comes next, where we factor in order to "combine like terms". We have
$$
2ax - 2bx = a^2 - b^2\\
2(a - b)x = a^2 - b^2
$$
Now, assuming $a\neq b$, we may divide both sides by $2(a-b)$ to get
$$
x = \frac 12 \frac{a^2 - b^2}{a - b}
= \frac 12 \frac{(a-b)(a+b)}{a - b}
= \frac 12 (a+b)
$$
As desired. I hope that clears things up.
|
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|
Multivaraible calculus,Global minima maxima How do I compute the global minimum or maximum of the function
$f(x,y)=-\sin x\cos y$.
Given it is on a square $(0\leq x\leq 2\pi)$ and $(0\leq y\leq2\pi)$
|
The inequality chain already mentioned by user7530 establishes the range of the function rather quickly:
$$ -1 \ \le \ \sin \ x \ \le \ +1 \ \ \Rightarrow \ \ -1 \ \le \ -\sin \ x \ \le \ +1 $$
$$ \Rightarrow \ \ -1 \ \le \ -\cos \ y \ \le \ -\sin \ x \ \cos \ y \ \le \ \cos \ y \ \le \ +1 \ \ . $$
It's a little more work to locate the points where the global extrema occur, at least by some methods, because of a small complicating factor. For the region $ \ (x, \ y) \ \in \ [0 \ , \ 2 \pi] \times [0 \ , \ 2 \pi] \ $ , we can work with the sinusoidal functions involved in various ways.
$ \ $
1. Since $ \ \sin \ x \ > \ 0 \ \ $ on $ \ \ (0 \ , \ \pi) \ \ , \ \ \sin \ x \ < \ 0 \ \ $ on $ \ \ (\pi \ , \ 2 \pi) \ \ , \ \ \cos \ x \ > \ 0 \ \ $ on $ \ \ (0 \ , \frac{\pi}{2}) \ \ , $ $ (\frac{3 \pi}{2} \ , \ 2 \pi) \ \ $ , and $ \ \ \cos \ x \ < \ 0 \ \ $ on $ \ \ (\frac{\pi}{2} \ , \ \frac{3 \pi}{2}) \ $ , our square is divided up into six sections, shown in the graph below, with the regions marked in green being those where $ \ f(x, y) \ = \ -\sin x \ \cos y \ > \ 0 \ $ and those where $ \ f(x, y) \ < \ 0 \ $ (the lines separating the regions indicate where $ \ f(x, y) \ = \ 0 \ $ ) .
We know from the symmetry of the sinusoidal functions that their extrema occur midway between their zeroes. Consequently, the global extrema occur at the centers of the squares of what would be the "infinite checkerboard" covering the plane, the maxima being marked by the green stars, the minima, by red stars.
$ \ $
2. We can apply a "product-to-sum formula" for the sinusoidal functions to write
$$ -\sin \ x \ \cos \ y \ = \ -\frac{1}{2} \ [ \ \sin(x + y) \ + \ \sin(x - y) \ ] \ \ . $$
We can see that this function will attain its global extrema among the points where $ \ \sin(x + y) \ = \ \pm 1 \ \ $ and $ \ \ \sin(x - y) \ = \ \pm 1 \ $ . This gives us the "candidate" results
$$ x \ + \ y \ = \ \frac{\pi}{2} \ (2m \ + \ 1) \ \ , \ \ x \ - \ y \ = \ \frac{\pi}{2} \ (2n \ + \ 1) \ \ , \quad \mathbf{[ 1 ]} $$
that is, the sum and difference of $ \ x \ $ and $ \ y \ $ must be odd integer multiples of $ \ \frac{\pi}{2} \ $ . The graph below shows the lines described by these equations as they pass through our square region.
The complication in this approach is that we want both sine-function terms in the sum for our function to have the same sign ; where they do not, the sum of terms is zero. Of the twelve candidates implied by the intersections of these lines, six of them are points where the terms have opposite signs [marked by " 0 "] ; eliminating these leaves the six points where the extrema are found [marked by asterisks].
