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Find the intersection points of two circles Find the intersection points of the circles $$k_1:(x-4)^2+(y-1)^2=9\\k_2:(x-8)^2+(y+4)^2=100$$
The intersections point (if they exist) will satisfy the equations of both the circles, so we can find their coordinates by solving the system $$\begin{cases}(x-4)^2+(y-1)^2=9\\(x-8)^2+(y+4)^2=100\end{cases}\iff\begin{cases}x^2-8x+y^2-2y=-8\\x^2-16x+y^2+8y=20\end{cases}$$
Substracting these equations, gives $4x-5y=-14\Rightarrow x=\dfrac{5y-14}{4}$. Substituting into the first, I got (if I didn't mess up the calculations) $$41y^2-332y+766=0$$ which has no real solutions (negative discriminant). Therefore the system has no solutions as well, so the circles don't intersect. Is there something else we can use in order to conclude that they do not intersect? Maybe something which requires less calculations? The center of the first is $O_1(4;1)$, the center of the second cirlce is $O_2(8;-4)$ and their radii are $r_1=3$ and $r_2=10$, respectively, if this somehow helps. The distance $O_1O_2=\sqrt{16+25}=\sqrt{41}$ and $r_1+r_2=13=\sqrt{169}$, but I am not sure how to interpret these findings.
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If you're familiar with implicit differentiation, we can locate the points where the slopes of the tangent lines to the two circles "match". We have
$$ (x \ - \ 4)^2 \ + \ (y \ - \ 1)^2 \ \ = \ \ 9 \ \ \Rightarrow \ \ 2·(x \ - \ 4) \ + \ 2·(y \ - \ 1)·y' \ \ = \ \ 0 \ \ \Rightarrow \ \ y' \ \ = \ \ \frac{4 \ - \ x}{y \ - \ 1 } \ \ , $$
$$ (x \ - \ 8)^2 \ + \ (y \ + \ 4)^2 \ \ = \ \ 100 \ \ \Rightarrow \ \ 2·(x \ - \ 8) \ + \ 2·(y \ + \ 4)·y' \ \ = \ \ 0 \ \ \Rightarrow \ \ y' \ \ = \ \ \frac{8 \ - \ x}{y \ + \ 4 } \ \ . $$
The slopes agree for $$ \frac{4 \ - \ x}{y \ - \ 1 } \ \ = \ \ \frac{8 \ - \ x}{y \ + \ 4 } \ \ \Rightarrow \ \ 24 \ - \ 5x \ \ = \ \ 4y \ \ . $$
This turns out to be the line [in green in the graph below] passing through the centers of the two circles ($ \ 5·4 + 4·1 \ = \ 5·8 + 4·(-4) $ $ = \ 24 \ $ ). Upon inserting $ \ y \ = \ 6 - \frac54·x \ $ into the equations of the circles, we obtain
$$ ( \ x \ - \ 4 \ )^2 \ + \ \left( \ \left[6 - \frac54·x \right] \ - \ 1 \right)^2 \ \ = \ \ 9 \ \ \Rightarrow \ \ \frac{41}{16}·x^2 \ - \ \frac{41}{2}·x \ + \ 32 \ = \ 0 $$
$$ \ \ \Rightarrow \ \ x \ \ = \ \ 4 \ \pm \ \frac{12}{\sqrt{41}} \ \ \ , \ \ \ y \ \ = \ \ 1 \ \mp \ \frac{15}{\sqrt{41}} \ \ \rightarrow \ \ ( \ \approx 2.13 \ , \ \approx 3.34 \ ) \ \ , \ \ ( \ \approx 5.87 \ , \ \approx -1.34 \ ) \ \ ; $$
$$ ( \ x \ - \ 8 \ )^2 \ + \ \left( \ \left[6 - \frac54·x \right] \ + \ 4 \right)^2 \ \ = \ \ 100 \ \ \Rightarrow \ \ \frac{41}{16}·x^2 \ - \ 41·x \ + \ 64 \ = \ 0 $$
$$ \ \ \Rightarrow \ \ x \ \ = \ \ 8 \ \pm \ \frac{40}{\sqrt{41}} \ \ \ , \ \ \ y \ \ = \ \ -4 \ \mp \ \frac{50}{\sqrt{41}} \ \ \rightarrow \ \ ( \ \approx 1.75 \ , \ \approx 3.80 \ ) \ \ , \ \ ( \ \approx 14.25 \ , \ \approx -11.81 \ ) \ \ . $$
From this, we see that along the line $ \ y \ = \ 6 - \frac54·x \ \ , \ $ which has negative slope, both of the "tangent points" for the smaller circle are "to the right and below" (larger $ \ x-$ and smaller $ \ y-$coordinates) of the leftmost "tangent point" of the larger circle and "to the left and above" its rightmost "tangent point". The smaller circle must then be entirely within the interior of the larger circle, so they have no intersections.
If one pair of these coordinates for each circle were the same, it would identify a point of mutual tangency, which would be the single intersection between them. With this "slope-matching" line having negative slope,
• if one pair of coordinates for the smaller circle were "to the left and above" one pair for the larger circle and the other pair were "to the right and below" the same pair for the larger circle, this would tell us that the smaller circle "straddles" the boundary of the larger circle and we would find two intersection points for the circle;
• if both pairs of coordinates for the smaller circle are "to the right and above" or "to the left or below" those of the larger circle, the smaller circle is entirely exterior to the larger circle, and again we should expect there to be no intersections.
We would make corresponding adjustments to this description if the "slope-matching" line had positive slope.
$$ \ \ $$
The "intersection line" $ \ 4x - 5y \ = \ -14 \ $ [in orange in the graph] that you found passes just "outside" of the larger circle, so there are no real intersection points for the circles. (If we translated the smaller circle until it made a point of mutual tangency with the larger one, we would see the intersection line also translate until it became tangent at that point.) Inserting either $ \ y \ = \ \frac{4x \ + \ 14}{5} \ $ or $ \ x \ = \ \frac{5y \ - \ 14}{4} \ $ into either circle equation produces for both of them
$$ 41·x^2 \ - \ 128·x \ + \ 256 \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ \frac{64}{41} \ \pm \ \frac{80}{41}·i \ \ \ , $$
$$ 41·y^2 \ - \ 332·y \ + \ 772 \ \ = \ \ 0 \ \ \Rightarrow \ \ y \ \ = \ \ \frac{166}{41} \ \pm \ \frac{64}{41}·i \ \ \ , $$
these coordinate values being consistent with $ \ 4x - 5y \ = \ -14 \ \ . \ $ [I am not getting your constant term in the $ \ y-$coordinate quadratic polynomial, but what you have does not affect your conclusion.] These intersection points can be better understood if we consider the "circle" equations in $ \ \mathbb{C}^2 \ \ , \ $ (where the equations describe two surfaces in four (real) dimensions), rather than just in $ \ \mathbb{R}^2 \ \ $ (but we won't go into that here).
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim_{x\to +\infty}\frac{\sqrt {2x^2+3}}{4x+2}$ and $\lim_{x\to -\infty}\frac{\sqrt {2x^2+3}}{4x+2}$. I was reading about limits of functions from Problems of Calculus in One Variable written by I.A Maron. There was an example given in the book:
Find $\lim_{x\to +\infty}\frac{\sqrt {2x^2+3}}{4x+2}$ and $\lim_{x\to -\infty}\frac{\sqrt {2x^2+3}}{4x+2}$.
The solution given in the book is as follows:
However, I dont get how they conclude "From this it follows, incidentally, that $\lim_{x\to \infty}\frac{\sqrt {2x^2+3}}{4x+2}$ does not exist." ? I am not quite getting it...
For a more accurate reference of the question, I am attaching a picture of how the question was presented in the book:
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$$
\text { Noting that } \sqrt{x^2}=\left\{\begin{array}{cl}
x & \text { if } x \geq 0 \\
-x & \text { if } x<0
\end{array}\right.
$$
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =\lim _{x \rightarrow+\infty} \frac{\frac{1}{\sqrt{x^2}} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =\lim _{x \rightarrow+\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =\frac{\sqrt{2}}{4} \\
& =\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{x \rightarrow -\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =-\lim _{x \rightarrow-\infty} \frac{-\frac{1}{x} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =-\lim _{x \rightarrow-\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =-\frac{\sqrt{2}}{4} \\
& =-\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
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|
$\lim_{(x,y)\to (0,0)}\frac{1-(\cos x)(\cos y)}{x^2+y^2} $ I need to find the limit for $$\lim_{(x,y)\to (0,0)}\frac{1-(\cos x)(\cos y)}{x^2+y^2} $$
whether exist.
I use many example (ex:line, interated limit, half angle formula, ...), and I always get the answer $1/2$. However, this does not mean the limit is $1/2$. Thus, I want to use definition to show that limit in fact be $1/2$. Unfortunately, I tried many times, and using the inequality I know to show, I failed. Can someone give me a useful inequality to solve it ? Or, point out that I miss somewhere, and show that the limit does not exist.
Thanks.
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Another way to calculate the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}\;.$
First of all, we will calculate the following limit :
$\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}\;.$
Let $\;\varphi:\,]\!-\!\infty,+\infty[\to\Bbb R\;$ be the function defined as :
$\varphi(t)=\begin{cases}\dfrac{\sin^2\!t}{t^2}\quad&\text{ for any }\,t\in\,]\!-\!\infty,+\infty[\,\setminus\,\{0\}\\\\\;\;1&\text{ for }\,t=0\end{cases}$
It follows that $\;\lim\limits_{t\to0}\varphi(t)=\varphi(0)=1\,.$
Moreover, for all $\,(u,v)\in\Bbb R^2\setminus\{(0,0)\}\;$ it results that
$\begin{align}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}&=\dfrac{\big[\varphi(u)\!-\!\varphi(v)\big]\!\left(u^2\!-\!v^2\right)\!+\!\big[\varphi(u)\!+\!\varphi(v)\big]\!\left(u^2\!+\!v^2\right)}{2\left(u^2+v^2\right)}=\\
&=\dfrac{\varphi(u)-\varphi(v)}2\cdot\dfrac{u^2-v^2}{u^2+v^2}+\dfrac{\varphi(u)+\varphi(v)}2\;.\end{align}$
Since $\;\left|\dfrac{u^2-v^2}{u^2+v^2}\right|\leqslant1\;$ for all $\,(u,v)\in\Bbb R^2\setminus\{(0,0)\}\;$ and
$\lim\limits_{(u,v)\to(0,0)}\varphi(u)=\lim\limits_{(u,v)\to(0,0)}\varphi(v)=1\;,\;$ it follows that
$\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=1\;.\quad\color{blue}{(*)}$
Now, we will calculate the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}\;.$
For all $\,(x,y)\in\Bbb R^2\setminus\{(0,0)\}\;$ it results that
$\begin{align}\dfrac{1-\cos x\cos y}{x^2+y^2}&=\dfrac{1-\cos(x+y)+1-\cos(x-y)}{2\left(x^2+y^2\right)}=\\
&=\dfrac{2\sin^2\left(\frac{x+y}2\right)+ 2\sin^2\left(\frac{x-y}2\right)}{4\left[\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}\right]}=\\
&=\dfrac12\!\cdot\!\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\;.\end{align}$
Consequently ,
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}=$
$=\dfrac12\!\cdot\!\!\lim\limits_{(x,y)\to(0,0)}\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\underset{\overbrace{\text{by letting }u=\frac{x+y}2\text{ and }v=\frac{x-y}2}}{=}$
$=\dfrac12\!\cdot\!\!\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=\dfrac12\!\cdot\!1=\dfrac12\;.$
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"url": "https://math.stackexchange.com/questions/4621053",
"timestamp": "2023-03-29T00:00:00",
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|
Calculating $\displaystyle{\lim_{x \to 0^+}}{\frac{1}{\sqrt{x}}\Big(e^x + \frac{2\log(\cos(x))}{x^2}}\Big)$ I am struggling to calculate this limit:
$$\displaystyle{\lim_{x \to 0^+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}}$$
I prefer not to use l'Hopital's rule, only when necessary. If possible, solving with the help of these limits:
$\displaystyle{\lim_{x \to 0^+}}{\frac{\log(x + 1)}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{e^x - 1}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{1 - \cos(x)}{x^2}} = \frac{1}{2}$.
Also when I tried solving it using only l'Hopital, it seemed to be very laborious, and also not sure if sufficient to solve it.
Also I prefer not using Taylor theorem and little/big o notation if possible at all.
Here's my (unsuccessful) attempt:
$$\displaystyle{\lim_{x \to 0+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}} = \displaystyle{\lim_{x \to 0^+}}{\frac{e^xx^2 + e^x2x + 2\log{\cos{x}}}{x^2\sqrt{x}}} = \frac{1}{5}\displaystyle{\lim_{x \to 0^+}}{\frac{e^x(x^2 + 2x) - 2tg{x}}{x\sqrt{x}}} = ...$$
Problem is, each time I use l'Hopital, it doesn't seem to simplify limit in any way.
Thanks.
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As the OP asks, an attempt to do it, using only
\begin{align}
\log(1+u)&=u+o(u),\\
e^x &= 1 + x + o(x),\\
\cos(x) &= 1 -\frac{x^2}{2} + o(x^2) .
\end{align}
Now, as $x \to 0$, the best we can deduce is:
\begin{align}
\cos x &= 1 - \frac{x^2}{2} + o(x^2)
\\
\log \cos x &= \log(1-(1-\cos x)) = -(1-\cos x) + o\big(1-\cos x\big)
\\ &= -1+\cos x + o\big(x^2\big)
= -1 + \big(1-\frac{x^2}{2} + o(x^2)\big)+ o\big(x^2\big)
\\ &=-\frac{x^2}{2}+o(x^2)
\\
\frac{2\log \cos x }{x^2} &= -1+o(1)
\\
e^x &= 1+x+o(x)
\\
e^x+\frac{2\log \cos x }{x^2} &= 1+x+o(x)-1+o(1) =o(1)
\\
\frac{e^x+\frac{2\log \cos x }{x^2}}{\sqrt{x}} &= o(x^{-1/2})
\end{align}
This is not enough to get the answer.
From this we can see how to get a counterexample. Instead of $\cos(x)$, use $c(x) := 1-\frac{x^2}{2}+x^{5/2}$. This still satisfies $c(x) = 1-\frac{x^2}{2}+o(x^2)$, but
$$
\frac{e^x+\frac{2\log c(x) }{x^2}}{\sqrt{x}} \to 2
$$
and not $0$.
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|
prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$ $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$
How to deal with this problem?
Observing that when $x=\pi/2$, the above inequality becomes equality.
Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have
$$-2(\sin x)^{-3}\cos x+2x^{-3}$$ Next, how to analysis the sign of $f'(x)$?
Any hints are wellcome! Thanks!
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The Laurent series expansion of $\frac{1}{\sin x} $, valid for $0<|x|< \pi$, is
$$\frac{1}{\sin x} = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{1520} x^5 + \cdots $$
with all coefficients positive. We conclude that also $\frac{1}{\sin^2 x}$ has a Laurent expansion with positive coefficients valid for $0< |x| < \pi$
$$\frac{1}{\sin^2 x} = \frac{1}{x^2} + \frac{1}{3}+ \frac{1}{15} x^2 + \frac{2}{189} x^4 + \frac{1}{675} x^6 + \cdots$$
From this several inequalities follow.
|
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
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I will assume your deduction is correct and I will try to prove the inequality to which you have reduced the original inequality.
Consider the sequences
$$(c_i)_{i=1}^{3}:=(\sqrt{bc},\sqrt{ac},\sqrt{ab}),\;\;(d_{i})_{i=1}=(\sqrt{a},\sqrt{b},\sqrt{c})$$
Notice that for any permutation $\sigma:\{1,2,3\}\to\{1,2,3\}$ if
$c_{\sigma(1)}\leq c_{\sigma(2)}\leq c_{\sigma(3)}$ we must have
$d_{\sigma(1)}\geq d_{\sigma(2)}\geq d_{\sigma(3)}$, since $c_{i}d_{i}=\sqrt{abc}$ for all $i$.
Notice also that $b\sqrt{a}+c\sqrt{b}+a\sqrt{c}=c_{3}d_{2}+c_{1}d_{3}+c_{2}d_{1}$, so it follows from
the rearrangement inequality that
$$ b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\geq \sum_{i}c_{i}d_{i}=3\sqrt{abc}. $$
After some obvious manipulations, we conclude that it suffices to show $abc+2-a-b-c$ for any $a,b,c\geq 1$. Replacing $a$ by $1+A$ on both sides this ends up reading as $ABC+\sum AB\geq 0$, which is obviously true for $A,B,C\geq 0$.
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Sine curve that passes through (1,1) (2,2) and (3,3) Question: Find a sine function in the form $a\sin(bx) + c$ that passes through points (1,1) (2,2) and (3,3)
Working so far:
*
*We have three points for three unknown variables in the function, so we can use simultaneous equations to solve for them.
*Simultaneous equations:
*
*$a\sin(b) + c = 1$
*$a\sin(2b) + c = 2$
*$a\sin(3b) + c = 3$
*Equation 1, solving for c:
*
*$c = 1 - a\sin(b)$
*Equation 2: solving for a:
*
*$a\sin(2b) + 1 - a\sin(b) = 2$
*$a\sin(2b) - a\sin(b) = 1$
*$a(\sin(2b) - \sin(b)) = 1$
*$a = \frac{1}{\sin(2b) - \sin(b)}$
*Equation 3: solving for b:
*
*$(\frac{1}{\sin(2b) - \sin(b)})(\sin(3b)) + 1 - (\frac{1}{\sin(2b) - \sin(b)})(\sin(b)) = 3$
*$\frac{\sin(3b)}{\sin(2b) - \sin(b)} - \frac{\sin(b)}{\sin(2b) - \sin(b)} = 2$
*$\frac{\sin(3b) - \sin(b)}{\sin(2b) - \sin(b)} = 2$
*$\sin(3b) - \sin(b) = 2\sin(2b) - 2\sin(b)$
*$\sin(3b) + \sin(b) = 2\sin(2b)$
*$3\sin(b) - 4\sin^{3}(b) + \sin(b) = 4\sin(b)\cos(b)$
*$4\sin(b) - 4\sin^{3}(b) = ±4\sin(b)\sqrt{1 - \sin^{2}(b)}$
*Substitute $u = \sin(b)$
*$4u - 4u^3 = ±4u\sqrt{1-u^2}$
*$u - u^3 = ±u\sqrt{1-u^2}$
*$1 - u^2 = ±\sqrt{1-u^2}$
*$1 - 2u^2 + u^4 = 1 - u^2$
*$u^4 - u^2 = 0$, $u = 0$
*$u^2 - 1 = 0$
*$u^2 = 1$
*$u = 1$, $u = -1$
*For $u = -1$:
*
*$\sin(b) = -1$
*$b = \arcsin(-1) = (\frac{4n - 1}{2})\pi$ and $(\frac{4n + 3}{2})\pi$, n is an integer.
*For $u = 0$:
*
*$b = \arcsin(0) = n\pi$, n is an integer.
*For $u = 1$:
*
*$b = \arcsin(1) = (\frac{4n + 1}{2})\pi$, n is an integer.
*$b = n\pi$ and $b = n\pi - \frac{\pi}{2}$
*Therefore $a = \frac{1}{\sin(2(n\pi)) - \sin((n\pi))}$
*$a = \frac{1}{0 - 0} = \frac{1}{0}$, let's see if the other solution works.
*$a = \frac{1}{\sin(2(n\pi - \frac{\pi}{2})) - \sin(n\pi - \frac{\pi}{2})}$
*$a = \frac{1}{1} = 1$ when n is an even integer and $a = \frac{1}{-1} = -1$ when n is an odd integer.
*$c = 1 - (1)(\sin(n\pi)) = 1$, $c = 1 - (-1)(\sin(n\pi)) = 1$, $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an even integer and $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an odd integer, $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an even integer and finally $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an odd integer
*With these results, let's try n = 0:
*
*$a = 1$, $b = 0$ and $b = -\frac{\pi}{2}$, $c = 1$, $c = 2$, $c = 0$
*$y = \sin(-\frac{\pi}{2}x) + 1$ touches none of the three points.
A way to get the curve to approximately touch the 3 points would be to equate $a$ to a very large number, and equate $b$ to the reciprocal of $a$. And have $c$ equal to 0, in this case the graph mimics the straight-line function $f(x) = x$ in the desired range. But this doesn't account for the possibility that the graph could pass through several 'waves' between $x = 0$ and $x = 3$ to pass through all three points in those waves.
What could the values of a, b and c be for the sine curve to pass through all three points?
|
The solutions are
$$ f(x) = -\sin\left(\left(\frac{\pi}{2} + 2\pi k\right)x\right) + 2$$
and
$$ f(x) = \sin\left(\left(\frac{3\pi}{2} + 2\pi k\right)x\right) + 2$$
We can add the first and third equations to get
$$a\ \sin(b) + a\ \sin(3b)+2c=4 \\ a\ \sin(2b-b)+a\ \sin(2b+B)+2c=4 \\ 2a\ \sin(2b)\cos(b) + 2c = 4 \\a\ \sin(2b)\cos(b) = 2 - c$$
But the second equation tells us that
$$a\ \sin(2b) = 2 - c \\ a\ \sin(2b) = a\ \sin(2b)\cos(b)\\ a\ \sin(2b)(1-\cos(b)) = 0$$
It is simple to verify that $a \ne 0$ and $\cos(b) \ne 1$ by plugging the values back into the original equations.