(A somewhat more analytical approach would be to solve the pair of equations [ 1 ] above for $ \ x \ $ and $ \ y \ $ , giving us
$$ x \ = \ (m \ + \ n \ + 1 ) \ \cdot \ \frac{\pi}{2} \ \ , \ \ y \ = \ (m \ - \ n ) \ \cdot \ \frac{\pi}{2} \ \ . $$
For $ \ m \ $ and $ \ n \ $ both odd or both even, $ \ x \ $ is then an odd multiple of $ \ \frac{\pi}{2} \ $ and $ \ y \ $ , an even multiple of $ \ \frac{\pi}{2} \ $ , or equivalently, any integer multiple of $ \ \pi \ $ ; these are the locations of the global extrema. For one of $ \ m \ $ or $ \ n \ $ being odd and the other even, $ \ x \ $ is the integer multiple of $ \ \pi \ $ and $ \ y \ $ , the odd multiple of $ \ \frac{\pi}{2} \ $ ; these are where the zeroes of $ \ f(x, y) \ $ lie.)
$ \ $
3. We could, of course, use calculus on this problem. The first partial derivatives of our function are
$$ f_x \ = \ - \cos \ x \ \cos \ y \quad , \quad f_y \ = \ \sin \ x \ \sin \ y \ . $$
Setting these derivatives equal to zero and solving for $ \ x \ $ and $ \ y \ $ leads to essentially the same situation as in the previous approach: twelve candidate points. (This is unavoidable since the function and its derivatives have factors which become zero at the same variable values.) We can construct the "discriminator"
$$ D \ = \ f_{xx} \ f_{yy} \ - \ (f_{xy})^2 \ $$
$$ = \ (\sin \ x \ \cos \ y) \ (\sin \ x \ \cos \ y) \ - \ (\cos \ x \ \sin \ y)^2 $$
$$ = \ \sin^2 x \ \cos^2 y \ - \ \cos^2 x \ \sin^2 y \ = \ \sin^2 x \ - \ \sin^2 y \ \ . $$
For those points where $ \ x \ $ is an integer multiple of $ \ \pi \ $ and $ \ y \ $ an odd multiple of $ \ \frac{\pi}{2} \ $ , we have $ \ D \ = \ 0^2 \ - \ 1^2 \ = \ -1 \ $ , which means that the zeroes of $ \ f(x,y) \ $ are saddle points. In the contrasting case, $ \ D \ = \ 1^2 \ - \ 0^2 \ = \ +1 \ $ , indicating that these points are relative (local) extrema. Specifically, we find
$$ \text{for} \ \ \left( \frac{\pi}{2}, \ 0 \ \right) \ , \ \left( \frac{\pi}{2}, \ 2 \pi \ \right) , \ \left( \frac{3 \pi}{2}, \ \pi \ \right) \ \ , \ \ f_{xx} \ = \ +1 \quad \text{[minima]} \ \ ; $$
$$ \text{for} \ \ \left( \frac{\pi}{2}, \ \pi \ \right) \ , \ \left( \frac{3 \pi}{2}, \ 0 \ \right) , \ \left( \frac{3 \pi}{2}, \ 2 \pi \ \right) \ \ , \ \ f_{xx} \ = \ -1 \quad \text{[maxima]} \ \ . $$
As has already been indicated by the bounded range of this function, the relative extrema listed here are also global extrema.
|
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Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? It seems like an expression from a simple induction proof problem that's missing its result.
|
It can be transformed into
$$
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{1+2+3+\cdots +n}}=
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{n(n+1)/2}}
$$
but whether this is a simplification I don't know.
|
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|
Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
|
Note that (can be easily seen with rule of Sarrus)$$
\begin{vmatrix}
x & y & z \\
z & x & y \\
y & z & x \\
\end{vmatrix}=x^3+y^3+z^3-3xyz
$$
On the other hand, it is equal to (if we add to the first row 2 other rows)
$$
\begin{vmatrix}
x+y+z & x+y+z & x+y+z \\
z & x & y \\
y & z & x \\
\end{vmatrix}=(x+y+z)\begin{vmatrix}
1 & 1 & 1 \\
z & x & y \\
y & z & x \\
\end{vmatrix}=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
$$
just as we wanted. The last equality follows from the expansion of the determinant by first row.
|
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|
$Z_1$, $Z_2$, and $Z_3$ are independent standard normal variables. Obtain the correlation matrix Define
$X_1= 2Z_1 - Z_2 +3Z_3$
$X_2=Z_1 + Z_2 - Z_3$
$X_3 = 3Z_1 + 2Z_2 + Z_3$
So would the correlation matrix of $(X_1, X_2, X_3)$ just be the coefficients of each variable?
|
Initially you have
$$
\mathbf{Z}=\begin{pmatrix}Z_1 \\ Z_2 \\ Z_3 \end{pmatrix}\sim N\left(\boldsymbol{0}, \mathbf{I}\right)
$$
Define $\mathbf{X}$:
$$
\mathbf{X}=\begin{pmatrix}X_1 \\ X_2 \\ X_3 \end{pmatrix}=\begin{pmatrix}2Z_1-Z_2+3Z_3 \\ Z_1+Z_2-Z_3 \\ 3Z_1+2Z_2+Z_3 \end{pmatrix}=\underbrace{\begin{pmatrix}2 & -1 & 3 \\1 & 1 & -1 \\ 3 & 2 & 1\end{pmatrix}}_{\mathbf{B}}\begin{pmatrix}Z_1 \\ Z_2 \\ Z_3 \end{pmatrix}
$$
For linear combinations of normal random variables, we have the following:
If $\mathbf{Z}\sim N(\boldsymbol{\mu}, \mathbf{\Lambda})$ and $\mathbf{X}=\mathbf{a}+\mathbf{BZ}$, then $\mathbf{X}\sim N(\boldsymbol{\mu}+\mathbf{a}, \mathbf{B}\boldsymbol{\Lambda}\mathbf{B}^\prime)$.
Hence, the covariance matrix of $\mathbf{X}$ is:
$$
\mathbf{BIB}^\prime=\begin{pmatrix}2 & -1 & 3 \\1 & 1 & -1 \\ 3 & 2 & 1\end{pmatrix}\begin{pmatrix}2 & 1 & 3 \\-1 & 1 & 2 \\ 3 & -1 & 1\end{pmatrix}=\begin{pmatrix}14 &-2 & 9\\ -2 & 3 &4 \\ 9 & 4 & 14\end{pmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/546969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?
|
Constrant:
$$q=x^2 +y^2-1=0$$
Goal:
$$ c= x+ y + \frac{1}{xy}$$
Now, by LaGrange multipliers , we can say that:
$$ \nabla q = \lambda \nabla c$$
Hence,
$$2x = 1 - \frac{1}{x^2y} \tag{1}$$
$$ 2y = 1- \frac1{y^2 x} \tag{2}$$
By multiplication of $x$ in eqtn (1) and $y$ in eqtn (2):
$$ 2x^2 = x - \frac{1}{xy} \tag{3}$$
$$ 2y^2 = y - \frac{1}{yx} \tag{4}$$
Substract the two equations:
$$ 2(x^2 - y^2) = (x-y)$$
Now, either $x=y$ or $x+y = \frac12$, substitute this into the equation $(x+y)^2 -2xy -1 = 0$ while considering the constraints
$$ 2y^2 =1 \to (x, y)=\{(\frac{1}{\sqrt{2} }, \frac{1}{\sqrt{2} }) \}\tag5$$
$$ (\frac12)^2 -2xy -1 = 0 \to \frac{1}{xy}=-\frac83 \tag6$$
We can check that it is indeed eqtn (5) where the function extremizes with $2+ \sqrt{2}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/547928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to get $\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}- \frac{1-z^n}{n}\bigg)$"? I asked a question here and got answer committing :
$$(1+yz^n)(1+y)^{n-1} - (1+yz)^n=\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}-
\frac{1-z^n}{n}\bigg) \tag{1}$$and$$\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}=
\frac{(z-1)^2}{n(n+1)}\bigg(
\sum_{k=0}^{n-1} (k+1)z^k
\bigg) \tag{2}$$
My try:
*
*Using binomial theorem, $$(1+yz^n)(1+y)^{n-1} - (1+yz)^n=(1+yz^n)\sum_{k=0}^{n-1}\binom{n-1}{k}y^{n-1-k}-\sum_{k=0}^n\binom{n}{k}(yz)^{n-k}$$
*Using this,$$\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}=(1-z)\cdot\bigg(\frac{1+z+\dots+z^{n-1}}{n}-\frac{1+z+\dots+z^{n-1}+z^n}{n+1}\bigg)$$
I can't go ahead in both the cases. Please help me.