Thus $\sin(2b) = 0$ which implies that $b = 0, \frac{\pi}{2}, \pi,$ or $\frac{3\pi}{2}\ (+\ 2\pi k)$.
For the same reasons as before, $b = 0$ and $\pi\ (+\ 2\pi k)$ do not work.
Luckily, $b = \frac{\pi}{2}$ and $\frac{3\pi}{2}\ (+\ 2\pi k)$ do work and plugging the values back into the original equation yields the other variables.
|
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}
|
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$
I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.
|
If you are comfortable with generalized harmonic number, you could consider first the partial sum
$$S_n=\sum_{r=1}^{n} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$$ and write first
$$\dfrac{r^2 - 1}{r^4 + r^2 + 1}=\frac{(r-1)(r+1)}{(r-a)(r-b)(r-c)(r-d)}$$
Using partial fraction decomposition, this is
$$\frac{a^2-1}{(a-b) (a-c) (a-d) (r-a)}+\frac{b^2-1}{(b-a) (b-c)
(b-d) (r-b)}+$$ $$\frac{c^2-1}{(c-a) (c-b) (c-d)
(r-c)}+\frac{d^2-1}{(d-a) (d-b) (d-c) (r-d)}$$ and use
$$\sum_{r=1}^n \frac 1{r-k}=H_{n-k}-H_{-k}$$
After simplification
$$S_n=\frac{(n-1) n}{2 \left(n^2+n+1\right)}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Bijective mapping/substitution - proof? Consider the ellipse
$$x^2/a^2 + y^2/b^2 = 1$$
(as a curve $L$) and its inner part $B$.
So $B$ is defined as
$$x^2/a^2 + y^2/b^2 < 1$$
Here $a,b$ are positive constants.
Both $L, B \subseteq \mathbb{R}^2$
Then consider just the inner part of this ellipse $B$ (excluding the contour/curve $L$).
Then let's consider the mapping $F$ from $(r, \theta)$ to $(x,y)$ defined as follows
$x = a \cdot r \cdot \cos(\theta)$
$y = b \cdot r \cdot \sin(\theta)$
for
$ 0 < r < 1$
$ 0 \le \theta < 2\pi $
How do we prove this $F$ is bijective and its image is $B$ (or maybe OK, it's $B$ without the origin point $(x,y) = (0,0)$)? How do we prove this is a bijection?
I am asking this because I noticed they often use this substitution when solving integrals involving ellipses or ellipsoids, but they somehow assume this mapping is bijective (don't prove it rigorously).
|
I can help you with injectivity, not yet sure about surjectivity.
Consider $(r_1,\theta_1) \neq (r_2,\theta_2) $ and $ f(r_1,\theta_1) = f(r_2,\theta_2) $, or,
$$\begin{align*}
(a \cdot r_1 \cos \theta_1, b\cdot r_1 \sin \theta_1) &= (a\cdot r_2 \cos \theta_2, b \cdot r_2 \sin \theta_2)
\end{align*}$$
This gives us the following two equations,
$$\begin{align*}
\frac { \cos \theta_1 }{\cos \theta_2} &= \frac { r_1 }{r_2} \tag{1}
\\ \frac { \sin \theta_1 }{\sin \theta_2} &= \frac { r_1 }{r_2} \tag{2}
\end{align*}$$
Equating (1) and (2), we get,
$$\begin{align*}
\sin \theta_2 \cdot \cos \theta_1 &= \sin \theta_1 \cdot \cos \theta_2
\\ \frac{ \sin(\theta_2 + \theta_1) + \sin(\theta_2 - \theta_1)}{2} &= \frac{\sin(\theta_1 + \theta_2) + \sin(\theta_1 - \theta_2) }{2} \tag{$\sin A \cos B$}
\\ \sin(\theta_2 - \theta_1) &= \sin(\theta_1 - \theta_2)
\end{align*}$$
WLOG now assume $\theta_2 > \theta_1$. Then $\sin(\theta_2 - \theta_1) = - \sin(\theta_1 - \theta_2)$. Same goes for $\theta_2 < \theta_1$. Thus, $\theta_1 = \theta_2$ is the only valid solution.
Using this in (1), we get $r_1 = r_2$ which completes the proof for injectivity.
Edit: Perhaps an easy way to show surjectivity is to show that the image is exactly $B\backslash\{(0,0)\}$, then by definition, a function is surjective to its image and the result will follow. To do so, we can bound our set as follows,
$$\begin{align*}
\frac{(a \cdot 0 \cdot \cos \theta)^2}{a^2} + \frac{(b \cdot 0 \cdot \sin \theta)^2}{b^2} < \frac{x^2}{a^2} &+ \frac{y^2}{b^2} < \frac{(a \cdot 1 \cdot \cos \theta)^2}{a^2} + \frac{(b \cdot 1 \cdot \sin \theta)^2}{b^2}
\\ 0 < \frac{x^2}{a^2} &+ \frac{y^2}{b^2} < 1
\end{align*}$$
Now we can argue since it is a continuous function with image between $0$ and $1$, it will take every value in between (exactly once due to injectivity) which gives us exactly the set $B \backslash \{(0,0)\}$. Hence, bijection is shown.
|
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"url": "https://math.stackexchange.com/questions/4644512",
"timestamp": "2023-03-29T00:00:00",
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|
Product of limit, sin, infinity, error? Hello I would like to know if there is a mistake :
I have to show that for any $t\geqslant0$ fixed
$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=0$$
That's what I said,
Since $\sin(\cdot)$ is continuous and $$\sqrt{t+4\pi n^{2}}=2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}$$ for $n\geqslant1$.
$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=\lim_{n\to \infty}\sin\left(2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)$$
$$=\sin\left(\lim_{n\to\infty}2n\pi\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=\\=\sin\left(\lim_{n\to\infty}2n\pi\cdot\lim_{n\to\infty}\sqrt{1+\frac{t}{4\pi n^{2}}}\right)=$$
$$=\sin\left(\lim_{n\to\infty}2n\pi\right)=\lim_{n\to\infty}\sin 2n\pi=\lim_{n\to\infty}0=0$$
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Actually it does not exist the limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}\;.$$
So, I think the OP intended to write the following limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi^2n^{2}}\;.\quad(\text{ where }n\in\Bbb N\;)$$
Indeed ,
$\lim\limits_{n\to \infty}\sin\sqrt{t+4\pi^2n^{2}}=$
$=\lim\limits_{n\to \infty}\sin\left(\sqrt{t+4\pi^2n^{2}}-2\pi n\right)=$
$=\lim\limits_{n\to \infty}\sin\left[\dfrac{\left(\sqrt{t+4\pi^2n^{2}}-2\pi n\right)\left(\sqrt{t+4\pi^2n^{2}}+2\pi n\right)}{\sqrt{t+4\pi^2n^{2}}+2\pi n}\right]=$
$=\lim\limits_{n\to \infty}\sin\left(\!\dfrac t{\sqrt{t+4\pi^2n^{2}}+2\pi n}\!\right)=$
$=\sin\left(\!\lim\limits_{n\to \infty}\dfrac t{\sqrt{t+4\pi^2n^{2}}+2\pi n}\!\right)=$
$=\sin 0=0\,.$
Addendum:
I am going to prove that it does not exist the limit:$$\lim_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}$$
without necessity of assuming that $\,n\,$ is any real number ( that is $\,n\,$ could be any positive integer ) .
If there existed the limit $\,\lim\limits_{n\to \infty}\sin\sqrt{t+4\pi n^{2}}=l\in\Bbb R\;,\;$ there would also exist the limit $\,\lim\limits_{n\to\infty}\sin\left(2\sqrt\pi n\right)=l\;,\;$ indeed
$\begin{align}
\lim\limits_{n\to\infty}&\,\sin\left(2\sqrt\pi n\right)=\\
&=\lim\limits_{n\to\infty}\sin\left(\sqrt{t+4\pi n^2}+2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)=\\
&=\lim\limits_{n\to\infty}\left[\sin\left(\sqrt{t+4\pi n^2}\right)\cos\left(2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)+\\
+\cos\left(\sqrt{t+4\pi n^2}\right)\sin\left(2\sqrt\pi n-\sqrt{t+4\pi n^2}\right)\right]=\\
&=\lim\limits_{n\to\infty}\left[\sin\left(\sqrt{t+4\pi n^2}\right)\cos\left(\!\!\dfrac{-t}{2\sqrt\pi n+\sqrt{t+4\pi n^2}}\!\!\right)+\\
+\underbrace{\cos\left(\sqrt{t+4\pi n^2}\right)}_{\text{it is bounded}}\;\underbrace{\sin\left(\!\!\dfrac{-t}{2\sqrt\pi n+\sqrt{t+4\pi n^2}}\!\!\right)}_{\text{it is an infinitesimal}}\right]=\\\\
&=l\cos0=l\;.
\end{align}$
Moreover,
$\begin{align}\lim\limits_{n\to\infty}&\cos\left(2\sqrt\pi n\right)=\\
&=\lim\limits_{n\to\infty}\dfrac{\sin\left[2\sqrt\pi(n+1)\right]-\sin\left[2\sqrt\pi(n-1)\right]}{2\sin\left(2\sqrt\pi\right)}=\\
&=\dfrac{l-l}{2\sin\left(2\sqrt\pi\right)}=0\,.
\end{align}$
On the other hand,
$0=\lim\limits_{n\to\infty}\cos\left(4\sqrt\pi n\right)=\lim\limits_{n\to\infty}\left[2\cos^2\left(2\sqrt\pi n\right)-1\right]=-1$
which is a contradiction.
Hence, there does not exist the limit $\,\lim\limits_{n\to\infty}\sin\sqrt{t+4\pi n^{2}}\,.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Investigate on the convergence of $I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $ for $a>0$ and its exact value in case of convergence. Inspired by the post,
I start to investigate the convergence of $$I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $$ for $a>0$.
For any $0\le a\le 4$, if $x\ge 1$, we have
$$
\frac{x}{\sqrt{1+x^a}}\ge\frac{x}{\sqrt{1+x^4}}=\frac{1}{\sqrt{x^2+\frac{1}{x^2}}}\ge \frac{1}{\sqrt{x^2+x^2}} =\frac{1}{x\sqrt2}
$$
Since $\int_0^{\infty} \frac{1}{x \sqrt{2}} d x$ is divergent, therefore $ I(a)$ is divergent for any $0\le a\le 4$.
For $a>4$, let $x=\tan ^{\frac{2}{a}} \theta$, then
$$
\begin{aligned}
I(a)& =\frac{2}{a} \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{2}{a}} \theta}{\sec \theta} \cdot \tan ^{\frac{2}{a}-1} \theta \sec ^2 \theta d \theta \\
& =\frac{2}{a} \int_0^{\frac{\pi}{2}} \tan ^{\frac{4}{a}-1} \theta \sec \theta d \theta \\
& =\frac{2}{n} \int_0^{\frac{\pi}{2}} \sin ^{\frac{4}{a}-1} \theta \cos ^{-\frac{4}{a}} \theta d \theta \\
& =\frac{1}{a} B\left(\frac{2}{a}, \frac{1}{2}-\frac{2}{a}\right) \\
&=\frac{\Gamma\left(\frac{2}{a}\right) \Gamma\left(\frac{1}{2}-\frac{2}{a}\right)}{a \sqrt{\pi}}
\end{aligned}
$$
which are convergent $ \Leftrightarrow \frac{1}{2}-\frac{2}{a}>0 \Leftrightarrow a>4$.
Comments and alternative solutions are highly appreciated!
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Similarly to the question you asked in, the Mellin transform of $f(x)=(1+x)^{-\rho}$ is
$$
\tilde{f}(s) = \mathcal{M}[f(x);s] = \frac{1}{\Gamma(\rho)}\Gamma(s)\Gamma(\rho-s)
$$
which is defined for $0<\Re(s)<\Re(\rho)$. Then, by the Mellin transform, one has $f(x^n) \rightarrow \frac{1}{|n|}\tilde{f}\left(\frac{s}{n} \right)$ that
$$
\int_0^{\infty} \frac{x^{s-1}}{(1+x^n)^{\rho}}dx = \frac{\Gamma\left(\frac{s}{n}\right)\Gamma\left(\rho-\frac{s}{n}\right)}{n\Gamma(\rho)}
$$
which is defined for $0<\Re(\frac{s}{n})<\Re(p)$, and in the specific case above gives that (with $s=2,n=a,\rho=1/2)$
$$
\int_0^{\infty} \frac{x}{\sqrt{1+x^a}} dx = \frac{\Gamma(\frac{2}{a})\Gamma(\frac{1}{2}-\frac{2}{a})} {a\sqrt{\pi}}
$$
which is defined for $0<\Re(\frac{2}{a})<\frac{1}{2}$. i.e. $a>4$ as you noted.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the infinite summation
Evaluate $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{1+n^3}$$
I tried to factor the denominator and then using partial fraction $$\frac{n^2}{1+n^3}=\frac{n^2}{(n+1)(n^2-n+1)}$$ $$=\frac{2n-1}{3(n^2-n+1)}+\frac{1}{3(n+1)}$$
So our question now becomes $$\frac13\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n-1}{(n^2-n+1)}+\frac{1}{(n+1)}$$
This is not a telescopic series. I wrote a few terms and have observed this. Now I'm stuck.
Any help is greatly appreciated.
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Let $S$ be the sum with the quadratic in the denominator. As user Ron Gordon points out in the linked question,
$$\begin{align*}
S &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} \\
&= -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}
\end{align*}$$
Take out the $k=0$ term and split up the sum by the sign of the index, then condense into a single sum over positive indices.
$$\begin{align*}
S &= \frac12 - \frac12 \left\{\sum_{k=-\infty}^{-1} + \sum_{k=1}^\infty\right\} (-1)^k \frac{2k-1}{k^2-k+1} \\
&= \frac12 - \frac12 \sum_{k=1}^\infty (-1)^{-k} \frac{-2k-1}{k^2+k+1} - \frac12 \sum_{k=1}^\infty (-1)^k \frac{2k-1}{k^2-k+1} \\
&= \frac12 - \frac12 \sum_{k=1}^\infty (-1)^k \frac{k^2-1}{k^4+k^2+1} \\
&= \frac12 + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{k^2-1}{k^4+k^2+1}
\end{align*}$$
Expand the summand into partial fractions.
$$\frac{k^2-1}{k^4+k^2+1} = \frac{e^{i\pi/3}}{e^{-i\pi/3}+k^2} + \frac{e^{-i\pi/3}}{e^{i\pi/3}+k^2}$$
Noting that $e^{\pm i\pi/3}=e^{\mp i2\pi/3}$ and $\dfrac12=\cos\left(\dfrac\pi3\right)=\dfrac{e^{i\pi/3}+e^{-i\pi/3}}2$, we can write
$$\begin{align*}
S &= \frac{e^{i\pi/3}+e^{-i\pi/3}}2 + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \left[\frac{e^{i\pi/3}}{e^{i2\pi/3}-k^2} + \frac{e^{-i\pi/3}}{e^{-i2\pi/3}-k^2}\right] \\
&= \left[\frac1{2e^{-i\pi/3}} + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{e^{i\pi/3}}{\left(e^{i\pi/3}\right)^2-k^2}\right] + \left[\frac1{2e^{i\pi/3}} + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{e^{-i\pi/3}}{\left(e^{-i\pi/3}\right)^2-k^2}\right]
\end{align*}$$
Comparing to the partial fraction expansion for $\csc$, the result follows,
$$S = \frac\pi2 \left[\csc\left(\pi e^{-i\pi/3}\right) + \csc\left(\pi e^{i\pi/3}\right)\right]$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expected average of differences between two points in a uniform distribution Let's say I have a discrete uniform distribution where a variable x can take any value between 1 to 100 (inclusive).
I spin up two values of x: x1 and x2 and they take the values 23 and 53 respectively.
The absolute difference between x1 and x2 is |x1 - x2| = 30.
The minimum absolute difference possible is 0 when x1 = x2 and the maximum is 100 when one of the two is 100 and the other is 0.
If I was to spin up pairs of (x1, x2) a gazillion times, what would be the expected average absolute difference?
|
I started with two die $(n=6)$ and made a table $$\begin{array}{|c|c|c|c|c|c|}\hline \text{die 1 / die 2 } & 1 &2 &3 &4 &5 &6 \\ \hline\hline \hline 1 & 0&1 &2 &3 &4 &5 \\ \hline 2 & 1 &0 &1 &2 &3&4 \\ \hline 3 &2 &1 &0 &1&2&3 \\ \hline 4 &3 &2 &1&0&1&2 \\ \hline 5 &4 &3&2&1&0 &1 \\ \hline 6 &5&4&3&2 &1 &0 \\ \hline\end{array}$$
For $|X_1-X_2|$ I got the following pmf
$f_{|X_1-X_2|}(x)=\begin{cases} \frac6{36}, \, x=0 \\ 2\cdot \frac{(6-x)}{36},\, x=\{1,2,3,4,5\} \\ 0, \, \textrm{elsewhere}\end{cases}$
From this smaller example I deduced the case $n=100$
$f_{|X_1-X_2|}(x)=\begin{cases} \frac{10}{100}, \, x=0 \\ 2\cdot \frac{(100-x)}{100},\, x=\{1,2,3\ldots , 99\} \\ 0, \, \textrm{elsewhere}\end{cases}$
Thus the expected value is
$$\mathbb E\left(|X_1-X_2| \right)=\sum_{x=1}^{99}x\cdot 2\cdot \frac{(100-x)}{100}=\frac{3333}{100}=33.33$$
To calculate it by hand it is good to know that $\sum\limits_{x=1}^{n-1} x^2=\frac16\cdot (n-1)\cdot n\cdot (2n-1)$. This is one approach to get the expected value.
For two discrete uniform random variables with outcomes from $1$ to $n$ we have
$$\mathbb E\left(|X_1^n-X_2^n| \right)=\frac{n^2-1}{3n}$$
|
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|
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
|
Grant Sanderson, aka 3Blue1Brown, has a good explanation of this in one of his Lockdown Math videos. His explanation, summarized:
*
*$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = f(1)$, where $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} -\ldots$
*$\frac{df}{dx} = 1 - x + x^2- x^3 + x^4 - \ldots = \frac{1}{1+x}$ (when $-1 ≤ x ≤ 1$)
*$f(x) = \ln(1+x)$
*$\therefore f(1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = \ln(2)$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:
$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$
For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations, one for each root.
Is there a general formula for solving equations of the following form [Quartic Equations]?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?
|
We can reduce the problem of solving the general quartic to merely solving a quadratic. Given,
$$x^4+ax^3+bx^2+cx+d=0$$
Then the four roots are,
$$x_{1,2} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}+\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag1$$
$$x_{3,4} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}-\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag2$$
where,
$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(v_1^{1/3}\zeta_3+\frac{b^2 - 3 a c + 12 d}{v_1^{1/3}\zeta_3}\right)\tag{3a}$$
or alternatively,
$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(\color{blue}{v_1^{1/3}+v_2^{1/3}}\right)\tag{3b}$$
with $v_1$ any non-zero root of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
and a chosen cube root of unity $\zeta_3^3 = 1$ such that $u$ is also non-zero. (Normally, use $\zeta_3=1$, but not when $a^3-4ab+8c = 0$.)
P.S. This is essentially the method used by Mathematica, though much simplified for aesthetics.
|
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|
How closely can we estimate $\sum_{i=0}^n \sqrt{i}$ By looking at an integral and bounding the error?
|
In case you were wondering, like me, Moron's excellent proof adapts easily to show that
$$1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} \sim \frac{3}{4}n^{4/3} + \frac{\sqrt[3]{n}}{2} + C,$$
for some constant $C.$ In this case $C = \zeta(-1/3) \approx -0.277343.$
Where, as before, $a_n \sim b_n$ means $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0.$
Similar to the previous proof, we use the AM-GM inequality to show
$$\sqrt[3]{x} \le \frac{x}{3k^{2/3}} + \frac{2k^{1/3}}{3}.$$
Summing from $k=1$ to $n-1$ and integrating we arrive at
$$ \frac{3}{4}n^{4/3} - \frac{3}{4} \le \sum_{k=1}^{n-1}\sqrt[3]{n} +
\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$
And so,
$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + n^{1/3} - \frac{3}{4} - \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$
Using $\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le 1+ \int_1^n x^{-2/3} dx,
\qquad n>1,$ we obtain
$$\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le \frac{1}{2}n^{1/3} - \frac{1}{3}.$$
And hence
$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + \frac{1}{2}n^{1/3} - \frac{5}{12}, \qquad n>1.$$
As in the previous argument, set
$$G_n = \sum_{k=1}^{n}\sqrt[3]{n} - \frac{3}{4}n^{4/3} - \frac{1}{2}n^{1/3}.$$
Then $G_n \ge -5/12,$ and we can show $G_n – G_{n+1} > 0$ by showing that
$$\frac{(n+1)^{4/3} – n^{4/3}}{(n+1)^{1/3} + n^{1/3}} > \frac{2}{3}.$$
So, as before, $G_n$ is convergent, since it is bounded below and monotonically decreasing.
I suspect this argument also adapts easily to the more general case
$\sum_{k=1}^{n}\sqrt[r]{n},$ for $r \in \mathbb{N},$ where I'm guessing, we'll find
$$1 + \sqrt[r]{2} + \dots + \sqrt[r]{n} \sim \frac{r}{r+1}n^{(r+1)/r} + \frac{\sqrt[r]{n}}{2} + C,$$
where $C = \zeta(-1/r).$ However, I confess to not having checked the details of this further generalisation.
|
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
Integrand $ =\dfrac {1}{\sin(x)\cos(x)} = 2 \csc 2x $
Its integral is obtained by direct application of listed standard trigonometric function integration formulae. Using chain rule for constant double angle:
$$ 2 \log (\tan \dfrac{2 x}{2}) \cdot \frac12= \log (\tan x ) + c $$
agrees with OP's second result when it is further simplified.
|
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|
Floor of Square Root Summation problem I have problem calculating the following summation:
$$
S = \sum_{j=1}^{k^2-1} \lfloor \sqrt{j}\rfloor.
$$
As far as I understand the mean of that summation it will be something like $$1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+\cdots$$
and I suspect that the last summation number will be $(k-1)^2$, but I really can't find the pattern of the equal simpler summation.
|
Since the last value for $j$ is $k^2-1$, none of the terms of the sum are $k$; they are all between $1$ and $k-1$.
How many $1$'s will be in the sum? Well, we'll get $1$ when $j$ is any number between $1^2$ and $2^2-1$; then we'll get $2$ for each number between $2^2$ and $3^2-1$. Then we'll get $3$ for each number between $3^2$ and $4^2-1$. Etc.
So, if $n\leq k-1$, how many times does it show up in the sum? It shows up exactly the number of times that there are integers between $n^2$ and $(n+1)^2-1$, inclusively. This is
$$(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n+1.$$
So your sum has $2(1)+1 = 3$ ones; $2(2)+1 = 5$ twos; $2(3)+1=7$ threes; etc. Up to $k-1$, which appears exactly $2(k-1)+1 = 2k-1$ times.
So we get that
$$S = \sum_{r=1}^{k-1} r(2r+1) = 2\left(\sum_{r=1}^{k-1}r^2\right) + \sum_{r=1}^{k-1}r.$$
|
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|
What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
|
So by Law of Sines we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$
From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.
This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is right-angled. Hurray!
NOTE: The keen idea when you see problems of this type is to use the Sine rule. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in Mathematics starting the solution is the difficult part. Once you figure that out, then the solutions simply flows.
|
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|
Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this
|
It is $x^2-2x-8 = 16$ my friend. So you get $x^2 - 2x -24 = 0$, which factors as $(x-6)(x+4) = 0$. Hence, $x=6$ or $x = -4$.
|
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|
How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
|
Since $\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}>0$, it's enough to prove that
$$\left(\sin\frac{3\pi}{11}+4\sin\frac{2\pi}{11}\cos\frac{3\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$\left(\sin\frac{3\pi}{11}+2\sin\frac{5\pi}{11}-2\sin\frac{\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$1-\cos\frac{6\pi}{11}+4-4\cos\frac{10\pi}{11}+4-4\cos\frac{2\pi}{11}+4\cos\frac{2\pi}{11}-4\cos\frac{8\pi}{11}-$$
$$-4\cos\frac{2\pi}{11}+4\cos\frac{4\pi}{11}-8\cos\frac{4\pi}{11}+8\cos\frac{6\pi}{11}=11+11\cos\frac{6\pi}{11}$$ or
$$\sum_{k=1}^5\cos\frac{2k\pi}{11}=-\frac{1}{2}$$ or
$$\sum_{k=1}^52\sin\frac{\pi}{11}\cos\frac{2k\pi}{11}=-\sin\frac{\pi}{11}$$ or
$$\sum_{k=1}^5\left(\sin\frac{(2k+1)\pi}{11}-\sin\frac{(2k-1)\pi}{11}\right)=-\sin\frac{\pi}{11}$$ or
$$\sin\frac{11\pi}{11}-\sin\frac{\pi}{11}=-\sin\frac{\pi}{11}.$$
Done!
|
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|
Evaluation of the sum $\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor$ Let $a, b$ and $c$ be positive integers. Recall that the greatest common divisor (gcd) function has the following representation:
\begin{eqnarray}
\textbf{gcd}(b,c) = 2 \sum_{k = 1}^{c- 1} \left \lfloor \frac{kb}{c} \right \rfloor + b + c - bc
\end{eqnarray}
as shown by Polezzi, where $\lfloor \cdot \rfloor$ denotes the floor function. In trying to generalize the formula I came across the following summation
\begin{eqnarray}
\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor.
\end{eqnarray}
I can prove the following identity for real $x$,
\begin{eqnarray}
\sum_{k = 0}^{c-1} \left \lceil \frac{x - kb}{c} \right \rceil = d \left \lceil \frac{x}{d} \right \rceil - \frac{(b-1)(c-1)}{2} - \frac{d-1}{2},
\end{eqnarray}
where $\lceil \cdot \rceil$ denotes the ceiling function and $d = \text{gcd}(b,c)$. (Note that in the first summation the upper index is in general independent of $c$.) Ideas or reference suggestions are certainly appreciated. Thanks in advance!
Update I can prove the identity
\begin{eqnarray}
\sum_{k = 0}^{c-1} \left \lfloor \frac{x - kb}{c} \right \rfloor = d \left \lfloor \frac{x}{d} \right \rfloor - \frac{(b+1)(c-1)}{2} + \frac{d-1}{2},
\end{eqnarray}
where $d = \text{gcd}(b,c)$. There is another identity which might be useful. If $n = c \ell +r$ with $0 \leq r < c$, then
\begin{eqnarray}
\sum_{k = 1}^{n} \left \lfloor \frac{k}{c} \right \rfloor = c \binom{\ell}{2} + (r + 1) \ell.
\end{eqnarray}
Update 2 Ok, so I can prove that for real $x, y > 0$,
\begin{eqnarray}
\sum_{k = 0}^{\lfloor y \rfloor} \left \lfloor x + \frac{k}{y} \right \rfloor = \lfloor xy + (\lceil y \rceil - y) \lfloor x + 1 \rfloor \rfloor + \chi_{\mathbb{N}}(y)(\lfloor x \rfloor + 1),
\end{eqnarray}
where $\chi_{\mathbb{N}}$ denotes the characteristic function of the positive integers. My original problem (and a nice generalization of it) will be in hand if I can evaluate the following minor generalization: For real $x, y > 0$ and $n \in \mathbb{Z}_{\geq 0}$,
\begin{eqnarray}
\sum_{k = 0}^{n} \left \lfloor x + \frac{k}{y} \right \rfloor.
\end{eqnarray}
Again, any help is certainly appreciated!
|
Here is an observation/partial result. For brevity write $t = \lfloor a/b \rfloor .$ When $\text{gcd}(b,c)=1$ and $c \, | \, (t+1) $ we have
$$ S = \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =
\frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-1}{2} \right \rbrace . $$
Proof:
Suppose
$$\begin{align}
a &= m_0 c + r_0 \\
a- b &= m_1 c + r_1 \\
a - 2b &= m_2 c + r_2 \\
\cdots &= \cdots \\
a - tb &= m_t c + r_t
\end{align}$$
for integer $m_i$ and $r_i$ where $ 0 \le r_i < c $ then, adding the above equations,
$$(t+1)a - \frac{t(t+1)}{2}b = Sc + \sum_{k=0}^t r_k . \quad (1)$$
Now if $\text{gcd}(b,c)=1$ and $k$ runs over a complete system of residues modulo $c$ then
$a-kb,$ where $a$ is any integer, also runs over a complete system of residues modulo $c$. So when $ c \, | \, (t+1) $ we have that $a-kb$ runs over $(t+1)/c$ complete residue systems modulo $c$.
Hence
$$\sum_{k=0}^t r_k = \frac{(t+1)(c-1)}{2}.$$
Substitute this into $(1)$ to obtain the result.
EDIT: Here is a generalisation for the case $\text{gcd}(b,c)>1,$ which reduces to the above when $b$ and $c$ are coprime.
Write $d=\text{gcd}(b,c)$ and suppose $ a \equiv \lambda \textrm { mod } d $ where
$ 0 \le \lambda < d.$
Now $a-kb$ runs through all the residues modulo $c$ that are congruent to $ \lambda $ modulo $c$ as $k$ runs through $0,1,2,\ldots,u-1$ where $u=c/d.$ When $ u \, | \, (t+1) $ we have that $r_0,r_1,\ldots,r_t$ runs through $(t+1)/u$ such residue systems. Hence
$$ \sum_{k=0}^t r_k =
\frac{t+1}{u} \left \lbrace \frac{du(u-1)}{2} + \lambda u \right \rbrace
= (t+1) \left \lbrace \frac{c-d}{2} + \lambda \right \rbrace .$$
And so
$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =
\frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-d}{2} - \lambda \right \rbrace . $$
EDIT2:
Here are a couple of numerical examples. With $a=91,b=15 \textrm{ and } c=21$ we have
$d=\text{gcd}(15,21)=3,$ $t=\lfloor 91/15 \rfloor = 6,$ $u=c/d=21/3=7$ and
$91 \equiv 1 \textrm{ mod } 3,$ and so $\lambda=1.$ Note that the condition $ u \, | \, (t+1)$ is satisfied. Our formula gives the sum as
$$\frac{7}{21} \left \lbrace 91 - \frac{6 \cdot 15}{2} - \frac{21-3}{2} - 1 \right \rbrace = 12.$$
This is small enough to check by hand.
$$ \sum_{k=0}^7 \left \lfloor \frac{91-15k}{21} \right \rfloor = 4+3+2+2+1+0+0=12.$$
With $a=703,b=35 \textrm{ and } c=49$ we have
$d=\text{gcd}(35,49)=7,$ $t=\lfloor 703/35 \rfloor = 20,$ $u=c/d=49/7=7$ and
$703 \equiv 3 \textrm{ mod } 7,$ and so $\lambda=3.$ Note that the condition $ u \, | \, (t+1)$ is satisfied. Our formula gives the sum as
$$\frac{21}{49} \left \lbrace 703 - \frac{20 \cdot 35}{2} - \frac{49-7}{2} - 3 \right \rbrace = 141.$$
One can verify with WolframAlpha, or similar, that
$$ \sum_{k=0}^{20} \left \lfloor \frac{703-35k}{49} \right \rfloor = 141.$$
|
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|
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
|
Another more general approach is to solve your equation for $x$. Since it is
linear in $\sin x$ and $\cos x$ it can be transformed into a quadratic
equation in $\tan \frac{x}{2}$ (see this answer):
$$\sin x+\cos x=\frac{1+\sqrt{3}}{2}\Leftrightarrow \frac{2\tan \frac{x}{2}}{%
1+\tan ^{2}\frac{x}{2}}+\frac{1-\tan ^{2}\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}%
=\frac{1+\sqrt{3}}{2}$$
$$\Leftrightarrow 2\tan \frac{x}{2}+1-\tan ^{2}\frac{x}{2}=\left( 1+\tan ^{2}%
\frac{x}{2}\right) \frac{1+\sqrt{3}}{2}.$$
Set $y=\tan \frac{x}{2}$
$$2y+1-y^{2}=\frac{1+\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}y^{2}\Leftrightarrow \left( 1+\frac{1+\sqrt{3}}{2}\right) y^{2}-2y-1+\frac{1+%
\sqrt{3}}{2}=0$$
and solve for $y$
$$y_{1}=\frac{1}{3}\sqrt{3},y_{2}=2-\sqrt{3}$$
Hence
$$x_{1}=2\arctan \frac{1}{3}\sqrt{3}=\frac{1}{3}\pi $$
or
$$x_{2}=2\arctan \left( 2-\sqrt{3}\right) =\frac{1}{6}\pi .$$
And finally
$$\tan \frac{1}{3}\pi +\cot \frac{1}{3}\pi =\frac{4}{3}\sqrt{3}$$
or
$$\tan \frac{1}{6}\pi +\cot \frac{1}{6}\pi =\frac{4}{3}\sqrt{3}.$$
|
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|
How to "Re-write completing the square": $x^2+x+1$ The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
|
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + \cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=\frac{1}{2}$.
But if you have $(x+\frac{1}{2})^2$, you get $x^2 + x + \frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$\frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $\frac{1}{4}$. So:
\begin{align*}
x^2 + x + 2 &= (x^2 + x + \cdots) + 1\\
&=\left( x^2 + 2\left(\frac{1}{2}\right)x + \cdots \right) + 1 &&\mbox{figuring out what $c$ is}\\
&= \left(x^2 + 2\left(\frac{1}{2}\right)x + \left(\frac{1}{2}\right)^2\right) -\frac{1}{4} + 1 &&\mbox{completing the square}\\
&= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}.
\end{align*}
|
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|
Find the vertical asymptote of a function For an assignment, I was asked to find the vertical asymptote of the function $$g(x)= \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84}.$$
According to my text, a reliable method of finding the asymptote is to factor the numerator and denominator, and what left in the denominator that was not cancelled out is the asymptote.
I factored the numerator to $\frac{1}{2}(x^2-6x)(x-2)$, and the denominator factored to $7(x-2)(x-6)$, therefore $(x-2)$ cancelled out, leaving $(x-6)$ in the denominator.
However, 6 was not accepted as the answer, and I would like to know why.
|
Note that $\displaystyle f(x) = \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84} = \frac{1}{2}\frac{x^3-8x^2+12x}{7x^2-56x+84} = \frac{1}{2}\frac{1}{7}\frac{x^3-8x^2+12x}{x^2-8x+12} = \frac{x}{14}$.
So the function is "almost" a straight line passing through origin with a slope $\frac{1}{14}$ except at $x=2$ and $x=6$.
The function is not defined at $x=2$ and $x=6$. But $\displaystyle \lim_{x \rightarrow 2} = \frac{1}{7}$, $\displaystyle \lim_{x \rightarrow 6} = \frac{3}{7}$.
There are no asymptotes for the problem.
|
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|
Computing $\sum_{m \neq n} \frac{1}{n^2-m^2}$ A series arising in perturbation theory in quantum mechanics:
$\sum_{m\neq n} \frac{1}{n^2 - m^2}$, where $n$ is a given positive odd integer and $m$ runs through all odd positive integers different from $n$. I have a hunch that residue methods are applicable here, but I don't know complex analysis.
|
You can write
$$ \frac{1}{n^2 - m^2} = \frac{1}{2n}
\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace . \quad (1)$$
Now if we sum up both sides over all odd $m \ne n ,$ taking into account that $n$ is odd, lots of cancelling goes on and we obtain
$$\sum_{m \ne n} \frac{1}{n^2 - m^2} = -\frac{1}{4n^2}.$$
At first sight it appears the cotangent identity could be useful but it's not actually needed.
As a numerical check try summing the following with wolframAlpha
$$1/24 + 1/16 + \sum_{k=3}^\infty 1/(5^2 - (2k+1)^2),$$
you will see that it is $-1/100,$ as expected.
Or try this:
$$1/48 + 1/40 + 1/24 + \sum_{k=4}^\infty 1/(7^2 - (2k+1)^2).$$
You will get $-1/196.$
EDIT: To clarify the cancellation taking place when we sum the RHS of $(1).$
We have
$$\sum_{m \ne n, \,\, m \textrm{ odd} }
\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace
= \sum_{m \ne n, \,\, m \textrm{ odd} }
\left\lbrace \frac{1}{n+m} + \frac{1}{n-m} \right\rbrace $$
$$= \left\lbrace \left( \frac{1}{n+1} + \frac{1}{n-1} \right) +
\left( \frac{1}{n+3} + \frac{1}{n-3} \right) +
\left( \frac{1}{n+5} + \frac{1}{n-5} \right) + \cdots
+ \left( \frac{1}{2n-2} + \frac{1}{2} \right) \right\rbrace $$
$$+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)
+ \left( \frac{1}{2n+4} - \frac{1}{4} \right)
+ \left( \frac{1}{2n+6} - \frac{1}{6} \right) + \cdots $$
and rearranging all the terms in the braces
$$= \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace
+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)
+ \left( \frac{1}{2n+4} - \frac{1}{4} \right) + \cdots $$
$$=\left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace
- \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right\rbrace
= - \frac{1}{2n}$$
and hence the result.
|
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Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} < 2n$? Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} < 2n$?
My attempt:
$${{2n}\choose{n}} = \frac{(2n)!}{n!n!}$$
Let $a_1$ be the highest of power of $(2n)!$
Let $a_2$ be the highest of power of $n!$
So the highest power of $\frac{(2n)!}{n!n!}$ = $a_1 - 2a_2$
where $a_1 <= 2n - 1$ and $2a_2 <= 2n - 2$
Therefore the highest power of $p$ that divides $\frac{(2n)!}{n!n!}$ is $2n - 1 - 2n + 2 = 1$.
Since $a <= 1 \implies p^{a} < 2n$
Am I in the right track? Any idea?
Update
Following Ross Millikan's hint:
Let $a$ be the highest power of $p$ such that $p^{a}|n!$
Then, $a$ = $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$
Let $b$ be the highest power of $p$ such that $p^{b}|(2n)!$
Then, $b$ = $\lfloor \frac{2n}{p} \rfloor + \lfloor \frac{2n}{p^2} \rfloor +\lfloor \frac{2n}{p^3} \rfloor + \ldots \lfloor \frac{2n}{p^q} \rfloor$
$\Longrightarrow b - 2a$ is the highest power of $p$ such that $p|\frac{(2n)!}{n!n!}$
Where $a, b \in N \implies b - 2a < b$
Besides, $p^{b} < 2n$
$\therefore p^{b - 2a} < p^{b} < 2n$
Am I in the right track now?
Thanks,
Chan
|
Hint: The highest power of a prime,$p$, that divides $n!$ is $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$. This is your $a_2$. Can you compare twice this with the expression for $2n$, your $a_1$?
|
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Help complete a proof of Dirichlet on biquadratic character of 2? I am stuck proving the theorem that there exists $x$, $x^4 \equiv 2 \pmod p$ iff $p$ is of the form $A^2 + 64B^2$.
So far I have got this (and I am not sure if it's correct)
Let $p = a^2 + b^2$ be an odd prime,
*
*$\left(\frac{a}{p}\right) = \left(\frac{p}{a}\right) = \left(\frac{a^2 + b^2}{a}\right) = \left(\frac{b^2}{a}\right) = 1$
since $p \equiv 1 \pmod 4$
*
*$\left(\frac{a+b}{p}\right) = \left(\frac{(a+b)^2-2ab}{a+b}\right) = \left(\frac{2}{a+b}\right) = (-1)^{((a+b)^2-1)/8}$
using the Jacobi symbol and second supliment of quadratic reciprocity.
*
*$(a+b)^{(p-1)/2} = (2ab)^{(p-1)/4}$
since $(a+b)^2 \equiv 2ab \pmod p$
and the last step which I'm stuck on now is for $p = a^2 + b^2$ let $x^2 \equiv -1 \pmod p$ then $2^{(p-1)/4} = x^{ab/2}$. And I don't see how to prove the theorem with this result.
|
We can asume that 2 is a quadratic residue mod $p$ and so that $p \equiv 1 \pmod 8$ and this implies that if we pick $a$ odd and $b$ even then $b$ is a multiple of 4. We have to prove that $b$ is a multiple of 8.
First observe that as $x^2 \equiv -1 \pmod{p}$ and $a^2 + b^2 = p$ we have
$$ \left(\frac{a+b}{p}\right) \equiv (-1)^{((a+b)^2-1)/8} \equiv x^{(p+2ab-1)/4} \pmod{p} $$
note that the exponent of $x$ is even because $b$ is multiple of 4, so we can chose the sign of the base as we wish.
In adition we also have
$$ -a^2 \equiv b^2 \pmod{p} $$
so $(xa)^2 \equiv b^2 \pmod{p}$ and
$$ax \equiv \pm b \pmod{p}$$
and picking the sign of $x$ we can asume that $ax \equiv b \pmod{p}$. So $ab \equiv a^2 x$ and
$$ (ab)^{(p-1)/4} = a^{(p-1)/2} x^{(p-1)/4} \equiv x^{(p-1)/4} \pmod{p} $$
because $a$ is a quadratic residue mod $p$.
The identity you obtained:
$$\left(\frac{a+b}{p}\right) = (a+b)^{(p-1)/2} \equiv (2ab)^{(p-1)/4} \pmod p $$
now becomes
$$x^{(p+2ab-1)/4} \equiv 2^{(p-1)/4} x^{(p-1)/4} \pmod p $$
and in consequence
$$2^{(p-1)/4} \equiv x^{ab/2} \pmod p $$
As $b$ is multiple of 4 say $b = 4b'$, we have
$$2^{(p-1)/4} \equiv (-1)^{ab'} \pmod p $$
so $2$ is a biquadratic residue iif $b'$ is even or what is the same thing iif $b$ is multiple of 8 and we are done.
|
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|
Limit of $\lim\limits_{x \to\infty} 3\left(\sqrt{\strut x}\sqrt{\strut x-3}-x+2\right)$ I have to compute this limit:
$$\lim_{x \to\infty} 3(\sqrt{\strut x}\sqrt{\strut x-3}-x+2)$$
wolfram alpha says that answer is $\frac{3}{2}$, but I can't get why. Does anyone know how to get this limit?
|
The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of $x$.
\begin{align*}
\lim_{x\to\infty}3\left(\sqrt{x}\sqrt{x-3} - x+2\right) &= 3\lim_{x\to\infty}\left(\sqrt{x^2-3x} - (x-2)\right)\\
&= 3\lim_{x\to\infty}\frac{(\sqrt{x^2-3x}-(x-2))(\sqrt{x^2-3x}+(x-2))}{\sqrt{x^2-3x}+ (x-2)} \\
&= 3\lim_{x\to\infty}\frac{(x^2-3x) - (x-2)^2}{\sqrt{x^2-3x}+(x-2)}\\
&= 3\lim_{x\to\infty}\frac{x^2 - 3x - x^2 + 4x - 4}{\sqrt{x^2-3x} + (x-2)}\\
&= 3\lim_{x\to\infty}\frac{x - 4}{\sqrt{x^2-3x}+(x-2)}\\
&= 3\lim_{x\to\infty}\frac{\frac{1}{x}(x-4)}{\frac{1}{x}(\sqrt{x^2-3x}+(x-2))}\\
&= 3\lim_{x\to\infty}\frac {1 - \frac{4}{x}}{\sqrt{1 - \frac{3}{x}} + 1 - \frac{2}{x})}\\
&= 3\left(\frac{1}{\sqrt{1}+1}\right) = \frac{3}{2}.