|
$1.$
\begin{align*}
(1+yz^n)(1+y)^{n-1} - (1+yz)^n &= (1 + yz^n) \sum_{k=0}^{n-1} \binom{n-1}{k}y^k - \sum_{k=0}^{n} \binom{n}{k}(zy)^k \\
&= \sum_{k=1}^{n-1} y^k \left(\binom{n-1}{k} + z^n \binom{n-1}{k-1} - \binom{n}{k} z^k \right )\\
&= \sum_{k=1}^{n-1} \binom{n-1}{k-1} y^k \left( \frac{n-k}{k } + z^n - \frac{n}{k} z^k \right ) \\
&= \sum_{k=1}^{n-1} \binom{n-1}{k-1} y^k \left( \frac{n}{k}(1-z^k) - (1-z^n) \right )\\
\end{align*}
$2.$ $\displaystyle \sum_{k=0}^{n-1} (k+1)z^k$ is Arithmetico-geometric sequence can be evaluated as
$\displaystyle \frac{n z^{n+1}-n z^n-z^n+1}{(z-1)^2}$ just by differentiating both sides of $\displaystyle \sum z^k = \frac{1-z^{n+1}}{1-z}$
Also to proceed from your step,
\begin{align*}
\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1} &= (1-z)\cdot\left( \frac{1+z+\dots+z^{n-1}}{n}-\frac{1+z+\dots+z^{n-1}+z^n}{n+1}\right ) \\
&= \frac{(1-z)}{n(n+1)} \cdot \left( 1 + z + \dots + z^{n-1} - n z^n \right )\\
&= \frac{(1-z)}{n(n+1)} \cdot \left( \sum_{k=0}^{n-1} (k+1)z^k - \sum_{k=0}^{n-1}k z^k - \left( \sum_{k=0}^{n-1} (k+1)z^{k+1} - \sum_{k=0}^{n-1} k z^{k} \right ) \right )\\
\end{align*}
|
{
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"url": "https://math.stackexchange.com/questions/548608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$ \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0$ implies that $ \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0$ $a ,b , c$ are real numbers such that $ \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0$ , then how do we prove
(without routine laborious manipulation) that $\dfrac a{(bc-a^2)^2} + \dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}=0$
|
Multiply out denominators, and rearrange the first equation to $(ab+bc+ca)^2=(ab+bc+ca)(a^2+b^2+c^2)$
|
{
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|
Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} = \begin{bmatrix}3 \\ 7 \\ 13\end{bmatrix}$
So we solve it like its a homogenous system. I end up getting to $\begin{bmatrix} 3 & 4& 1 & 2\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}$.So would $x_2$ and $x_3$ be free variables? I can set them to $x_2 = \alpha$ and $x_3 = t$, $x_4 = 0$ then we set up:
$\begin{cases}3x_1 + 4x_2 + x_3 + 2x_4 = 0\\0+0+0+x_4 = 0 \\0+0+0+0 = 0\end{cases}$
So, $\underline{v} = \begin{pmatrix}3 \\ 7 \\ 13\end{pmatrix} + \begin{pmatrix}\frac{1}{3}(-4\alpha -t - 2_4) \end{pmatrix}$
I don't think this is right. What would $x_4$ be? I'm confused
|
Try row reducing the augmented matrix to reduced row echelon form:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 4 & 1
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & -3
\end{array}\right]
$$
From the last row, we find that $0=-3$, a contradiction. So the system has no solution. Perhaps you made a typo?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{mi}{n}+\frac{1}{2}\right\rfloor$ is an even number Let $m$, $n$ be positive odd numbers such that $\gcd(m,n)=1$. Show that
$$\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\dfrac{mi}{n}+\dfrac{1}{2}\right\rfloor$$
is an even number, where $\lfloor{x}\rfloor$ is the largest integer not greater than $x$.