\end{align*}
|
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|
Inscribed kissing circles in an equilateral triangle
Triangle is equilateral (AB=BC=CA), I need to find AB and R.
Any hints?
I was trying to make another triangle by connecting centers of small circles but didn't found anything
|
Let $a$ be the side of the triangle. If $A$ denotes the area and $P$ denotes the perimeter, then the radius of the incircle is given by $R = \frac{2A}{P} = \frac{2\sqrt{3} a^2/4}{3a} = \frac{\sqrt{3} a}{6}$
Let $x$ be the distance of the center of the smaller circle to the nearest vertex.
The altitude is $x + 8 + 2R = \frac{\sqrt{3}a}{2}$.
From similar triangle, we also have $\frac{x+4}{4} = \frac{x+8+R}{R} \Rightarrow \frac{x}{4} = \frac{x+8}{R}$
You have three equations in $x$,$R$ and $a$ solve them to get your $R$ and $a$.
EDIT
$x+8 = \frac{\sqrt{3}a}{2} - 2R = \frac{\sqrt{3}a}{2} - \frac{\sqrt{3}a}{3} = \frac{\sqrt{3}a}{6} = R$. Hence, $\frac{x}{4} = 1 \Rightarrow x = 4$.
$R = x + 8 = 12 \Rightarrow a = 2 \sqrt{3} R = 24 \sqrt{3}$.
|
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Find limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$, with the initial conditions: $a>b>0$ With the initial conditions: $a>b>0$;
I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$
I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.
|
Here is a short solution based on standard inequalities.
Our first inequality is obvious since $b^n>0$
$$(1)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\leq a^n.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$
Next we note that
$$a^n-b^n = a\cdot a^{n-1}-b\cdot a^{n-1}+ b\cdot a^{n-1}- b\cdot b^{n-1}
=(a-b)a^{n-1} + b(a^{n-1}-b^{n-1})$$
which together with $a^{n-1}- b^{n-1}\ge0$ leads to
$$(2)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\geq (a-b)a^{n-1}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $$
Combining (1) with (2) and taking the $n$:th-root we get
$$
a \ge (a^n-b^n)^{1/n}\ge(a-b)^{1/n}\cdot a^{(n-1)/n}= a\cdot(a-b)^{1/n}\cdot a^{-1/n}
$$
where the right hand side tends to $a$ as $n\to\infty$, and hence
we reach $$\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$$
|
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|
$2$ equations with $4$ variables Given the equations,
$$v_1+v_2=a-1$$
$$v_1v_2=b$$
for what ranges of $a$ and $b$, can I be sure to find $0<v_1<1$ and $0<v_2<1$.
Also, for what ranges, can I be sure to find at least one $v_i$ such that $0<v_i<1$
|
The solution is $v_1=\frac{1}{2}(-1+a-\sqrt{(a-1)^2-4b}), v_2=\frac{1}{2}(-1+a+\sqrt{(a-1)^2-4b})$ or the same with $v_1, v_2$ reversed. So we need $\frac{(a-1)^2}{4} \gt b$ to keep the square roots real. Then we need $1 \lt a-\sqrt{(a-1)^2-4b}$, which with $b \gt 0$ gives $a \gt 1$. We also need $a+\sqrt{(a-1)^2-4b} \lt 3$, which leads to $b \gt a-2$. So the solution area is $a \gt 1$, $b \lt 1$ and $\frac{(a-1)^2}{4} \gt b \gt a-2$.
|
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|
Multiplication in Permutation Groups Written in Cyclic Notation I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if
$$
a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4),
$$
then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?
|
You are thinking of the permutations as functions, so when you write "$ab$", you mean that you perform the permutation $b$ first, and the permutation $a$ second.
Here's one way to do it: write the disjoint cycle expressions for both $a$ and $b$, in the given order:
$$(1,3,5,2)(2,5,6)$$
Now, moving from right to left, see what happens to each number in each cycle.
For instance, start with $1$, so we write $1$ down:
$$(1,$$
The first cycle, $(2,5,6)$, does nothing to $1$, so it stays $1$. Then the next cycle, $(1,3,5,2)$, sends $1$ to $3$. So, in total, $1$ is sent to $3$. We write
$$(1,3,$$
Now consider $3$. The first cycle, $(2,5,6)$, does nothing to $3$. The second maps $3$ to $5$. So the product maps $3$ to $5$. So now we have
$$(1,3,5,$$
Now $5$. The first cycle, $(2,5,6)$, sends $5$ to $6$; the second cycle does nothing to $6$, so in total, $5$ is sent to $6$. So for the product we now have
$$(1,3,5,6,$$
Next, what happens to $6$? The first cycle sends $6$ to $2$; and then the next cycle sends $2$ to $1$. So $6$ is sent to $1$, and because we started out with $1$, this now closes the cycle we have; thus, we also close the bracket. So the product so far is
$$(1,3,5,6)$$
Now we consider the "next" number that hasn't been described yet, $2$. The first cycle, $(2,5,6)$, sends $2$ to $5$; then we check what the next cycle does to $5$, which is that it sends it back to $2$. So $2$ maps to $2$, and since we started out with $2$, it again closes the cycle. So now we have
$$(1,3,5,6)(2)$$
Finally we check what happens $4$, as it's the remaining number: $(2,5,6)$ fixes $4$ (it doesn't do anything to it – it remains as it is), as does $(1,3,5,2)$, so $4$ is overall fixed. So now finally we have:
$$ab = (1,3,5,2)(2,5,6)=(1,3,5,6)(2)(4) = (1,3,5,6)$$
$$\therefore (1,3,5,2)(2,5,6)=(1,3,5,6)$$
It's similar with $ac$. Here we have:
$$(1,3,5,2)(1,6,3,4).$$
First consider $1$: the first cycle maps it to $6$, the second cycle fixes $6$. So $1\mapsto 6$. Then $6$ is sent to $3$ by the first cycle, and $3$ to $5$ by the second cycle (reading right to left, remember), so $6\mapsto 5$. Then $5$ is fixed by the first cycle and sent to $2$ by the second cycle, so $5\mapsto 2$. Then $2$ is fixed by the first cycle and sent to $1$ by the second, which means $2\mapsto 1$, closing the cycle: we have $(1,6,5,2)$. The next number not already covered is $3$; $3$ is mapped to $4$ by the first cycle (by $b$), and $4$ is fixed by $a$, so $3\mapsto 4$. Then $4$ is sent to $1$ by the first cycle, and $1$ is sent to $3$ by the second cycle, so this closes this second cycle as $(3,4)$. Putting the two together we get
$$(1,3,5,2)(1,6,3,4) = (1,6,5,2)(3,4)$$
as given.
|
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Positive integer solutions of $x^2+21y^2=z^4 $ Can one find all positive integer solutions of
$$x^2+21y^2=z^4 ?$$
I am not sure if this is possible. I just saw this problem and this problem came to my mind.
|
This is an old post, but I believe a clarification to the other answer might be useful. The equation,
$$x^2+dy^2 = z^k\tag{1}$$
for $k = 2$ has a complete solution in terms of the single formula,
$$((p^2-dq^2)u)^2+d(2pqu)^2=((p^2+dq^2)u)^2\tag{2}$$
where $u$ is a scaling factor. However, the situation when $k>2$ is different. One can factor $(1)$ over $\sqrt{-d}$ and find formulas for all positive integer $k$. For example,
$$\big((p^3-3dpq^2)u^3\big)^2 + d((3p^2q-dq^3)u^3\big)^2 = ((p^2+dq^2)u^2)^3\tag{3}$$
But Pepin found that additional solutions can be given, like,
$$(13u^3+60u^2v-168uv^2-144v^3)^2 + 47(u^3-12u^2v-24uv^2+16v^3)^2 = 2^3(3u^2+2uv+16v^2)^3\tag{4}$$
not covered by $(3)$. For $k=4$, the impulse is to apply the complete formula $(2)$ onto itself,
$$p = (r^2-ds^2)v$$
$$q = (2rs)v$$
to get,
$$\big((r^4 - 6 d r^2 s^2 + d^2 s^4) u v^2\big)^2 + d\big(4 r s (r^2 - d s^2) u v^2\big)^2 = (r^2 + d s^2)^4 u^2 v^4\tag{5}$$
For the RHS to be a 4th power, one has to make the assumption that $u$ in the original triple $(2)$ is a square, hence it loses its generality. For example, for $d=21$, there is no rational ${r,s,u,v}$ that yields,
$$170^2 + 21\cdot51^2 = 17^4\\
17^2\big(10^2 +21\cdot3^2 = 17^2\big)
\tag{6}$$
This instead has the form,
$$\big((p^2 - d q^2) u^2 z\big)^2 + d\big(2 p q u^2 z\big)^2 = (u z)^4\tag{7}$$
where $z = p^2 + d q^2$. Notice that the terms of $(5)$ can be co-prime if $u=v=1$, but $(7)$ is never co-prime.
In summary, I am not sure if $(5)$ and $(7)$ cover all cases of $k=4$. (Compare it to $k = 2$, which has only one formula whether terms are co-prime or not.)
|
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If $\alpha$ is an acute angle, show that $\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$ If $\alpha$ is an acute angle, show that $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
My attempt:
Write $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\alpha} = (x+\cos{\alpha})^2+\sin^2{\alpha}$, we have:
$\displaystyle \begin{aligned}\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} & = \int_0^1 \frac{dx}{(x+\cos{\alpha})^2+\sin^2{\alpha}} = \bigg[\frac{1}{\sin{\alpha}}\tan^{-1}\left(\frac{x+\cos{\alpha}}{\sin{\alpha}}\right)\bigg]_{x=0}^1 \\ & = \frac{1}{\sin{\alpha}}\bigg[\color{blue}{\tan^{-1}\left(\frac{1+\cos{\alpha}}{\sin{\alpha}}\right)-\tan^{-1}\left(\frac{\cos{\alpha}}{\sin{\alpha}}\right)}\bigg] \end{aligned}$
I'm not sure, however, how the blue bit reduces to $\frac{1}{2}\alpha$. Any hints/suggestions? Thanks.
|
Or you can use this: $$\tan^{-1}{x} - \tan^{-1}{y} = \tan^{-1}\biggl(\frac{x-y}{1+xy}\biggr)$$
I am doing just the calculation part. We have
\begin{align*} \frac{1}{1+ \frac{\cos\alpha+\cos^{2}\alpha}{\sin^{2}\alpha}}\times \frac{1 + \cos\alpha}{\sin\alpha} - \frac{\cos\alpha}{\sin\alpha} &= \frac{1}{\sin\alpha} \times \frac{\sin^{2}\alpha}{\sin^{2}\alpha + \cos^{2}\alpha + \cos\alpha} \\ &= \frac{\sin\alpha}{1 + \cos \alpha} \\ &=\frac{ \displaystyle 2 \sin\frac{\alpha}{2}\cdot \cos\frac{\alpha}{2}}{2 \cos^{2}\frac{\alpha}{2}} \\ &=\tan\frac{\alpha}{2}
\end{align*}
Now take the inverse and then get the answer.
|
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
|
From your computations, it's enough to show that $n^3+2n=n(n^2+2)\equiv 0\pmod{3}$. If $n\equiv 0\pmod{3}$, you are done, else $n\equiv 1,2\pmod{3}$, in which case $n^2+2\equiv 0\pmod{3}$.
Reducing it to modular arithmetic may save you a lot of messy multiplication.
|
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Problem: Sum of absolute values of polynomial roots Can you please give me some hints as to how I might approach this problem? Thanks!
Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$
Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$
|
HINT
$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$
Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equality holds implies $|x_1| = |x_2| = |x_3| = 1$
We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and $x_1 x_2 x_3$ to be real, and further $x=-1$ satisfies the equation.
Hence $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And $|x_1| = 1$ and $|x_2| = 1$. Hence, $x_1 = e^{i \theta}$ and $x_2 = e^{i \phi}$.
$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$
$x_1 x_2 = 1 \implies \phi = -\theta$
$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$
Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.
Hence, $a = 4 \cos(\theta) - 2$ and the roots are $-1,e^{i \theta},e^{-i \theta}$
|
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Minimal polynomial of $\frac{\sqrt{2}+\sqrt[3]{5}}{\sqrt{3}}$ I am struggling to find the minimal polynomial for $\displaystyle \frac{\sqrt{2}+\sqrt[3]{5}}{\sqrt{3}}$.
Does anyone have any suggestions?
Thanks,
Katie.
|
One can compute the minimal polynomial using resultants or Grobner bases. But that is a bit overkill here since it can be done fairly straightforwardly by hand. Namely, let $\rm\ y = \sqrt{2}+\sqrt[3] 5\:.\ $ Then $\rm\: (y-\sqrt 2)^3 = 5\:,\:$ i.e. $\rm\:y^3 + 6\ y - 5 - (3\ y^2 + 2)\ \sqrt{2} = 0\:.\:$ Multiplying that by its conjugate yields $\rm\: y^6 - 6\ y^4 -10\ y^3 + 12\ y^2 -60\ y + 17 = 0\:.\:$ Putting $\rm\ y = \sqrt{3}\ x\:,\:$ then multiplying that by its conjugate yields $\rm\:729\ x^{12} -2916\ x^{10} + 4860\ x^8 - 5670\ x^6 -11340\ x^4 - 9576\ x^2 + 289 = 0\:.$
|
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Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the next section. (Simmons' Calculus)
The hint given with the problem is to integrate another series. This series must be $$1-\frac{ 1}{2} x^2+\frac{ 1\cdot 3}{2\cdot 4} x^4-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^6+\cdots$$
But where does it come from? [How would I know to integrate that series, if it weren't for the hint?]
The derivative of $\ln(x+\sqrt{1+x^2})$ is $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$,
but I can't see how to get from $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$ to that series.
The derivative can be written as $\displaystyle \frac{ 1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, but that looks even worse.
|
Do you know how to expand $1/\sqrt{1+x^2}$? See for instance this Wikipedia article.
Updated to add: You don't need calculus to derive this series: $$\frac{1}{\sqrt{1-y}} = 1+\frac{ 1}{2} y+\frac{ 1\cdot 3}{2\cdot 4} y^2+\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} y^3+\cdots$$ Just square the right-hand side, and you will get (miraculously) $$1 + y + y^2 + y^3 +\cdots$$ which you recognise as $1/(1-y)$ (right?). Now put $y = -x^2$ to obtain your expression for $1/\sqrt{1+x^2}$.
|
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|
Solving trigonometric equation involving summation For $ 0 <\theta<\frac{\pi}{2}$ find the solution of
$$\sum\limits_{m=1}^{6}\csc\left(\theta+\frac{(m-1)\pi}{4}\right)\cdot\csc\left(\theta+\frac{m\pi}{4}\right)=4\sqrt{2}$$
I thought of solving this as the angles form an A.P , But the given sum does not
come under any standard type such as the sum of the sines or cosines of
the angles in an A.P.So I am unable to proceed further.
|
The left side of the equation can be rewritten as:
$$ \Delta = \sum_{1 3 5} \csc\left(\theta+\frac{m\pi}{4}\right) \cdot \left( \csc\left(\theta+\frac{(m-1)\pi}{4}\right) + \csc\left(\theta+\frac{(m+1)\pi}{4}\right) \right) $$
Now using the formulas
$$ \sin u +\sin v = 2\sin\left(\frac {u + v} 2\right) \cdot \cos \left(\frac {u - v} 2\right)$$
and
$$ \sin u \sin v = \frac 1 2 \left(\cos (u - v) - \cos (u + v)\right)$$
we have
$$ \begin{aligned}\csc\left(\theta+\frac{(m-1)\pi}{4}\right) + \csc\left(\theta+\frac{(m+1)\pi}{4}\right) &= \frac {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) + \sin\left(\theta+\frac{(m+1)\pi}{4}\right)} {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) \cdot \sin\left(\theta+\frac{(m+1)\pi}{4}\right)}\\ &= \frac {4 \sin\left(\theta + \frac {m\pi} {4}\right) cos\left( \frac \pi 4\right)}{\cos\left(\frac \pi 2 \right) - \cos\left(2\theta + \frac {m\pi} {2} \right)}\end{aligned}$$
So $\Delta$ becomes
$$\Delta = -2\sqrt 2\sum_{1 3 5} \frac 1 {\cos\left(2\theta + \frac {m\pi} {2} \right)} = -2\sqrt 2 \left( -\frac 1 {\sin 2\theta} +\frac 1 {\sin 2\theta} -\frac 1 {\sin 2\theta}\right) = 2\sqrt 2 \frac 1 {\sin 2\theta}$$
I hope there are no typos... ;)
|
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|
Blockwise inversion case when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular What means in blockwise matrix inversion when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular but $\textbf{A}$ is not? is that necessary and sufficient for the whole composed matrix be singular as well? are there any cases where the whole matrix is not singular but blockwise matrix inversion fails because $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular?
|
It cannot happen. You are starting with a block matrix of the form
$$\mathbf{M}=\left(\begin{array}{cc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}\right),$$
which we are assuming is invertible; that is, $\det(\mathbf{M})\neq 0$.
But if $\mathbf{A}$ is invertible, then we also have
$$\det(\mathbf{M}) = \det(\mathbf{A})\det(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}),$$
so $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ cannot be singular.
To see this, note first that
$$\det\left(\begin{array}{cc}
\mathbf{X}& \mathbf{Y}\\
\mathbf{0} & \mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W})$$
and
$$\det\left(\begin{array}{cc}
\mathbf{X}&\mathbf{0}\\
\mathbf{Z}&\mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W}).$$
If $\mathbf{A}$ is invertible, then we have
$$\mathbf{M}=\left(\begin{array}{cc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}\end{array}\right) = \left(\begin{array}{cc}
\mathbf{A} & \mathbf{0}\\
\mathbf{C} & \mathbf{I}
\end{array}\right)\left(\begin{array}{cc}
\mathbf{I} & \mathbf{A}^{-1}\mathbf{B}\\
\mathbf{0} & \mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}
\end{array}\right),$$
where $\mathbf{I}$ is a suitably sized identity matrix. Since the determinant of a product is the product of the determinants, we get the desired formula.
So if $\mathbf{M}$ and $\mathbf{A}$ are both nonsingular, then $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ is nonsingular as well.
|
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|
Expansion concerning the binomial theorem The question goes:
Expand $(1-2x)^{1/2}-(1-3x)^{2/3}$ as far as the 4th term.
Ans: $x + x^2/2 + 5x^3/6 + 41x^4/24$
How should I do it?
|
Substituting $r=1/2$ and $y=-2x$ in the binomial series
$$
(1 + y)^r = 1 + \frac{r}{{1!}}y + \frac{{r(r - 1)}}{{2!}}y^2 + \frac{{r(r - 1)(r - 2)}}{{3!}}y^3 + \frac{{r(r - 1)(r - 2)(r - 3)}}{{4!}}y^4 + \cdots
$$
gives
$$
(1-2x)^{1/2} = 1 - x - x^2/2 -x^3/2 - (5/8)x^4 + \cdots,
$$
while substituting $r=2/3$ and $y=-3x$ gives
$$
(1-3x)^{2/3} = 1 - 2x - x^2 - (4/3)x^3 - (7/3)x^4 + \cdots.
$$
Now take the difference to get the desired result.
|
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|
How to prove that $\lim\limits_{(x,y) \to (0,0)} \frac{\left | x \right |^{\frac{3}{2}}y^{2}}{x^{4} + y^{2}} \rightarrow 0$ How can I prove that
$$\lim_{(x,y)\to (0,0)} \frac{\left | x \right |^{\frac{3}{2}}y^{2}}{x^{4} + y^{2}} \rightarrow 0\;?$$
Thanks!
|
For $y \neq 0$,
$$
0 \le \frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} = \frac{{|x|^{3/2} }}{{x^4 /y^2 + 1}} \le \frac{{|x|^{3/2} }}{1} = |x|^{3/2} \to 0.
$$
EDIT: In retrospect, simply note that $0 \le \frac{{y^2 }}{{x^4 + y^2 }} \le 1$, for $(x,y) \neq (0,0)$, to conclude that $\frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} \to 0$ as $x \to 0$, independently of $y$.
|
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|
Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$
|
divide that formular like this.
$$\sum_{k=1}^{\infty}\left (\frac{k}{3^k}+\frac{1}{3^k}\right )$$
then $$\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{\frac{1}{3}}{1-\frac{1}{3}}=\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{1}{2}$$
power series
$$ \sum_{k=1}^{\infty}\frac{k}{3^k}$$
let $$ \sum_{k=1}^{\infty}\frac{k}{3^k}=S$$
Then,
$$ S=1\times \frac{1}{3}+2\times \frac{1}{3^2}+3\times \frac{1}{3^3}+\cdots \cdots $$
$$\frac{1}{3}S=1\times \frac{1}{3^2}+2\times \frac{1}{3^3}+3\times \frac{1}{3^4}+\cdots \cdots $$
$$S-\frac{1}{3}S=1\times \frac{1}{3}+(2-1)\times \frac{1}{3}+(3-2)\times \frac{1}{3}\cdots \cdots =\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$$
Answer is
$$\therefore \frac{2}{3}S=\frac{1}{2},S=\frac{3}{4}$$
$$\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$$
|
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|
A "fast" way to ,find the maximum value of $(x^2) \times (y^3)$, if $3x+4y=12$ for $x,y \ge 0$ If $3x+4y=12$ $\forall x,y \ge 0$, the maximum value of $(x^2) \times (y^3)$ is
*
*$6 \times (6/5)^5$
*$3 \times (6/5)^5$
*$ (6/5)^5 $
*$7 \times (6/5)^5$
How to approach this problem? I thought of using the approach for finding maxima-minima for two independent variable but I am not sure about that as the $x$ and $y$ vanishes after the first partial derivative itself.
|
Lagrange multipliers are pretty fast, and don't require seeing any tricks:
$$2xy^3 = 3\lambda$$
$$3x^2y^2 = 4\lambda$$
Divide the two equations, and you get ${\displaystyle {2 \over 3} {y \over x} = {3 \over 4}}$ or ${\displaystyle y = {9 \over 8} x}$. Putting back into the equation you get
$$3x + {9 \over 2} x = 12$$
This solves as $x = 24/15 = 8/5$ and $y = 9/8\times 8/5 = 9/5$. So the minimum value of $x^2y^3$ is $(8/5)^2(9/5)^3$.
|
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|
Precision of operations on approximations If $ x $ and $ y $ have $ n $ significant places, how many significant places do $ x + y $, $ x - y $, $ x \times y $, $ x / y $, $ \sqrt{x} $ have?
I want to evaluate expressions like $ \frac{ \sqrt{ \left( a - b \right) + c } - \sqrt{ c } }{ a - b } $ to $ n $ significant places, where $ a $, $ b $, $ c $ are nonnegative integers. I was thinking about doing it recursively, i.e., if want to evaluate $ x / y $ to $ n $ places, I need to evaluate $ x $, $ y $ to $ m $ places, if want to evaluate $ x - y $ to $ n $ places, I need to evaluate $ x $, $ y $ to $ m $ places...
What book should I be reading?
|
$ \left[ a , b \right] + \left[ c , d \right] = \left[ a + c , b + d \right] $
$ \left[ a , b \right] - \left[ c , d \right] = \left[ a - d , b - c \right] $
$ \left[ a , b \right] \times \left[ c , d \right] = \left[ \min \left( a \times c , a \times d , b \times c , b \times d \right) , \max \left( a \times c , a \times d , b \times c , b \times d \right) \right] $
If $ 0 \notin \left[ c , d \right] $, then $ \left[ a , b \right] / \left[ c , d \right] = \left[ \min \left( a / c , a / d , b / c , b / d \right) , \max \left( a / c , a / d , b / c , b / d \right) \right] $
If $ a \geq 0 $, then $ \sqrt { \left[ a , b \right] } = \left[ \sqrt a , \sqrt b \right] $
$ a \approx a ^ { \prime } $ to $ n $ decimal significant places after the period if and only if $ a \in \left[ a ^ { \prime } - 5 \times 10 ^ { - \left( n + 1 \right) }, a ^ { \prime } + 5 \times 10 ^ { - \left( n + 1 \right) } \right) $, assuming I'm rounding half up.
If $ a \approx a ^ { \prime } $, $ b \approx b ^ { \prime } $ to $ n + 1 $ decimal significant places after the period, then $ a + b \approx a ^ { \prime } + b ^ { \prime } $, $ a - b \approx a ^ { \prime } - b ^ { \prime } $ to $ n $ decimal significant places after the period because $ a + b $ $ \in \left[ a ^ { \prime } - 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } + b ^ { \prime } - 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } , a ^ { \prime } + 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } + b ^ { \prime } + 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } \right) $ $ = \left[ a ^ { \prime } + b ^ { \prime } - 10 ^ { - \left( n + 1 \right) } , a ^ { \prime } + b ^ { \prime } + 10 ^ { - \left( n + 1 \right) } \right) $ $ \subset \left[ a ^ { \prime } + b ^ { \prime } - 5 \times 10 ^ { - \left( n + 1 \right) }, a ^ { \prime } + b ^ { \prime } + 5 \times 10 ^ { - \left( n + 1 \right) } \right) $, $ a - b $ $ \in \left[ a ^ { \prime } - 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } - \left( b ^ { \prime } + 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } \right) , a ^ { \prime } + 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } - \left( b ^ { \prime } - 5 \times 10 ^ { - \left( \left( n + 1 \right) + 1 \right) } \right) \right) $ $ = \left[ a ^ { \prime } - b ^ { \prime } - 10 ^ { - \left( n + 1 \right) } , a ^ { \prime } - b ^ { \prime } + 10 ^ { - \left( n + 1 \right) } \right) $ $ \subset \left[ a ^ { \prime } - b ^ { \prime } - 5 \times 10 ^ { - \left( n + 1 \right) }, a ^ { \prime } - b ^ { \prime } + 5 \times 10 ^ { - \left( n + 1 \right) } \right) $. So, to calculate $ a + b $ to $ n $ decimal significant places after the period, I need to first calculate $ a $, $ b $ to $ n + 1 $ decimal significant places after the period.
I'm still working on multiplication, division, root. My brain is frying...
|
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|
Exponential Diophantine perfect square problem I need to find all positive integer solutions $(a,b,c)$ to $1+4^a+4^b=c^2$. I am certain that the only solutions are $(t,2t-1, 2^{2t-1}+1) $ and $ (2t-1, t, 2^{2t-1}+1) $ , $t\in \mathbb{N}$ but I am having some trouble confirming this. Could someone help me please? Thank you.
|
By symmetry it is enough to deal with the case $1 \le a \le b$.
Rewrite the equation as
$$4^a(1+4^{b-a})=c^2-1=(c-1)(c+1).$$
Suppose first that $b=a$. Then
$4^a(1+4^{a-b})=2^{2a+1}$. But $(c-1)(c+1)$ is a power of $2$ only if $c=3$. Thus $a=b=1$. From now on we can assume that $b>a$.
Since $a \ge 1$, $c$ must be odd, say $c=2d+1$.
Substitute and divide by $4$.
We get
$$4^{a-1}(1+4^{b-a})=d(d+1).$$
We cannot have $a=1$, since if $a=1$ the left side is odd and the right side is even. Thus $a>1$.
Since $b>a$, $1+4^{b-a}$ is odd. But one of $d$ and $d+1$ is even. Whichever one this is, it must be divisible by $4^{a-1}$. We first examine the possibility that $d$ or $d+1$ is exactly $4^{a-1}$.
Then either (i) $4^{a-1}=d$ and $4^{b-a}+1=d+1$, or (ii) $4^{a-1}=d+1$ and $4^{b-a}+1=d$.
Possibility (ii) is easy to dismiss: two non-trivial powers of $4$ cannot differ by $2$.
If possibility (i) holds, we have $b-a=a-1$, so $b=2a-1$. It is easy to verify that this always does give a solution of the original system, namely your solutions with $t>1$.
Finishing the argument:
We found that one of $d$ or $d+1$ is divisible by $4^{a-1}$, and dealt fully with the case of equality. So now we ask: could we have $d$ or $d+1$ equal to $4^{a-1}k$, where $k>1$?
If that is the case, then since $k \ge 3$ we have $4^{b-a}+1\ge 9\cdot 4^{a-1}$. It follows that $2b-2a>2a-2+3$, and hence $b\ge 2a+1$.
But this is impossible. For note that since $1+4^a+4^b$ is a perfect square, we must have
$$4^b+4^a+1 \ge (2^b+1)^2 =4^b+2^{b+1}+1$$
However, if $b \ge 2a+1$ then $4^{a}<2^{b+1}$, contradicting the above inequality.
|
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Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$ Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$
|
We can use the Vasc's EV Method.
See here: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Indeed, let $a+b+c+d+e=constant.$
Thus, by corollary 1.9, case 1(b) ($p=0$,$q=-1$)
the expression $\sum\limits_{cyc}\frac{1}{a}$ gets a minimal value for equality case of four variables.
Id est, it remains to prove our inequality for $b=c=d=e$ and $a=\frac{1}{e^4}$, which gives
$$(e-1)^2(40e^8+80e^7+120e^6+160e^5-132e^4-89e^3-46e^2-3e+40)\geq0,$$
which is true because $$40e^8+80e^7+120e^6+160e^5-132e^4-89e^3-46e^2-3e+40=$$
$$=40(e^4+e^3+e^2-3e+1)^2+e(320e^4-12e^3+71e^2-486e+237)\geq$$
$$\geq e((e^2+40.5)(3e-2)^2+311e^4-297.5e^2+75)>0$$ and the last inequality is true because $$297.5^2-4\cdot311\cdot75=297.5^2-311\cdot300<0.$$
|
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Finding limit of a quotient I can't find this for some reason. I know I asked about 6 of these before, and I was able to finish my homework but now I went back to review and I can't do a single one of these problems on my own. Even the ones I did figure out on my own. I spent probably a total of 14 hours on the homework, not sure what is wrong with me but I don't know how to factor or do basic algebra.
Anyways I need to find
$$\lim_{x\to -2}\frac{x+2}{x^3+8}.$$
I have spent at least an hour on it and I can't figure it out.
|
For $x=-2$,
$$\begin{eqnarray*}
x+2 &=&0 \\
x^{3}+8 &=&0,
\end{eqnarray*}$$
which means $x=-2$ is a root of both equations. Hence $x^{3}+8$ may be
factored$^1$ as
$$x^{3}+8=(x-(-2))Q(x)=(x+2)Q(x).\qquad(*)$$
To compute $Q(x)$
*
*you may perform the long division $(x^{3}+8):(x+2)$ (or apply the so called Riffini's rule or sinthetic division) and
find
$$Q(x)=x^{2}-2x+4.$$
*Alternatively you write $x^{3}+8$ as
$$x^{3}+8=(x+2)(ax^{2}+bx+c),$$
(you could have made $a=1$, because the coefficient of $x^3$ is $1$) expand the RHS
$$(x+2)(ax^{2}+bx+c)=ax^{3}+bx^{2}+cx+2ax^{2}+2bx+2c,$$
group the terms of the same degree
$$(x+2)(ax^{2}+bx+c)=ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c$$
and equate to $x^{3}+8$
$$ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c=x^{3}+8.$$
The two sides are the same polynomial if and only if their coefficients are
equal
$$\begin{eqnarray*}
a &=&1 \\
b+2a &=&0 \\
c+2b &=&0 \\
2c &=&8,
\end{eqnarray*}$$
which means $a=1,b=-2,c=4$.
So
$$x^{3}+8=(x+2)(x^{2}-2x+4).\qquad(**)$$
Thus, for $x\neq -2$, we have
$$
\frac{x+2}{x^{3}+8}=\frac{x+2}{\left( x+2\right) \left( x^{2}-2x+4\right) }=
\frac{1}{x^{2}-2x+4},\qquad(***)$$
i.e $\frac{x+2}{x^{3}+8}$ is equal to $\frac{1}{x^{2}-2x+4}$, except for $x=-2
$. Now you can safely compute the limit
$$\lim_{x\rightarrow -2}\frac{x+2}{x^{3}+8}=\lim_{x\rightarrow -2}\frac{1}{
x^{2}-2x+4}=\dfrac{1}{\displaystyle\lim_{x\rightarrow -2}
x^{2}-2x+4}=\dfrac{1}{(-2)^{2}-2(-2)+4}=\dfrac{1}{12}.$$
--
$^1$ If a polynomial of degree $n$,
$$P(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots +a_{n-1}x+a_{n}\quad
(a_{0}\neq 0),$$
has as roots $n$ different numbers $\alpha _{1},\alpha _{2},\ldots
,\alpha _{n}$ , then it may be factored as
$$P(x)=a_{0}(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n}).$$
|
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|
Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.
|
For every $k\in\mathbb{N}$
$$(k+1)^4=k^4+4k^3++6k^2+4k+1$$
therefore
$$\sum_{k=1}^n(k+1)^4=\sum_{k=1}^nk^4+4\sum_{k=1}^nk^3+6\sum_{k=1}^nk^2+4\sum_{k=1}^nk+\sum_{k=1}^n1$$
which is equivalent to
$$\sum_{k=1}^nk^4+(n+1)^4-1=\sum_{k=1}^nk^4+4\sum_{k=1}^nk^3+6\sum_{k=1}^nk^2+2n(n+1)+n$$
After simplifications we obtain
$$4\sum_{k=1}^nk^3=(n+1)^4-1-2n(n+1)-n-6\sum_{k=1}^nk^2=n^4+4n^3+4n^2+n-6\sum_{k=1}^nk^2$$
Using
$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\hspace{0.2cm}\text{and}\hspace{0.2cm}\sum_{k=1}^nk=\frac{n(n+1)}{2}$$
we get
$$4\sum_{k=1}^nk^3=n^4+4n^3+4n^2+n-6\sum_{k=1}^nk^2\\=n^4+4n^3+4n^2+n-n(n+1)(2n+1)\\=n^4+2n^3+n^2=n^2(n+1)^2$$
Finally
$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}=\Big(\frac{n(n+1)}{2}\Big)^2=\Big(\sum_{k=1}^nk\Big)^2$$
|
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|
Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
|
This picture shows that $$1^2=1^3\\(1+2)^2=1^3+2^3\\(1+2+3)^2=1^3+2^3+3^3\\(1+2+3+4)^2=1^3+2^3+3^3+4^3\\$$ this is handmade of mine
|
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|
Asymptotic behaviour of sums of consecutive powers Let $S_k(n)$, for $k = 0, 1, 2, \ldots$, be defined as follows
$$S_k(n) = \sum_{i=1}^n \ i^k$$
For fixed (small) $k$, you can determine a nice formula in terms of $n$ for this, which you can then prove using e.g. induction. For small $k$ we for example get
$$\begin{align}
S_0(n) &= n\\
S_1(n) &= \frac{1}{2}n^2 + \frac{1}{2}n \\
S_2(n) &= \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \\
S_3(n) &= \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 \\
S_4(n) &= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3 - \frac{1}{30}n
\end{align}$$
The coefficients of these polynomials are related to the Bernoulli-numbers, and getting arbitrary coefficients in these polynomials (i.e. the coefficient of $n^m$ in $S_k(n)$ for large $k,m$) is not so easy. However, th first two coefficients follow a simple pattern: the coefficient of $n^{k+1}$ is $\frac{1}{k+1}$ and the coefficient of $n^k$ (for $k > 0$) is always $\frac{1}{2}$. My main question now is:
How can we prove that $S_k(n) = \frac{1}{k+1}n^{k+1} + \frac{1}{2}n^k + O(n^{k-1})$ for $k > 0$?
The first coefficient can be explained intuitively, as
$$S_k(n) = \sum_{i=1}^n \ i^k \approx \int_{i=1}^n i^k di \approx \frac{n^{k+1}}{k+1}$$
Maybe you could make this more rigorous, but I don't see how you will get the term $\frac{1}{2}n^k$ with this.
Also, while the coefficient of $n^{k+1}$ can be explained intuitively, it's not clear to me why the coefficient of $n^k$ is $\frac{1}{2}$, and why this one is fixed while e.g. the coefficient of $n^{k-1}$ is different for different $k$. If someone could explain that, that would be appreciated as well.
Thanks.
|
This is a classic application of the Euler-Maclaurin formula for approximating a sum by an integral. Euler-Maclaurin says
$$\sum_{i=0}^n f(i) = \int_0^n f(x) dx + \frac{f(n)+f(0)}{2} + \sum_{i=1}^{\infty} \frac{B_{2i}}{(2i)!} \left(f^{(2i-1)}(n)-f^{(2i-1)}(0)\right),$$
where $B_i$ is the $i$th Bernoulli number.
If we take $f(x) = x^k$, the infinite series is actually finite. The first two terms plus the first two terms from the series give us, for $k \geq 2$, $$\sum_{i=0}^n i^k = \int_0^n x^k dx + \frac{n^k}{2} + \frac{B_2}{2}k n^{k-1} + O(n^{k-3}) = \frac{n^{k+1}}{k+1} + \frac{n^k}{2} + \frac{k n^{k-1}}{12} + O(n^{k-3}),$$
which is what you're asking for.
And, of course, if you want a better asymptotic approximation you can just take more terms from the series.
This also explains why the $n^k$ term is the only one whose coefficient does not change; it's the only one in the formula in which the function $f(x) = x^k$ rather than its integral or one of its derivatives is being evaluated at $n$.
|
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|
Solving the equation $x + \sqrt{2x+1} = 7$ I can't solve
$x + \sqrt{2x+1} = 7$.
Well, I know the answer is 4, but that is from just reasoning it out. I can't algebraically solve it.
Thus, a step by step is what I really need.
Thanks in advance!!
|
If $x + \sqrt{2x+1} = 7$, multiply both sides by $x - \sqrt{2x+1}$
to get $x^2 - (2x+1) = 7(x - \sqrt{2x+1})$, or
$$\sqrt{2x+1} = x - (x^2 - (2x+1))/7 = (-x^2 + 9x+1)/7.$$
Substituting in the original equation,
$7 = x + (-x^2 + 9x+1)/7 = (-x^2 + 16x+1)/7$
or $x^2 - 16x + 48 = 0$.
Solving this, we get $x = 4$ and $x = 12$.
$x = 4$ satisfies the original equation with a positive square root,
and $x = 12$ satisfies it with a negative square root.
You also have to make sure that the
multiplying by $x - \sqrt{2x+1}$
does introduce any extraneous root(s).
Since doing this throws away the sign of the square root,
this introduces the value $x = 12$.
|
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|
How to show that $91$ is a pseudoprime to the base $3$? The given problem:
Use Lemma 2.3.3 together with Fermat's little theorem to show that 91
is a pseudoprime to the base 3.
Lemma 2.3.3.
Let $m_1 \dots m_r \in $ N. If $a \equiv b \pmod {m_i}$, $\forall i =1, \dots, r$, then $a \equiv b \pmod {{\rm lcm}(m_1, \dots, m_r)}$. (If $\gcd(m_i, m_j)=1$ when $i \neq j$, so $a \equiv b \pmod {m_1 \dots m_r}$.)
Theorem 2.4.5 (Fermat's little theorem II).
Let $p$ prime and $a \in$ Z such that $p \nmid a$. Then
$a^{p-1} \equiv 1 \pmod p$.
My own attempt:
Because $91=7 \cdot 13$, the number is composite.
According to Theorem 2.4.5 $3^6 \equiv 1 \pmod 7$. On the other hand $90 = 6 \cdot 15 $, so
$3^{90} = 3^{6 \cdot 15} = (3^6)^{15} \equiv 1^{15} \equiv 1 \pmod 7$
Then let's assume according to Theorem 2.4.5 that
$3^{12} \equiv 1 \pmod {13}$. On the other hand $90=12 \cdot 7+6$.
So $3^{90}=3^{12\cdot7+6}=3^{12\cdot7}\cdot 3^6 = (3^{12})^7\cdot3^6 \equiv 1^7 \cdot 3^6 \equiv 1 \pmod {13}$. So according to Lemma 2.3.3 we have $3^{90} \equiv 1 \pmod {91}$ so $3^{91} \equiv 3 \pmod {91}$.
So 91 would be a pseudoprime to the base 3.
|
Powers of $3$ cycle as follows
$$
\begin{array}{|c|c|c|}
n&3^n&3^n \mod 91\\
\hline\\
1 & 3 & 3\\
2 & 9 & 9\\
3 & 27 & 27\\
4 & 81 & 81\\
5 & 243 & 61\\
6 & 729 & 1\\
\hline
\end{array}
$$
Therefore, since $90 = 6 \times 15$
$$
\begin{align*}
3^{90} &\equiv 1 \hspace{4pt} (\mod 91)\\
\Rightarrow 3^{91} &\equiv 3 \hspace{4pt} (\mod 91)
\end{align*}
$$
|
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|
A coin is flipped when dice hits 6 (conditional probability) Prof gave us homework on conditional probability that is due on the day of the lecture on conditional probability. Yeah, this has been a bad week and I've no idea what I'm doing.
Q: 3 dice are rolled, then, a coin is flipped as many times as the number 6 is obtained.
a) find the probability of getting less than 2 heads.
b) knowing this experiment results in less than 2 heads, what is the conditional probability that exactly 2 sixes were obtained?
I don't even know where to start...
Attempt:
a)
The only way I figure is:
Rolling one six, probability of head < 2 = $ \displaystyle \frac{3}{6^3} \cdot \frac{2}{2}$
Rolling two sixes: $\displaystyle \frac{3}{6^3} \cdot \frac{2}{2^2}$
Rolling three sixes: $\displaystyle \frac{1}{6^3} \cdot \frac{4}{2^3}$
Then adding those.
b)
$\displaystyle \frac{\frac{1}{6^3} \cdot \frac{4}{2^3}}{\frac{1}{6^3}}$
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a) The computation takes a while. It may be useful to draw a tree in order not to lose track of the possibilities. Initially, we toss $3$ dice. We get $0$, $1$, $2$, or $3$ $6$'s. Then, depending on the outcome, we toss a certain number of coins. So from the "start" position, there are $4$ branches, corresponding to the number of $6$'s obtained.