My try:
$$\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\dfrac{mi}{n}+\dfrac{1}{2}\right\rfloor=\left\lfloor\dfrac{m}{n}+\dfrac{1}{2}\right\rfloor+\left\lfloor\dfrac{2m}{n}+\dfrac{1}{2}\right\rfloor+\cdots+\left\lfloor\dfrac{m(n-1)}{2n}+\dfrac{1}{2}\right\rfloor$$
Then I can't it. Thank you for your help.
|
Note $[x+\frac{1}{2}]=[2x]-[x]$
$$\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\dfrac{mi}{n}+\dfrac{1}{2}\right\rfloor=\sum_{i=1}^{\frac{n-1}{2}}(\left\lfloor2x\right\rfloor-\lfloor x\rfloor)\\=\sum_{i\ge \frac{n-1}{2},i\equiv0\pmod{2}}\left\lfloor x \right\rfloor-\sum_{i\le \frac{n-1}{2},i\equiv1\pmod{2}}\left\lfloor x \right\rfloor\\=\sum_{i\ge \frac{n-1}{2},i\equiv0\pmod{2}}\left\lfloor x \right\rfloor+\sum_{i\le \frac{n-1}{2},i\equiv1\pmod{2}}\left\lfloor -x+1 \right\rfloor \\ \equiv 2\sum_{i\ge \frac{n-1}{2},i\equiv0\pmod{2}}\left\lfloor x \right\rfloor \pmod{2}\\ \equiv0\pmod{2}$$
I hope you don't mind me trying to clarify what you have above:
$$
\begin{align}
\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{mi}{n}+\frac12\right\rfloor
&=\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{2mi}{n}\right\rfloor
-\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{mi}{n}\right\rfloor\\
&=\sum_{\substack{i=2\\i\text{ even}}}^{n-1}\left\lfloor\frac{mi}{n}\right\rfloor
-\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{mi}{n}\right\rfloor\\
&=\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor\frac{mi}{n}\right\rfloor
-\sum_{\substack{i\lt n/2\\i\text{ odd}}}\left\lfloor\frac{mi}{n}\right\rfloor\\
&=\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor\frac{mi}{n}\right\rfloor
-\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor\frac{m(n-i)}{n}\right\rfloor\\
&=\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor\frac{mi}{n}\right\rfloor
-\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor m-\frac{mi}{n}\right\rfloor\\
&=\sum_{\substack{i\gt n/2\\i\text{ even}}}\left\lfloor\frac{mi}{n}\right\rfloor
-\sum_{\substack{i\gt n/2\\i\text{ even}}}\left(m-1-\left\lfloor\frac{mi}{n}\right\rfloor\right)\tag{$\ast$}\\
&=2\sum_{\substack{i\gt n/2\\i\text{ even}}}\left(\left\lfloor\frac{mi}{n}\right\rfloor-\frac{m-1}{2}\right)
\end{align}
$$
$(\ast)$ is true as long as $\frac{mi}{n}\not\in\mathbb{Z}$.
|
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"url": "https://math.stackexchange.com/questions/552860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Implicit derivative - $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Let $y$ be a function of $x$ determined by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Find $\Large\frac{dy}{dx}$ and $\Large\frac{d^2y}{dx^2}$
I've obtained $\Large\frac{dy}{dx} = \frac{-xb^2}{ya^2}$ and $\Large\frac{d^2y}{dx^2}=\frac{-b^2a^2y^2 - x^2b^4}{a^4y^3}$, but the book gives $\Large\frac{d^2y}{dx^2} = \frac{-b^2}{a^4y^3}$ (the answer of the first one is correct)
I don't know what I am doing wrong:
$\Large\frac{d^2y}{dx^2} = \frac{-b^2(ya^2) +xb^2 a^2 \frac{dy}{dx}}{a^4y^2}$. Substituting $\Large\frac{dy}{dx}$ I take that answer
Thanks in advance!
|
Your answer is correct, just unsimplified. Note:
$$
-b^2a^2y^2-x^2b^4=-b^2(a^2y^2+b^2x^2)=-b^2
$$
The second equality holds when you cross multiply the equation for the ellipse.
|
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"url": "https://math.stackexchange.com/questions/553004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the complex $z$ such $\max{(|1+z|,|1+z^2|)}$ is minimum find the complex $z$,such
$$\max{(|1+z|,|1+z^2|)}$$ is minimum
My try: let $z=a+bi$,then
$$|1+z|=\sqrt{(a+1)^2+b^2}$$
$$|1+z^2|=|1+a^2+2abi-b^2|=\sqrt{(1+a^2-b^2)^2+4a^2b^2}$$
Then I can't,Thank you
|
go further, there is two cases:
case I: $|1+z|\ge|1+z^2|$
$(a+1)^2+b^2\ge(1+a^2-b^2)^2+4a^2b^2 \iff a^4+2 a^2 b^2+a^2-2 a+b^4-3 b^2 \le 0 \iff (b^2-x_1)(b^2-x_2)\le 0, x_1=\dfrac{3-2a^2-\sqrt{D}}{2},x_2=\dfrac{3-2a^2+\sqrt{D}}{2},D=-16a^2+8a+9 \iff x_1\le b^2 \le x_2$
it is trivial that $D\ge 0$ for this case $\implies \dfrac{(1-\sqrt{10}}{4} \le a \le\dfrac{ (1+\sqrt{10}}{4})$,
$\min{|1+z|^2}=\min{(1+a)^2+b^2},(1+a)^2+b^2\ge (1+a)^2+x_1$, so the problem because to find:
the minimum of $f(a)=(1+a)^2+\dfrac{3-\sqrt{D}}{2}-a^2=2a+\dfrac{5-\sqrt{-16a^2+8a+9}}{2} \to f_{min}=3-\sqrt{5}, a=\dfrac{1-\sqrt{5}}{4},b^2=\dfrac{11-3\sqrt{5}}{4}$
caseII:$|1+z|<|1+z^2|$
I left this case for op. this case is a little more complex. and final answer is case I.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
|
By putting $x=\sinh(z)^2$ we have:
$$ I = \int_{0}^{+\infty}2\tanh z\operatorname{arccot}\left(\sinh z-2\cosh z\right)dz; $$
integrating by parts we have:
$$ I = \int_{0}^{+\infty} 2\frac{-1+2\tanh z}{5\cosh z-4\sinh x}\log\left(\cosh z\right)dz;$$
now putting $z=\log t$ we get:
$$ I = 4\int_{1}^{+\infty} \frac{t^2-3}{(t^2+1)(t^2+9)}\log\left(\frac{t+t^{-1}}{2}\right)dt.$$
By putting $y=\frac{2}{t+t^{-1}}$ we can split the integral in two parts, i.e.
$$ I_1 = 24\int_{0}^{1}\frac{\log x}{9+16x^2} dx,$$
$$ I_2 = -4\int_{0}^{1}\frac{(3-8x^2)\log x}{(9+16x^2)\sqrt{1-x^2}}dx.$$
By considering the Taylor series of $\frac{1}{9+16x^2}$ and integrating term-by-term we get $I_1=\Im\operatorname{Li}_2\frac{3i}{4}$.
The really mysterious thing is that $I_2=\pi\log\frac{3}{4}$. Not so mysterious, anyway. We have that:
$$\int_{0}^{1}\frac{\log x}{\sqrt{1-x^2}}=\int_{0}^{\pi/2}\log\cos x\, dx=-\frac{1}{2}\int_{0}^{+\infty}\frac{\log(1+t^2)}{1+t^2}=-\frac{\pi}{2}\log 2,$$
(through $x=\arctan t$) in virtue of the magic formula:
$$\int_{\mathbb{R}}\frac{\log\left(A^2 x^2 + B^2\right)}{x^2+1}dx=2\pi\log(A+B),\tag{1}$$
that can be proved through standard complex-analytic techniques. So we have only to deal with:
$$\int_{0}^{1}\frac{\log x}{(16x^2+9)\sqrt{1-x^2}}dx=\int_{0}^{\pi/2}\frac{\log\cos\theta}{16\cos^2\theta+9}d\theta,$$
or, by putting $\theta=\arctan t$,
$$\int_{0}^{\infty}\frac{\log(t^2+1)}{25+9t^2}dt=\frac{1}{15}\int_{0}^{+\infty}\frac{\log(25t^2+9)-\log(9)}{t^2+1}dt,$$
and the magic formula tell us that the last expression is equal to $\frac{1}{15}\left(\pi\log 8-\pi\log 3\right)$, completing the proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
|
Let me add a more general answer.