If $3$ dice are rolled, then the probability of $k$ $6$'s is equal to
$$\binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3-k}.$$
In particular, the probability of $0$ $6$'s is $\frac{125}{216}$, the probability of exactly $1$ $6$ is $\frac{75}{216}$, the probability of exactly $2$ $6$'s is $\frac{15}{216}$, and the probability of $3$ $6$'s is $\frac{1}{216}$.
If we have $0$ $6$'s, we toss $0$ coins, so for sure (probability $1$) we get fewer than $2$ heads. If we get $1$ $6$, then we toss $1$ coin, and again for sure we get fewer than $2$ heads. If we get $2$ $6$'s, we toss $2$ coins, and the probability of fewer than $2$ heads is easily computed to be $3/4$. If we get $3$ $6$'s, we toss $3$ coins. By symmetry the probability of fewer than $2$ heads is then $1/2$. It follows that the overall probability of fewer than $2$ heads is
$$\frac{125}{216}\cdot 1 + \frac{75}{216}\cdot 1 +\frac{15}{216}\cdot \frac{3}{4}+\frac{1}{216}\cdot\frac{1}{2}.$$
This may simplify to $\dfrac{847}{864}$. But check the arithmetic!
b) Let $E$ be the event that exactly $2$ sixes were obtained, and let $L$ be the event we got fewer than $2$ heads. We want $P(E|L)$.
There are various formulas that we could now use. We go for the simplest one, though using it is a bit less mechanical than the full Bayes formula. Note that
$$P(E|L)P(L)=P(E \cap L).$$
This formula comes straight from the the definition of conditional probability.
From the above formula, we see that we will be essentially finished once we find $P(E\cap L)$ and $P(L)$.
The probability $P(L)$ that we get fewer than $2$ heads has been computed in part (a). So we only need to find $P(E \cap L)$.
In the solution of part (a), we already saw that
$$P(E)=\binom{3}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^1=\frac{15}{216}=\frac{5}{72}.$$
Given that we got exactly $2$ $6$'s, we got to flip the coin twice, and thus the probability of fewer than $2$ heads is $3/4$, as shown by @Yuval Filmus. It follows that
$$P(E\cap L)=P(L|E)P(E)=\left(\frac{3}{4}\right)\left(\frac{5}{72}\right)=\frac{5}{96}.$$
Now put the pieces together. I think that $P(E|L)=\dfrac{45}{847}$.
Do check the arithmetic!
|
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probablity of random pick up three points inside a regular triangle which form a triangle and contain the center what is the probablity of random pick up three points inside a regular triangle
which form a triangle and contain the center of the regualr triangle
the three points are randomly picked within the regular triangle and then form a new triangle
and the new triangle have to contain the center of the original regular triangle
what is the probability
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Let $R$ be a convex region of the plane with unit area, and choose $n\ge 3$ points at random from the region. We will derive a general expression for the probability $P_{n}$ that the convex hull of these points contains a particular point $\tau \in R$ (which we will take as the origin of our coordinate system). We can then evaluate that expression for the particular question posed (in which $n=3$, $R$ is an equilateral triangle, and $\tau$ is its center).
Suppose the hull does not contain $\tau$, and assume that $\tau$ is not collinear with any two of the points (this is true with probability 1). Then we can draw a ray from the center point through one of the $n$ points (the "leftmost" point) such that the remaining $n-1$ points are to the right of the ray. (Note that this condition is both necessary and sufficient. Note also that if the convex hull does contain $\tau$, then there is no "leftmost" point.) We can calculate the desired probability by integrating over all such configurations. In particular, take $f(\varphi)$ to be the area of the subregion that lies to the right of the radial ray at angle $\varphi$, and take $r(\varphi)$ to be the distance from $\tau$ to the boundary of $R$ along that ray. Then the probability that a particular point will lie in the angular interval $[\varphi, \varphi+d\varphi]$ is $da = \frac{1}{2}r(\varphi)^2 d\varphi$, and the probability that each of the other $n-1$ points will lie to its right is $f(\varphi)$. Since there are $n$ points to choose as the leftmost, the overall probability of not capturing $\tau$ in the convex hull is
$$
1-P_{n} = n \int f(\varphi)^{n-1} da = n \int_{-\pi}^{\pi} \frac{1}{2} f(\varphi)^{n-1} r(\varphi)^2 d\varphi,
$$
We can make a few observations at this point. First, if the region $R$ is symmetric under reflection through $\tau$, then $f(\varphi)$ is identically $1/2$: each radial line splits the region in half. The result then is
$$
P_{n} = 1 - \frac{n}{2^{n-1}},
$$
which gives the known result $P_{3} = 1/4$ for polygons with an even number of sides. Second, if the region $R$ instead has bilateral symmetry across the $x$-axis, then $r(\varphi)$ is an even function, and $f(\varphi) = 1/2 + f^{-}(\varphi)$, where $f^{-}$ is an odd function. Then terms in the integral with odd powers of $f^{-}$ must vanish: in particular, we have
$$
P_{3} = \frac{1}{4} - 3 \int{{f^{-}(\varphi)}^2 da} = \frac{1}{4} - 3\int_{-\pi}^{\pi}\frac{1}{2}{f^{-}(\varphi)}^2 r(\varphi)^2 d\phi \le \frac{1}{4},
$$
and interestingly the relation $P_{4} = 2P_{3}$ continues to hold.
It remains to evaluate $f^{-}(\varphi)$, $r(\varphi)$, and the resulting integral in the case of an equilateral triangle. The figure above shows an equilateral triangle with vertices at $(-x,0)$ and $(x/2, \pm x\sqrt{3}/2)$. For $\varphi \in [0,\pi/3]$, the function $f^{-}(\varphi)$ is the area of the blue triangle minus the area of the red triangle,
$$
f^{-}(\varphi) = \frac{x^2}{8}\tan\varphi - \frac{x^2}{2}\frac{1}{\cot\varphi + \cot\frac{\pi}{6}} = \frac{x^2}{8}\tan\varphi \left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right);
$$
and the radius $r(\varphi) = \frac{1}{2}x\sec\varphi$ over the same domain. By symmetry, the full integral is just six times its value over $[0,\pi/3]$:
$$
\begin{eqnarray}
P_3 &=& \frac{1}{4} - 18\int_{0}^{\pi/3}\frac{x^6}{512}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{36\sqrt{3}}\int_{0}^{\pi/3}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{12}\int_{0}^{1} u^2\left(1 - \frac{4}{1+3u}\right)^2 du,
\end{eqnarray}
$$
where we have used $x = 2/\sqrt{3\sqrt{3}}$ (in order for the triangle to have unit area) and introduced the transformed variable $u = \tan\varphi / \sqrt{3}$ (so $du = (\sec^2\varphi / \sqrt{3}) d\varphi$). The final integral is a straightforward exercise for the reader; the result is
$$
P_3 = \frac{1}{4} - \frac{1}{324}\left(57 - 80\ln{2}\right) = \frac{2}{81}\left(3 + 10\ln{2}\right) = 0.2452215...,
$$
in agreement with the brute-force and numerical results already given. This result can be generalized easily to the regular $m$-gon for any odd $m$ (changing only the values of some constants), and to the case where $n>3$ (complicating the final integral).
|
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|
How many ordered pairs of positive integers
For a prime integer p, how many ordered pairs of positive integers
(a, b) are there that satisfy $$\frac{1}{a} + \frac{1}{b} =\frac{1}{p}$$
For example, for p = 5, $$\frac{1}{6} + \frac{1}{30}$$ and
$$\frac{1}{30} + \frac{1}{6}$$ are two different ways of getting
$\frac{1}{5}$.
Ok, so I tried this for prime numbers from 2 to 19. It seems like that for any prime number there are only 3 such ordered pairs.
(Question: $\frac{1}{10} + \frac{1}{10}$ is just one ordered pair for p=5, right?)
But I don't know how to go about proving this. I can see that:
After $\frac{1}{p}$, take the next smallest fraction of the form $\frac{1}{a}$. Now we always get a fraction $\frac{1}{a} + \frac{1}{b}$. Also, now b is the upper bound for the numbers we need to check. But there is always just one other integer between a and b which satisfies for this property, this integer is is 2p
The question does not specifically ask for a prove, but they always expect a prove for everything.
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$1/a+1/b=1/p$, $bp+ap=ab$, $(a+b)p=ab$, so $p$ divides $a$ or $p$ divides $b$ (or both). Let's say $p$ divides $a$, so $a=cp$ for some $c$. Now $cp+b=bc$, $b=bc-pc=(b-p)c$, so $c$ divides $b$. Let $b=cd$. Then $d=cd-p$, so $p=cd-d=(c-1)d$. So either $c-1=p$ and $d=1$, or $c-1=1$ and $d=p$. Now work your way back up to $a$ and $b$.
|
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|
$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
\geq \dots \geq
=2 \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.
What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.
|
Hint:
$$
(n+1)! = (n+1) n! \leq (n+1) \left( \frac{n+1}{2} \right)^n = 2 \left( \frac{n+1}{2} \right)^{n+1}.
$$
You can check that $2 \left( \frac{n+1}{2} \right)^{n+1} \leq \left( \frac{n+2}{2} \right)^{n+1}$, by proving that
$$
2 \leq \left( \frac{n+2}{n+1} \right)^{n+1}.
$$
|
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|
Integrate $\int\frac{1}{x^6} \sqrt{(1-x^2)^3} ~ dx$ How to integrate the following?
$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$
|
Hint: Do the substitution $x = \sin(\alpha)$, $\alpha \in [-\frac{\pi}{2},\frac{\pi}{2}]$. Then you get
$$\int \frac{\cos^4(\alpha)}{\sin^6(\alpha)} d\alpha.$$
Now consider the following identities:
$$D\left(\frac{\cos^3(\alpha)}{\sin^5(\alpha)}\right) = \frac{-3\cos^2(\alpha)\sin^6(\alpha) - 5\sin^4(\alpha)\cos^4(\alpha)}{\sin^10(\alpha)} = -3\frac{\cos^2(\alpha)}{\sin^4(\alpha)} - 5 \frac{\cos^4(\alpha)}{\sin^6(\alpha)}$$
$$D\left(\frac{\cos(\alpha)}{\sin^3(\alpha)}\right) = \frac{-\sin^4(\alpha)-3\sin^2(\alpha)\cos^2(\alpha)}{\sin^6(\alpha)} = -\frac{1}{\sin^2(\alpha)} - 3 \frac{\cos^2(\alpha)}{\sin^4(\alpha)}$$
By using these it boils down to being able to integrate $\frac{1}{\sin^2(\alpha)}$.
|
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|
Solving $(2y-4)(2y+1) = (2y-2)^2$ I'm getting different answer from answer key.
Solving $$(2y-4)(2y+1) = (2y-2)^2$$
FOIL left side $$4y^2+2y-8y-4 = (2y-2)^2$$
Right side $$4y^2+2y-8y-4 = 4y^2+4 $$
Subtract $4y^2$ from both sides
$$2y-8y-4 = 4 $$
Combine $y$
$$6y-4 = 4$$
add 4 to both sides
$$6y = 8$$
But the answer key has $y=4$
|
The error is in the "Right Side" step.
You essentially wrote
$$(2y-2)^2 = 4y^2 + 4.$$
That's incorrect.
Remember: $(a-b)^2 = a^2 - 2ab + b^2$. So
$$(2y-2)^2 = 4y^2 - 8y + 4.$$
The third displayed equation should thus be
$$4y^2 +2y - 8y - 4 = 4y^2 -8y + 4.$$
You will find that this leads to $2y = 8$, from which you get $y=4$.
|
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|
quotient rule difficulties I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have
$$\begin{align*}
\frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac {d}{dr} \sqrt{r^2+1}} {r^2+1} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac{1}{2}(r^2+1)^{-1/2}2r} {(r^2+1)}\\\\\\\\
&= \frac {\sqrt{r^2+1} - r^2 (r^2+1)^{-1/2}} {(r^2+1)}\\\\\\\\
&= (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}
\end{align*}$$
However, the correct answer is just $$(r^2+1)^{-3/2} $$
I've been over it a number of times now, but I can't see the error. I'm pretty new to the quotient rule.
|
There's nothing wrong with your application of the quotient rule. You just need to simplify your answer further:
$$
\begin{eqnarray*}
(r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}
&=& (r^2+1) \cdot (r^2+1)^{-3/2} - r^2 \cdot (r^2+1)^{-3/2}
\\ &=& (r^2+1)^{-3/2} \cdot \left((r^2+1) - r^2 \right)
\\ &=& (r^2+1)^{-3/2} \cdot 1
\\ &=& (r^2+1)^{-3/2},
\end{eqnarray*}
$$
which is the official answer.
|
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|
If $\gcd(a,b)=1$ , and $a$ is even and $b$ is odd then $\gcd(2^{a}+1,2^{b}+1)=1$? How to prove that:
$\gcd(a,b)=1 \Rightarrow \gcd(2^{a}+1,2^{b}+1)=1$ ,where $a$ is even and $b$ is odd natural number
For example:
$\gcd(2^8+1,2^{13}+1)=1 , \gcd(2^{64}+1,2^{73}+1)=1$
I know that Knuth showed that:
$\gcd(2^{a}-1,2^{b}-1)=2^{\gcd(a,b)}-1$
so: $\gcd(a,b)=1\Rightarrow \gcd(2^{a}-1,2^{b}-1)=1$
but I don't see whether this fact is useful.
|
This is a minor tweak of my answer to an earlier post by user952949. That one asked for a proof that if $a$ and $b$ are relatively prime and odd, then $\gcd(2^a+1,2^b+1)=3$.
A very useful fact: If $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$.
We can arrange for $x$ to be $\ge 0$ and $y$ to be $\le 0$. So there are non-negative integers $u$ and $v$ such that $au=bv+1$. Thus
$$2^{au}=2^{bv+1}=2\cdot 2^{bv}. \qquad\text{(Equation 1)}$$
Let $m$ be any common divisor of $2^a+1$ and $2^b+1$. Then $2^a \equiv -1\pmod{m}$ and $2^b\equiv -1 \pmod m$. It follows that
$$2^{au} =(2^a)^u \equiv (-1)^u \pmod{m} \qquad\text{and}\qquad 2^{bv} =(2^b)^v \equiv (-1)^v \pmod{m}.$$
From Equation $1$, we conclude that
$$(-1)^u \equiv 2\cdot (-1)^v\pmod{m}.$$
If $(-1)^u=(-1)^v$, we find that $1\equiv 0\pmod{m}$, so the only positive common divisor of $2^a+1$ and $2^b+1$ is $1$. If $(-1)^u=-(-1)^v$, we find that $3\equiv 0\pmod{m}$.
If we only know that $3\equiv 0\pmod{m}$, then all we can say is that any common divisor of $2^a+1$ and $2^b+1$ divides $3$.
We need to rule out the possibility that $3$ divides both of $2^a+1$ and $2^b+1$. This is easy. One of $a$ or $b$ is even, say $a$. Then $2^a\equiv 1 \pmod{3}$, so $2^a+1$ is not divisible by $3$.
|
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|
Proving that $f(n)$ is an integer using mathematical induction I want to prove that
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$
is an integer for every integer $n \geq 1$.
I define P(n) to be:
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$ is an integer.
For my basis step, P(1) is true because
$$\frac{1^3}{3}+\frac{1^5}{5}+\frac{7}{15}=1$$
which is an integer.
The inductive step is what's tripping me up...
Let k be an arbitrary positive integer. Assume that P(k) is true, that is,
$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}$$
is an integer.
So based on that assumption, I need to now show that P(k+1) is true, i.e., that
$$\frac{(k+1)^3}{3} +\frac{(k+1)^5}{5} +\frac{7 (k+1)}{15}$$
is an integer.
At this point, I am stuck as to where to go next...
I have tried rewriting the assumption:
$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}=15 m$$
for some integer m. Then I solve for m:
$$\frac{1}{15} \left(\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}\right)=m$$
But this looks like a dead-end, seems there's nothing I can do with this to the "to prove" equation.
I have also tried re-writing the "to show" equation as this, but I get a dead end there and am not sure where to go next:
$$\frac{1}{15} \left(5 (k+1)^3+3 (k+1)^5+7 (k+1)\right)$$
|
Why do you think that $P(k) = 15m$ for some integer $m$ if it does not hold for, say $k=1$? If you assume that $P(k)$ is integer then the strategy is to show that
$$
P(k+1) - P(k) \in\mathbb Z
$$
and let us do it:
$$
P(k+1) - P(k) = \frac{1}{5}((n+1)^5-n^5)+\frac13((n+1)^3 - n^3)+\frac7{15} =
$$
$$
= \frac15(5n^4+10n^3+10n^2+5n+1) +\frac13(3n^2+3n+1)+\frac7{15}
$$
$$
= n^4+2n^3+2n^2+n +\frac15+n^2+n+\frac13+\frac{7}{15}
$$
$$
= n^4+2n^3+3n^2+2n+1
$$
$$
=(n^2+n+1)^2\in \mathbb Z
$$
and you're done.
|
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|
Determine $z^n+z^{-n}$ if $z+\frac{1}{z}=-2\cos{x}$ Determine $z^n+z^{-n}$ if $z+\frac{1}{z}=-2\cos{x}$ with $z \in \mathbb{C}$.
|
Since $z + \frac{1}{z} = - 2 \cos(x)$ is equivalent to $z^2 + 2 z \cos(x) + 1 = 0$, it is solved by $z_{1,2} = -\cos(x) \pm i \sqrt{1-\cos^2(x)}$. Since $1-\cos^2(x) = \sin^2(x)$, these also solve the equation $\tilde{z}_{1,2} = -\cos(x) \mp i \sin(x) = -\exp(\pm i x)$.
Now to find $z^n+z^{-n}$ for $z$ being the solution of $z+\frac{1}{z} = -2 \cos(x)$ subsitute the $z = \tilde{z}_{1,2}$.
|
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|
integrate square of $\arctan x$. Tricky $$\int \left(\frac{\tan^{-1}x}{x-\tan^{-1}x}\right)^{2}dx$$
I ran across an integral I am having a time solving.
The solution merely works out to $\displaystyle\frac{1+x\tan^{-1}x}{\tan^{-1}x-x}$, but for the life of me I can not find a suitable method to tackle it.
Does anyone have any hints on a good strategy, substitution, parts, etc?. Thanks much.
|
With $x = \tan(u)$, $\mathrm{d}x = \frac{1}{\cos^2(u)}\mathrm{d} u$, thus
$$
\int \left(\frac{\tan^{-1}x}{x-\tan^{-1}x}\right)^{2} \mathrm{d}x =
\int \left( \frac{u}{\tan(u) - u} \cdot \frac{1}{\cos(u)} \right)^2 \mathrm{d} u =
\int \left( \frac{u}{\sin(u) - u \cdot \cos(u)} \right)^2 \mathrm{d} u
$$
Now, observe that
$$
\mathrm{d} \left( \frac{1}{\sin(u)-u \cos(u)} \right) = \frac{-u \sin(u) \mathrm{d} u }{(\sin(u)-u \cos(u))^2}
$$
This allows to integrate by parts:
$$
\begin{eqnarray}
I &=& \int \left( \frac{u}{\sin(u) - u \cdot \cos(u)} \right)^2 \mathrm{d} u =
\int \left( -\frac{u}{\sin(u)} \right) \mathrm{d}\left( \frac{1}{\sin(u)-u \cos(u)}
\right) \\
&=& -\frac{u}{\sin(u)\left(\sin(u)-u \cos(u) \right)} + \int \frac{1}{\sin(u)-u \cos(u)} \mathrm{d} \left( \frac{u}{\sin(u)} \right) \\
&=& -\frac{u}{\sin(u)\left(\sin(u)-u \cos(u) \right)} + \int \frac{\mathrm{d}u}{\sin^2(u)} \\
&=& -\frac{u}{\sin(u)\left(\sin(u)-u \cos(u) \right)} - \frac{1}{\tan(u)} + C \\
&=& -\frac{\arctan(x)}{\frac{x}{\sqrt{1+x^2}} \left(\frac{x}{\sqrt{1+x^2}} -\arctan(x) \cdot \frac{1}{\sqrt{1+x^2}} \right)} - \frac{1}{x} + C \\
&=& -\frac{1}{x} \left( 1+ \frac{(1+x^2) \arctan(x) }{x - \arctan(x)}\right) + C =
-\frac{ 1 + x \arctan(x) }{x - \arctan(x)} + C
\end{eqnarray}
$$
|
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|
chances of getting three of one kind and four of another out of seven dice There are several questions similar to this one but after reading those,
I am still very confused.
I also did a similar problem of this one and I think I got it, but then I got stuck again.
So if four dice are rolled, the chance of getting three of a kind is:
$ \binom{6}{1} \frac{1}{6}* \binom{5}{1} \frac{1}{6}*\binom{4}{1} \frac{1}{6} *\frac{1}{6}$
so if seven dice are rolled, in my understanding,
the chance of getting three of a kind and four of another would be:
$ \binom{7}{1} \frac{1}{6}*\binom{6}{1} \frac{1}{6}*\binom{5}{1} \frac{1}{6} *\binom{4}{1} \frac{1}{6} *\binom{3}{1} \frac{1}{6} *\binom{2}{1} \frac{1}{6} *\binom{1}{1} \frac{1}{6} $
however, the answer in the book is $ \frac{6 *5*\binom{7}{4} }{6^7} $ and I am totally lost.
Please help!
additional problems
The answers you guys gave kind of make sense to me but they also make me very confused.
can I think of it using the way I did above?
for example, another part of the questions asked about the chance of getting two fours, two fives and three sixes.
I think of it as:
$ \binom{7}{2} \frac{1}{6}^2*\binom{5}{2} \frac{1}{6}^2*\binom{3}{3} \frac{1}{6}^3 $ which matches the solution in the book.
|
Think of filling in 7 slots; in each you have the value of a die roll.
There are $6\cdot 5$ ways to choose the values for the three and, different valued, four of a kind (for example,
the three of a kind is three '2's and the four of a kind is four '5's).
There are $7\choose 4$ ways to select ''slots'' in which to place your 4 of a kind.
The remaining three slots will then contain the three of a kind. So, there are
$6\cdot5\cdot{7\choose 4}$ different ways to obtain a three of a kind and 4 of a kind. Since outcomes are equally likely here, the chance of obtaining a three of a kind and 4 of a kind is as you stated.
|
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|
Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:
*
*When $b$ is huge, for instance $5^{844325}\bmod 21$?
*When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$?
*When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?
Are there any other tricks for evaluating exponents in modular arithmetic?
|
Here are two examples of the square and multiply method for $5^{69} \bmod 101$:
$$ \begin{matrix}
5^{69} &\equiv& 5 &\cdot &(5^{34})^2 &\equiv & 37
\\ 5^{34} &\equiv& &&(5^{17})^2 &\equiv& 88 &(\equiv -13)
\\ 5^{17} &\equiv& 5 &\cdot &(5^8)^2 &\equiv& 54
\\ 5^{8} &\equiv& &&(5^4)^2 &\equiv& 58
\\ 5^{4} &\equiv& &&(5^2)^2 &\equiv& 19
\\ 5^{2} &\equiv& &&(5^1)^2 &\equiv& 25
\\ 5^{1} &\equiv& 5 &\cdot &(1)^2 &\equiv& 5
\end{matrix} $$
The computation proceeds by starting with $5^{69}$ and then working downward to create the first two columns, then computing the results from the bottom up. (normally you'd skip the last line; I put it there to clarify the next paragraph)
As a shortcut, the binary representation of $69$ is $1000101_2$; reading the binary digits from left to right tell us the operations to do starting from the value $1$: $0$ says "square" and $1$ says "square and multiply by $5$".
The other way is to compute a list of repeated squares:
$$ \begin{matrix}
5^1 &\equiv& 5
\\ 5^2 &\equiv& 25
\\ 5^4 &\equiv& 19
\\ 5^8 &\equiv& 58
\\ 5^{16} &\equiv& 31
\\ 5^{32} &\equiv& 52
\\ 5^{64} &\equiv& 78
\end{matrix} $$
Then work out which terms you need to multiply together:
$$ 5^{69} \equiv 5^{64 + 4 + 1} \equiv 78 \cdot 19 \cdot 5 \equiv 37 $$
|
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|
Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} = \frac{5 \times \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,
|
The sum $S= \sum \limits_{i=1}^n \frac{1}{2^i}=1-\frac{1}{2^n}$ is geometric, thus easy to calculate.
Here is a simple elementary way of calculating
$$T=\sum_{i=1}^n \frac{i}{2^i} \,.$$:
$$T=\sum_{i=1}^n \frac{i}{2^i} =\frac{1}{2}+ \sum_{i=2}^n \frac{i}{2^i} =\frac{1}{2}-\frac{n+1}{2^{n+1}}+ \sum_{i=2}^{n+1} \frac{i}{2^i}
\,.$$
Changing the index in the last sum yields:
$$T= \frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i+1}{2^{i+1}}=\frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i}{2^{i+1}}+\sum_{i=1}^{n} \frac{1}{2^{i+1}} \,.$$
Thus
$$T= \frac{2^n-n-1}{2^{n+1}}+\frac{1}{2}T+ \frac{1}{2}-\frac{1}{2^{n+1}}$$
Thus
$$\frac{1}{2}T=\frac{2^{n+1}-n-2}{2^{n+1}}\,.$$
Hence
$$\sum_{i=1}^n \frac{i}{2^i} =\frac{2^{n+1}-n-2}{2^{n}} \,.$$
|
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|
$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$ $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$
Work out the values of $\frac{1}{x+y}$
|
My solution is quite elementary.
So we consider the equations $$5^x + 2^y = \frac {7}{10} = 2^x +5^y $$
Note that $5^x + 2^y = \frac {7}{10} ( eq.1 ) $ and $\frac {7}{10} = 2^x +5^y (eq.2) $ are inverse functions.
By inspection, we can see that $(-1, -1 )$ is a solution. Now, If we show that this is the only solution then we can clearly state that $$\frac {1}{x+y} = -\frac {1}{2}$$.
Let us now find the domain and the range of the functions. Since the two functions are inverses , the the domain of the first function is the range of the other function.
Consider only the function $5^x + 2^y = \frac {7}{10} $.
Solving for $x$ and $y$ , we get $x = log_{5}(\frac {7}{10} - 2^y )$ and $y = log_{2}(\frac {7}{10} - 5^x )$ respectively.
We know that given $y = log x$ ,y is defined if $x> 0$. So, we have the equations $\frac {7}{10} - 2^y > 0 $ and$ \frac {7}{10} - 5^x >0$
Solving for x and y,$ y < log_{2}\frac {7}{10} $ and $x < log_{5}\frac {7}{10}$.
Clearly, the domain and the range are $$\{ x|x < log_{5}\frac {7}{10}\} $$ and $$\{ y|y < log_{2}\frac {7}{10}\}$$.
Therefore, the asymptotes of the first equation are $x = log_{5}\frac {7}{10}$ and $ y = log_{2}\frac {7}{10}\ $, the asympotes of the second equation are $y = log_{5}\frac {7}{10}$ and $ x = log_{2}\frac {7}{10}\ $, and we know that $ log_{5}\frac {7}{10} < log_{2}\frac {7}{10}$.
From this, We can conclude that as the $ x \rightarrow -\infty $ from $-1$. then for each y, $y_{ eq 1} < y_{ eq 2}$ and as $ y \rightarrow -\infty $ from $-1$. then for each x, $ x_{ eq 1} > x_{ eq 2} $. This is clear due to the asymptotes and the idea that they are inverses.
Furthermore, If we show that indeed $ x \rightarrow -\infty $ from $-1$. then, for each y, $y_{ eq 1} < y_{ eq 2}$, then it follows that $ y \rightarrow -\infty $ from $-1$. then for each x, $ x_{ eq 1} > x_{ eq 2} $.
Hence, to show this, notice that $$5^x + 2^y = 5^y +2^x =\frac {7}{10} = \frac {2+5}{10} =\frac {2}{10}+\frac {5}{10} =\frac {1}{5}+\frac {1}{2} = 5^{-1} + 2^{-1} $$
For the equality to be correct as we make $ x \rightarrow -\infty $ from $-1$, and since $5^x < 2^x$ since $x< -1$, then the decrease in the equation 1 is larger and it requires smaller value of y to be equal to $5^{-1} + 2^{-1}$.
thank you.
|
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|
Existence of the limit of a sequence? I solved this limit problem by following this way, but I'm not exactly sure about ....
can anyone help me and tell me if it is correct?
the problem is:
Let $k>1$. If it exists, calculate the limit of the sequence $(x_n)$,
$$x_n := \Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n.$$
my solution:
From trigonometry we know that:
$$ 0< \sin x < x <\tan x$$
for $$0 < x < \pi/2 ;$$
then $$\sin x < x \Rightarrow \sin \frac{1}{n^2}<\frac{1}{n^2};$$
the cosine function we know to be a bounded function; then:
$$\left | \cos x \right |\le1\Rightarrow -1\le\cos x\le 1 \Rightarrow -\frac{1}{k}\le
\frac{1}{k}\cos n \le\frac{1}{k} $$
so we have that:
$$k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \le \Biggl(k\frac{1}{n^2}+\frac{1}{k}\Biggr) \Rightarrow\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n\le \Biggl(k\frac{1}{n^2}+\frac{1}{k}\Biggr)^n,$$
then
$$x_n \le \Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n;$$
On the other hand, we have also that:
$$\left | \sin x \right |\le1\Rightarrow -1\le\sin x\le 1 \Rightarrow -{k}\le\ k \cdot \sin (\frac{1}{n^2}) \le{k},$$
and then in particular
$$-\frac{1}{k}\le\cos n \qquad,\qquad -{k}\le\ k \cdot \sin (\frac{1}{n^2});$$
well:
$$\Biggl(-k-\frac{1}{k}\Biggr)\le\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr) \Rightarrow -\Biggl(k+\frac{1}{k}\Biggr)^n\le\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n,$$
end then:
$$-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n;$$
combining the two results we obtained:
$$-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n\le\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n.$$
at this point we have found that $a_n$ $b_n$ such that:
$$a_n \le x_n \le b_n,$$
an so $x_n$, if it admits limit, must be admitted to the same limit $a_n$ and $b_n$;
so:
$$\lim_{n\to \infty}a_n=\lim_{n\to \infty}-\Biggl(k+\frac{1}{k}\Biggr)^n=-\lim_{n\to \infty}\Biggl(\frac{k^2+1}{k}\Biggr)^n=-\infty$$
$$\lim_{n\to \infty}b_n=\lim_{n\to \infty}=\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n=\Biggl[\frac{n^2\cdot(1+\frac{k}{n^2})}{kn^2}\Biggr]^n =\frac{\Biggl[(1+\frac{k}{n^2})^{n^2}\Biggr]}{k^n}^\frac {1}{n}=\frac {e^\frac {k}{n}}{k^n}=\frac {1}{k^n}=0$$
as seen , the two limits are different, so the sequence $ x_n $ admits no limit. In conclusion:
$$\lim_{n\to \infty}\Biggl(k \sin (\frac{1}{n^2}) + \frac{1}{k}\cos n \Biggr)^n=\nexists$$
|
The limit is zero. You can argue as follows. If $n$ is large enough that $1/n^2 \leq \pi/2$, then $\sin(1/n^2)\leq 1/n^2$. Then we have that
$$|x_n| \leq \left(\frac{k}{n^2} + \frac{1}{k}\right)^n.$$
Then choose $n$ large enough so that $\frac{k}{n^2} + \frac{1}{k} \leq 1-\delta$ for some positive $\delta<1$. This can be done as $k > 1$ so $1/k < 1$. Then you have that
$$|x_n| \leq (1-\delta)^n$$
for large enough $n$, and this limit is zero.
|
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|
random variable transformation I'm having trouble with the following random variable transformation:
$Y = X^2 + X$
I am looking for the pdf of Y. I tried the following method:
$p_Y(y) = \int_{X} p_{Y|X=x}(y)\cdot p_{X}(x)dx$
and we know that $(Y|X=x) \sim (x^2+x) \Rightarrow p_{Y|X=x} = \delta_{x^2+x}(y)$
thus: $p_Y(y) = \int_{X} \delta_{x^2+x}(y)\cdot p_{X}(x)dx$
But I don't see a way to reduce this further.
Then I tried a different approach:
$p_Y(y) = DF(Y < y) = DF(X^2 + X < y) = ...$
But then I don't see a way to find the inverse of $X^2 + X$.
Can anyone help me further on this?
|
Let $Y = X^2 + X = \left( X+\frac{1}{2} \right)^2 - \frac{1}{4}$. Then
$$
F_Y(y) = \mathbb{P}(Y \le y) = \mathbb{P}\left( \left( X+\frac{1}{2} \right)^2
\le y + \frac{1}{4} \right)
$$
Assume, additionally, that $y+\frac{1}{4} > 0$. Then
$$ \begin{eqnarray}
F_Y(y) &=& \mathbb{P}\left( -\sqrt{y + \frac{1}{4}} \le X+\frac{1}{2}
\le \sqrt{y + \frac{1}{4}} \right) \\ &=& F_X\left( -\frac{1}{2} + \sqrt{y + \frac{1}{4}} \right) - F_X\left( -\frac{1}{2} -\sqrt{y + \frac{1}{4}} \right) + \mathbb{P}\left( X = -\frac{1}{2} - \sqrt{y + \frac{1}{4}} \right)
\end{eqnarray}
$$
If $X$ is continuous rv, the last term is zero.
Differentiating with respect to $y$, we get
$$
f_Y(y) = \frac{1}{\sqrt{4y+1}} \left( f_X\left(-\frac{1}{2} + \sqrt{y + \frac{1}{4}} \right) + f_X\left( -\frac{1}{2} -\sqrt{y + \frac{1}{4}} \right) \right) \cdot \mathbf{1}\left(y > -\frac{1}{4}\right)
$$
|
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|
Probability that a family with $n$ children has exactly $k$ boys Let the probability $p_n$ that a family has exactly $n$ children be $\alpha p^n$ when $n\geq1$, and $$p_0=1-\alpha p(1+p+p^2+\cdots).$$ Suppose that all the sex distributions have the same probability. Show that for $k\geq1$ the probability that a family has exactly $k$ boys is $2\alpha p^k/(2-p)^{k+1}$.
|
Extended hint: We sketch an argument that uses only basic notions.
Note that the probability $b_k$ of $k$ boys is, by a conditional probability argument, given by
$$b_k=\sum_{n=1}^\infty \alpha p^n \binom{n}{k} \left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}.$$
This simplifies to
$$b_k=\sum_{n=1}^\infty \alpha \binom{n}{k}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast)$$
(We define $\binom{n}{k}$ to be $0$ if $n<k$.)
There are many tools for evaluating $(\ast)$. We do it using not much machinery.
Recall the combinatorial identity
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.$$
Substitute for $\binom{n}{k}$ in $(\ast)$.
We obtain
$$b_k=\sum_{n=1}^\infty \alpha \binom{n-1}{k}\left(\frac{p}{2}\right)^n +\sum_{n=1}^\infty \alpha \binom{n-1}{k-1}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast\ast)$$
The first term in $(\ast\ast)$ is just $\dfrac{p}{2}b_k$. The second term is $\dfrac{p}{2}b_{k-1}$.
So we have derived the recurrence
$$b_k=\frac{p}{2}b_k+\frac{p}{2}b_{k-1}$$
or equivalently
$$b_k=\frac{p}{2-p}b_{k-1}.$$
This almost settles things: each time we increment $k$ by $1$, the probability gets multiplied by $\dfrac{p}{2-p}$. To get the process started, we need $b_1$. We have
$$b_1=\alpha\sum_{n=1}^\infty n \left(\frac{p}{2}\right)^n.$$
There is a trick for finding $\sum_{n=1}^\infty n x^{n}$. Using the fact that for $|x|<1$,
$$1+x+x^2+x^3+ x^4+ \cdots=\frac{1}{1-x},$$
we find, by differentiating, that
$$\frac{1}{(1-x)^2}=1+2x+3x^2+ 4x^3+\cdots.$$
It follows that
$$x+2x^2+3x^3+4x^4+\cdots =\frac{x}{(1-x)^2}.$$
Comment: Alternately, one could evaluate the sum $(\ast)$ by a repeated differentiation argument that generalizes the method we used for $b_1$.
The recurrence $b_k=\dfrac{p}{2}b_k+\dfrac{p}{2}b_{k-1}$ can also be obtained directly, bypassing the series manipulation.
|
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|
How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
|
I will try this way:
Since you are dividing a 3rd degree polynomial by a 2nd degree polynomial, WLOG, we may assume
$$-2x^3+4x^2-3x+5=(4x^2+5)(ax+b)+cx+d\quad(1)$$
Now, comparing the coefficients of $x^3$ and $x^2$ readily give $a=-\frac{1}{2}$ and $b=1$. Comparing coefficients of $x$, we have $5a+c=-3\Rightarrow c=-\frac{1}{2}$. Finally comparing the constant term, we get $5b+d=5\Rightarrow d=0$.
|
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|
How do you factor $x^3-3x^2+3x-1$? $$x^3-3x^2+3x-1?$$
I know this may seem trivial, but I, for the life of me, I cannot figure out how to factor this polynomial, I know that the root is $$(x-1)^3=0$$ because of wolframalpha, but I don't know how to get there. any help would be greatly appreciated. and also if you have any recommended web sites that help with higher order polynomial factoring that would be extremely helpful.
|
$$
\begin{align*}
x^3-3x^2+3x-1 &=x^3-x^2-2x^2+3x-1
\\ &=x^2(x-1)-2x^2+2x+x-1
\\ &=x^2(x-1)-2x(x-1)+1(x-1)
\\ &=(x-1)(x^2-2x+1)
\\ &=(x-1)(x-1)^2
\\ &=(x-1)^3
\end{align*}$$
|
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|
Finding complex solutions of an equation How does one solve this equation. I would like to see the solution of this problem in steps.
$z\cdot\bar{z}=\left|3\cdot z \right|$
EDIT: Is it possible to solve this by converting to the form $z=a+b\cdot i$
What about the solution of this equation.
$z\cdot\bar{z}-z^{2}=1-i$
EDIT2:
$a^2+b^2-(a+b\cdot i)(a+b\cdot i) = 1 - i$
$a^2+b^2-a^2-ab\cdot i - ab\cdot i + b^2=1-i$
$2b^2-2ab\cdot i = 1-i$
And we keep in mind that two imaginary numbers are equal if their real and imaginary parts are the same.
$2b^2 = 1$ and $-2ab=-1$
So $b = \pm \frac{1}{\sqrt{2}}$
and $a=\frac{1}{2b}\Rightarrow a=\pm \frac{\sqrt{2}}{2}$.
Is this correct?
|
First, note that for any complex number $z=a+bi$, we have
$$z\cdot \bar{z}=(a+bi)\cdot(a-bi)=a^2+abi-abi+b^2(i)(-i)=a^2+b^2=\left(\sqrt{a^2+b^2}\right)^2=|z|^2.$$
Now note that for any complex number $z=a+bi$ and real number $t$, we have
$$|t\cdot z|=|t(a+bi)|=|(ta)+(tb)i|=\sqrt{(ta)^2+(tb)^2}=\sqrt{(t^2)(a^2+b^2)}=$$
$$\sqrt{t^2}\sqrt{a^2+b^2}=|t|\sqrt{a^2+b^2}=|t|\cdot|z|$$
(In fact, it is true that for any two complex numbers $w$ and $z$, we have $|w\cdot z|=|w|\cdot|z|$.)
These are both important facts to know in general.
Thus, starting from the equation
$$z\cdot \bar{z}=|3\cdot z|$$
we get
$$|z|^2=3\cdot|z|.$$
Now treat $|z|$ as a real number to be solved for - that is, think of it as if we are solving
$$x^2=3x.$$ Note that there are two solutions, i.e. two possible values for $|z|$. Do you see what they are?
Finally, note that the set of complex $z$ for which $|z|=c$ forms a circle of radius $c$ in the complex plane; using polar coordinates, i.e. $z=re^{i\theta}$, we have that $|z|=c$ if and only if $$|z|=|re^{i\theta}|=|r|\cdot|e^{i\theta}|=|r|\cdot1=|r|=r=c,$$
so the complex $z$ for which $|z|=c$ are the complex numbers of the form $ce^{i\theta}$ for some $\theta$.
|
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|
How to calculate $\sqrt{\frac{-3}{4} - i}$
Possible Duplicate:
How do I get the square root of a complex number?
I know that the answer to $\sqrt{\dfrac{-3}{4} - i}$ is $\dfrac12 - i$. But how do I calculate it mathematically if I don't have access to a calculator?
|
For $a \ne 0$, there are two numbers $z$ such that $z^2=a$.
We look at the given example, using only basic tools. We want to solve the equation
$$z^2=-\frac{3}{4}-i.$$
Because of a general discomfort with negative numbers, we look at the equivalent equation
$$z^2=-\frac{1}{4}(3+4i).$$
In order to deal with simpler numbers, we first consider the equation
$$w^2=3+4i.$$
Let $w=a+ib$. Then $w^2=(a^2-b^2)+2ab i$.
So we want to solve the system of two equations
$$a^2-b^2=3, \qquad 2ab=4.$$
The solutions can be found by inspection. However, we continue, in order to show how to proceed when the numbers are less simple. Rewrite the second equation as $b=2/a$. Substitute in the first equation. We get
$$a^2-\left(\frac{2}{a}\right)^2=3,$$
which after some simplification becomes
$$a^4-3a^2-4=0.$$
This is a quadratic equation in $a^2$. By using the Quadratic Formula, we find that the roots are $a^2=4$ and $a^2=-1$. The second equation has no real solution, so $a=\pm 2$. We get the two solutions $a=2$, $b=1$ and $a=-2$, $b=-1$.
Thus $w=2+i$ or $w=-(2+i)$. So find $z$, multiply these two values of $w$ by $\dfrac{i}{2}$.
Another way: Any complex number $z^2$ can be written as $r(\cos \theta+i\sin\theta)$ where $r$ is non-negative. Then
$$z^2=r^2[(\cos^2\theta)+i(2\cos \theta\sin\theta)].$$
We can rewrite this as $r^2(\cos(2\theta)+i\sin(2\theta)$. We want $z^2=-\frac{3}{4}-i$. The norm of $-\frac{3}{4}-i$ is the square root of $(-3/4)^2+(-1)^2$, so $r^2=(9/16)+1=25/16$ and $r=5/4$. We know that $\cos(2\theta)=(4/5)(-3/4)$ and $\sin(2\theta)=(4/5)(-1)$.