Note that
$$
{1 \over {\left( {x + 1} \right)\left( {x + 2} \right) \cdots \left( {x + m} \right)}}
= {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }}
= {{\Gamma \left( {x + 1} \right)} \over {\Gamma \left( {x + 1 + m} \right)}}
= x^{\,\underline {\, - m\,} }
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.
In virtue of the definition through the Gamma function, the above is valid for $x,m \in \mathbb C$.
Then we apply the Indefinite Sum concept, by which
$$
\eqalign{
& \Delta _{\,x} \;x^{\,\underline {\,q\,} }
= \left( {x + 1} \right)^{\,\underline {\,q\,} } - x^{\,\underline {\,q\,} }
= qx^{\,\underline {\,q - 1\,} } \quad \Rightarrow \cr
& \Rightarrow \quad \Delta _{\,x} ^{\left( { - 1} \right)} \;x^{\,\underline {\,q\,} }
= \sum\nolimits_x {x^{\,\underline {\,q\,} } } = \left\{ {\matrix{
{{1 \over {q + 1}}\;x^{\,\underline {\,q + 1\,} } + c} & { - 1 \ne q} \cr
{\psi (x + 1) + c} & { - 1 = q} \cr
} } \right. \cr}
$$
which is to recall that it is valid for $x,q,c \in \mathbb C$.
So in particular, for $q=-m$ and $m \ne 1$, the Indefinite sum becomes
$$
\sum\nolimits_x {x^{\,\underline {\, - m\,} } } \quad \left| {\,1 \ne m} \right.
= {1 \over { - m + 1}}x^{\,\underline {\, - m + 1\,} } + c
$$
which means that for the definite sum we get
$$
\eqalign{
& \sum\limits_{x = a}^b {x^{\,\underline {\, - m\,} } }
= \sum\nolimits_{x = a}^{b + 1} {x^{\,\underline {\, - m\,} } } \quad \left| \matrix{
\,m,a,b \in C \hfill \cr
\;1 \ne m \hfill \cr} \right. = \cr
& = {1 \over { - m + 1}}\left( {\left( {b + 1} \right)^{\,\underline {\, - m + 1\,} }
- a^{\,\underline {\, - m + 1\,} } } \right) = \cr
& = {1 \over {m - 1}}\left( {a^{\,\underline {\, - \left( {m - 1} \right)\,} }
- \left( {b + 1} \right)^{\,\underline {\, - \left( {m - 1} \right)\,} } } \right) = \cr
& = {1 \over {m - 1}}\left( {{1 \over {\left( {a + 1} \right)^{\,\overline {\,m - 1\,} } }}
- {1 \over {\left( {b + 2} \right)^{\,\overline {\,m - 1\,} } }}} \right) \cr}
$$
and taking the limit $b \to \infty$
$$
\sum\limits_{x = a}^\infty {x^{\,\underline {\, - m\,} } } \quad \left| {\,\;1 < m \in R} \right.
= {1 \over {\left( {m - 1} \right)m!}}
$$
Finally, in your particular case, the summand, the indefinite, definite and infinite sums are
$$
\eqalign{
& {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }}
= \left( {n - 1} \right)^{\,\underline {\, - 3\,} } \cr
& \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
= \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr
& = - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} } + c
= - {1 \over {2n^{\,\overline {\,2\,} } }} + c = - {1 \over {2n\left( {n + 1} \right)}} + c \cr
& \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
= \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr
& = - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}
- {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2}
- {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right) \cr
& \sum\limits_{n = 1}^\infty {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
= {1 \over 4} \cr}
$$
|
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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.