Simplify. We get $\cos(2\theta)=-\frac{3}{5}$ and $\sin(2\theta)=-\frac{4}{5}$. Now we could proceed by calculator, finding $2\theta$, then $\theta$, then $\cos\theta$ and $\sin\theta$. Or else we can proceed algebraically, using the fact that $\cos^2\theta-\sin^2\theta=-\frac{3}{5}$ and $2\sin\theta\cos\theta=-\frac{4}{5}$. If we do this, the rest is much like the first solution. We have $\sin\theta=(-4/5)/(2\cos\theta)$. Substitute in $\cos^2\theta-\sin^2\theta=-\frac{4}{5}$. After simplifying, we obtain a quadratic equation in $\cos^2\theta$, and the rest is routine. There will be two values of $\cos\theta+i\sin\theta$ that work, and they will be the negatives of each other, and the roots are $\pm\sqrt{r}(\cos\theta+i\sin\theta)$.
|
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|
Conditions for intersection of parabolas? What are the conditions for the existence of real solutions for the following equations:
$$\begin{align}
x^2&=a\cdot y+b\\
y^2&=c\cdot x+d\end{align}$$
where $a,b,c,d $ are real numbers.
These represent two parabolas; how might we find out the conditions for the existence of $0,2,4$ real solutions of the equations?
|
Assume that $(x,y)$ is a point common to both parabolas. If we add the two equations together and complete the square we get the circle equation
\begin{equation*}
\left( x - (c/2) \right)^2 + \left(y - (a/2) \right)^2 = (a/2)^2 + (c/2)^2 + b + d
\end{equation*}
So one condition which is necessary for a solution is that the right-hand side be $\geq 0$.
If we instead subtract the second equation from the first we get
\begin{align*}
(x-y)(x+y) &= ay - xc + b - d \newline
&= \frac{a-c}{2}(x+y) - \frac{a+c}{2}(x-y) + b - d
\end{align*}
Let $w=(x+y)$, $z=(x-y)$, $k_1 = (a-c)/2$, $k_2=(a+c)/2$. Then
\begin{equation*}
wz = k_1w - k_2z + b-d
\end{equation*}
\begin{equation*}
(w+k_2)(z-k_1) = (b-d) -k_1 k_2
\end{equation*}
So the point $(w,z)$ lies on a certain hyperbola. Unfortunately I can't think of anything else that one can say with this line of reasoning, but I thought it was interesting so I decided to post this incomplete answer anyway.
|
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|
Find $ n\geq1 $ such that 7 divides $n^n-3$ Find $ n\geq1 $ such that 7 divides $n^n-3$.
Here is what I found:
$ n\equiv 0 \mod7, n^n\equiv 0 \mod7,n^n-3\equiv -3 \mod7$ no solution.
$ n\equiv 1 \mod7, n^n\equiv 1 \mod7,n^n-3\equiv -2 \mod7 $ no solution.
$ n\equiv 2 \mod7, n^n\equiv 2^n \mod7, n^n-3\equiv 2^n-3 \mod7$. Is $2^n\equiv 3 \mod7 $ possible?
$ n=7k+2,..., 2^{7k+2}\equiv 2^{k+2} \mod7$. Is $2^{k+2}\equiv 3 \mod7$ possible?
Studying $k$ modulo 6: $2^{6q+2}\equiv 4 \mod7, 2^{6q+3}\equiv 1 \mod7, 2^{6q+4}\equiv 2 \mod7, 2^{6q+5}\equiv 4 \mod7, 2^{6q+6}\equiv 1 \mod7, 2^{6q+7}\equiv 2 \mod7 $
The congruence is never 3 so there is no solution for $n\equiv 2 \mod7$.
$ n\equiv -2 \mod7,..., n=42q+5 $
$ n\equiv-1 \mod7, n^n-3\equiv (-1)^n-3 \mod7 $: no solution.
$ n\equiv 3 \mod7, n^n-3 \equiv 3^n-3 \mod7$. Is $3^n\equiv 3 \mod 7$ possible?
$n=7k+3,..., 3^{7k+3}\equiv -3^k \mod7 $. Is $3^k \equiv 4 \mod 7 $ possible?
Studying $k$ modulo 6:
$ 3^{6q} \equiv 1 \mod7, 3^{6q+1} \equiv 3 \mod7, 3^{6q+2} \equiv 2 \mod7, 3^{6q+3} \equiv -1 \mod7, 3^{6q+4} \equiv 4 \mod7 , 3^{6q+5} \equiv -2 \mod7 $
So $ n=7k+3=7(6q+4)+3=42q+31 $ is a solution.
$ n\equiv -3 \mod7 $,..., there is no solution.
$42q+31 \equiv 3 \mod 7, (42q+31)^{42q+31} \equiv 3^{42q+31} \mod 7 \equiv 3 \mod7$ OK
$ 42q+5 \equiv 4 \mod7, (42q+5)^{42q+5} \equiv 5^{42q+5} \mod 7 \equiv 3 \mod7$ OK
So 7 divides $ n^n-3 $ if and only if $ n\equiv 31 \mod 42$ or $ n\equiv 5 \mod 42$
|
$$
5^5-3=2\cdot 7\cdot 233
$$
Other solutions are
$$n=31, 47, 73, 89, 115, 131, 157, 173, 199, 215, 241, 257, 283, 299, 325$$
All of them are odd and congruent to $3$ or $5$ modulo $7$.
|
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|
Functional inverse of $\sin\theta\sqrt{\tan\theta}$ What is the functional inverse of $f(\theta) = \sin\theta\sqrt{\tan\theta}$? Or, equivalently, what is the inverse of
$$f(\theta)=\sin^2\,\theta\tan\,\theta=\frac{\sin^3\,\theta}{\cos\,\theta}$$
It comes from a physics setup involving two equivalently massed and charged pith balls separated by a certain distance, and the equation simplifies to $q = 4L\sin\theta\sqrt{\pi\epsilon_0mg\tan{\theta}}$, where $\pi$, $g$, and $\epsilon_0$ are the obvious physical constants and $L$ and $m$ will be fixed. The question asked for $\theta$ in terms of $q$, however, so I'm wondering if there is a way to rearrange this. I can't seem to find anything on the internet, and Wolfram refuses to reveal the steps for its complex rearranged formula.
|
I will assume you are interested in finding $\theta = f^{-1}(x)$ for $x \geq 0$ with the range $0 \leq \theta < \frac{\pi}{2}$.
$$
x^2 = \left(f(\theta)\right)^2 = \sin^2(\theta) \tan(\theta) = \frac{\tan^3(\theta)}{1+\tan^2(\theta)}
$$
Hence $\theta = \arctan(y(x))$, where $y$ is the positive root of $y^3 = x^2 (1 + y^2)$.
Using Cardano's formula:
$$
\theta(x) = \arctan\left( \frac{1}{3} \left(x^2+\frac{\sqrt[3]{2} x^{10/3}}{\sqrt[3]{2 x^4+3 \left(\sqrt{12
x^4+81}+9\right)}}+\frac{\sqrt[3]{2 x^4+3 \left(\sqrt{12 x^4+81}+9\right)}
x^{2/3}}{\sqrt[3]{2}}\right) \right)
$$
|
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|
Something interesting that I found about some numbers - and would like to see if it's known Well I am quite sure it's known (I mean number theory exists thousands of years), warning beforehand, it may look like numerology, but I try not to go to mysticism.
So I was in a bus, and from boredom I started just adding numbers in the next way:
$$1+1=2$$
$$2+2=4$$
$$4+4=8$$
$$....$$
etc. up to $32,768$ (it was quite boring, I can tell... :-)), I didn't have calculator.
And notice that if I keep adding the digits until I get a number from 1 to 10, I get that for example for $8+8=16$, $1+6=7$, now seven steps after this, at $512+512=1024$, which $1+2+4=7$, and again after $7$ steps $32768+32768=65536$, and adding $6+5+5+3+6=10+12+3=25$, $2+5=7$. So this led me to the conjecture that this repetition may occur endlessly.
Now, of course I can program some code that will check for large numbers, but I am tired, long day. So, if this indeed the case (which could be disproved, but even then I would wonder when this repetition stops) then why?
As I said, I am tired, it may make no sense, and I might have done mistakes in my calculations, and it may be trivial.
Either way, if you have some answer, I would like to hear it.
|
A repetition of this sort was bound to happen, and it always happens even under more general circumstances.
First, as others have pointed out, the sum of the base-10 digits of a number $N$ is congruent to $N$ modulo $9$. The reason for this is that $10\equiv 1 \bmod 9$, and so
$$\begin{align*} N &= a_t \cdot 10^t + a_{t-1} \cdot 10^{t-1} +\cdots +a_2 \cdot 10^2 + a_1 \cdot 10 +a_0 \\
& \equiv a_t\cdot 1^t + a_{t-1}\cdot 1^{t-1} +\cdots +a_2\cdot 1^2 + a_1 \cdot 1 +a_0 \bmod 9\\
& \equiv a_t + a_{t-1} +\cdots +a_2 + a_1 +a_0 \bmod 9.
\end{align*}$$
Thus, if you add all the digits of $N$ and obtain $N_1$, then $N\equiv N_1\bmod 9$. If now we add all the digits of $N_1$ and obtain $N_2$, then $N\equiv N_1\equiv N_2 \bmod 9$. In this way we may create a sequence $N>N_1>N_2> \cdots$, and since all $N_i$ are natural numbers, we end up at some $1\leq N_t \leq 9$, such that $N_t\equiv N\bmod 9$, so $N_t$ is simply the remainder of division of $N$ by $9$.
Let $a>1$ be any natural number relatively prime to $3$, whose sum of base-10 digits is $b$, and let $s$ be the order of $b\bmod 9$, i.e., $s$ is the smallest positive number such that $b^s\equiv 1 \bmod 9$. Then:
*
*The sum of the base-10 digits of $a$ is $b\bmod 9$.
*The sum of the base-10 digits of $a^{1+sk}$ is also $b\bmod 9$, for all $k\geq 0$, because
$$a^{1+sk}\equiv b^{1+sk}\equiv b\cdot b^{sk}\equiv b\cdot (b^s)^k\equiv b \cdot 1\equiv b \bmod 9.$$
*For a fixed $t\geq 1$, the sum of the base-10 digits of $a^{t+sk}$ is $b^t\bmod 9$, for all $k\geq 0$, for similar reasons as above.
*For a fixed $t\geq 1$, the sum of the base-10 digits of
$$a^{t+sk} + a^{t+sk}$$
is $2\cdot b^t \bmod 9$, for all $k\geq 0$.
*And more generally, for fixed $r\geq 1$ (relatively prime to $3$) and $t\geq 1$, the sum of the base-10 digits of
$$a^{t+sk} + \cdots + a^{t+sk} = r\cdot a^{t+sk},$$
where we have added $r$ copies of $a^{t+sk}$, is $r\cdot b^t \bmod 9$, for all $k\geq 0$.
Your example is the case where $a=2$, $b=2$, $s=6$, $t=3$ and $r=2$. According to the formula above, the sum of the digits must be
$$r\cdot b^t \equiv 2\cdot 2^3\equiv 7 \bmod 9.$$
But any other choice works just as well. Pick $a=11$, $b\equiv a\equiv 2 \bmod 9$, $s=6$, $t=2$ and $r=5$. Then, the sum of the digits of the numbers
$$11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k},$$
for all $k\geq 0$, is congruent to
$$r\cdot b^t \equiv 5\cdot 2^2\equiv 2 \bmod 9.$$
For instance:
*
*$11^{2}+11^{2}+11^{2}+11^{2}+11^{2}=605,$ and $6+5=11$ and $1+1=2$.
*$11^{8}+11^{8}+11^{8}+11^{8}+11^{8}=1071794405,$ and
$$1+0+7+1+7+9+4+4+5=38$$ and $3+8=11$, and $1+1=2$.
*Etc.
|
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|
Minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ I am attempting to compute the minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ over $\mathbb Q$. So far, my reasoning is as follows:
The Galois conjugates of $2^{1/3}$ are $2^{1/3} e^{2\pi i/3}$ and $2^{1/3} e^{4\pi i /3}$. We have $4^{1/3} = 2^{2/3}$, so the image of $4^{1/3}$ under an automorphism $\sigma$ fixing $\mathbb Q$ is determined by the image of $2^{1/3}$: it must equal the square of $\sigma(2^{1/3})$. Therefore, the Galois conjugates of $1 + 2^{1/3} + 4^{1/3}$ are $1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3}$ and $1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}$. Therefore, the minimal polynomial is $(x-a)(x-b)(x-c)$, where
$$\begin{align*}
a&=1 + 2^{1/3} + 4^{1/3},\\
b&=1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3},\text{ and}\\
c&=1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}.
\end{align*}$$
However, this polynomial does not seem to have coefficients in $\mathbb Q$! What am I doing wrong?
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By expanding and using the relation $1+e^{2\pi i/3}+e^{4\pi i/3}=0$ heavily I got that
$$
(x-a)(x-b)(x-c)=x^3-3x^2-3x-1.
$$
Looks like rational coefficients to me.
Another way of seeing this is to compute
$$
(a-1)^3=(2^{1/3}+4^{1/3})^3=2+3\cdot 2^{4/3}+3\cdot 2^{5/3}+4=6+6(2^{1/3}+4^{1/3})=6+6(a-1)=6a.
$$
Hence
$$
0=a^3-3a^2+3a-1-6a=a^3-3a^2-3a-1.
$$
|
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|
How to prove $\lim\limits_{n \to \infty} (1+\frac1n)^n = e$?
How to prove the following limit?
$$\lim_{n \to \infty} (1+1/n)^n = e$$
I can only observe that the limit should be a very large number!
Thanks.
|
Actually, the way things work out in mathematics usually is that we only prove that $x_n = (1+1/n)^n$ is a convergent sequence, and we define its limit to be $e$. Some people use other definitions for $e$ and show that it is equivalent to this definition, but there are many ways to do this that are logically equivalent.
(Just for the record, you don't "prove" limits, you compute them.)
One way to show that this sequence is convergent is to show that it is increasing and bounded above. This is a bit technical but works out quite well. To prove that $x_n$ is increasing, we show that $x_{n+1} \ge x_n$, which is equivalent to $x_{n+1}/x_n \ge 1$ :
\begin{align}
\frac{x_{n+1}}{x_n} & = \frac{ \left( 1 + \frac 1{n+1} \right)^{n+1} }{\left( 1 + \frac 1n \right)^n } = \left( 1 + \frac 1{n+1} \right) \left( \frac{ 1 + \frac 1{n+1} }{1 + \frac 1n} \right)^n = \left( \frac {n+2}{n+1} \right) \left( \frac{(n+2)n}{(n+1)^2} \right)^n \\
& = \left( \frac{n+2}{n+1} \right) \left( 1 + \frac{(n+2)n - (n+1)^2 }{(n+1)^2} \right)^n \\
& = \left( \frac{n+2}{n+1} \right) \left( 1 - \frac{1}{(n+1)^2} \right)^n \\
& \ge \left( \frac{n+2}{n+1} \right) \left( 1 - \frac n{(n+1)^2} \right) \\
& \ge \left( \frac{n+2}{n+1} \right) \left( \frac{n^2 + n + 1}{n^2 + 2n + 1} \right) = \frac{n^3 + 3n^2 + 3n + 2}{n^3 + 3n^2 + 3n + 1} \ge 1.
\end{align}
The first inequality holds by Bernoulli's inequality ( I'm referring to $(1+x)^n \ge 1+nx$ for $x \ge -1$).
To prove that $x_n$ is bounded above, we use the binomial theorem :
\begin{align}
\left(1+\frac 1n\right)^n & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac 1{n^k} = \sum_{k=0}^n \frac 1{k!} \frac{n!}{n^k(n-k)!} \\
& \le 1 + \sum_{k=1}^n \frac 1{k!} \le 1 + \sum_{k=1}^n \frac 1{2^{k-1}} = 1 + \frac{1- \frac 1{2^n}}{ 1- \frac 12} < 1 + 2 = 3. \\
\end{align}
Note that I used the fact that $k! \ge 2^{k-1}$, which implies $\frac 1{2^{k-1}} \ge \frac 1{k!}$. (To see this, just write $k! = k \times \dots \times 2 \times 1 \ge 2 \times \dots \times 2 \times 1 = 2^{k-1}$.)
Another way would be to show that this sequence converges to $\sum_{k=0}^{\infty} \frac 1{k!}$ (which is another definition for $e$) by expanding $(1+1/n)^n$ with the binomial theorem.
Hope that helps,
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|
Approximate $\int_{0}^{\infty} \frac{\text{d} x}{1 + x^4}$ Now, I have been given this integral. And need to approximate it. My first idea was to use a Taylor series, but this series explodes, as x reaches infinity.
Does anyone know how to approximate improper integrals, (and this one in particular)?
I know I can use contour-integration to evaluate it exact, but I want to estimate it.
Someone mentioned something about a Taylor expansion at infinity, but alas I have not learned about this.
$$ \int_{0}^{\infty} \frac{\text{d} x}{1 + x^4} $$
|
One way is to split integration range at $x=1$ and use geometric series approximation:
$$\begin{eqnarray}
\int_0^\infty \frac{\mathrm{d} x}{1+x^4} &=& \int_0^1 \frac{\mathrm{d} x}{1+x^4} +\int_1^\infty \frac{\mathrm{d} x}{1+x^4} \stackrel{x -> 1/x \text{ in second}}{=}
\\ &=& \int_0^1 \frac{\mathrm{d} x}{1+x^4} +\int_0^1 \frac{x^2 \mathrm{d} x}{1+x^4} =
\int_0^1 \frac{1+x^2}{1+x^4}\mathrm{d} x \\
&=& \int_0^1 \left( 1+ x^2-x^4 - x^6+x^8 + \cdots\right) \mathrm{d}x \\
&=& 1 + \frac{1}{3}-\frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \cdots = \frac{347}{315} + \cdots
\end{eqnarray}
$$
Now $347/315$ is approximately $1.10159$, while the exact answer is $\frac{\pi}{2\sqrt{2}} \approx 1.11072$.
|
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|
Solving $y'' - xy'+(3x-2)y=0$ using power series I am trying to solve this equation using the series $$\sum_0^\infty a_nx^n$$
$$y'' - xy'+(3x-2)y=0$$
How to do that? I mean that I can replace the variables using the series but then I cannot add this thing cause the limits of the sums are not the same. Maybe I am doing something wrong here. I tried to make all sums start from $0$ with $x^{n+1}%$. This will leave $2a_2 - 2a_0 + \sum\dots = 0$ and I don't know what to do. I can't just say that $2a_2 = 2a_0 = 0$ cause it may be $2a_2 - 2a_0 = 0$. Well I am really confused.
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Let
$$
y=\sum\limits_{n=0}^\infty a_n x^n
$$
then by a straightforward computation we get
$$
y''-xy'+(3x-2)y=\sum\limits_{n=2}^\infty n(n-1)a_n x^{n-2}-x\sum\limits_{n=1}^\infty n a_n x^{n-1}+3x\sum\limits_{n=0}^\infty a_n x^n-2\sum\limits_{n=0}^\infty a_n x^n=
$$
$$
\sum\limits_{n=0}^\infty (n+2)(n+1)a_{n+2} x^n-\sum\limits_{n=0}^\infty n a_n x^n+\sum\limits_{n=0}^\infty 3a_n x^{n+1}-2\sum\limits_{n=0}^\infty a_n x^n
$$
Now note that if we take by definition $a_{-1}=0$ we get
$$
\sum\limits_{n=0}^\infty 3a_n x^{n+1}=\sum\limits_{n=0}^\infty 3a_{n-1} x^n
$$
So
$$
y''-xy'+(3x-2)y=
\sum\limits_{n=0}^\infty (n+2)(n+1)a_{n+2} x^n-\sum\limits_{n=0}^\infty n a_n x^n+\sum\limits_{n=0}^\infty 3a_{n-1} x^n-2\sum\limits_{n=0}^\infty a_n x^n=
$$
$$
\sum\limits_{n=0}^\infty\left((n+2)(n+1)a_{n+2}-(n+2)a_n+3a_{n-1}\right)x^n
$$
And we get a recurrence equation equation for determining $a_n$:
$$
(n+2)(n+1)a_{n+2}-(n+2) a_n+3a_{n-1}=0
$$
where $a_{-1}=0$ and $a_0=y(0)$, $a_1=y'(0)$. Now taking $n=0$ you can determine $a_2$, taking $n=1$ you can determine $a_3$ et cetera...
Honestly, I don't think that this recurrence equations have an explicit solution. Even Mathematica doesn't have any idea
On the other hand, Mathematica gives an explicit solution for the original differential equation:
$$
y(x)=\frac{e^{3 x}}{96
\sqrt{(x-6)^2}}
$$
$$\begin{multline}\Biggl(\sqrt{2 \pi } c_2 \left(x^6-36 x^5+519
x^4-3816 x^3+15009 x^2-29772 x+23115\right)(x-6)^2
\text{erfi}\left(\frac{\sqrt{(x-6)^2}}{\sqrt{2}}\right)\\
+2 \sqrt{(x-6)^2} \left(384
\sqrt{2} c_1 \left(x^7-42 x^6+735 x^5-6930 x^4+37905 x^3-119826 x^2+201747
x-138690\right)-c_2 e^{\frac{1}{2} (x-6)^2} \left(x^6-36 x^5+520 x^4-3840
x^3+15207 x^2-30420 x+23820\right)\right)\Biggr)
\end{multline}
$$
After looking at this monster I think one can not get an explicit solution for that recurrence equation.
|
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|
Groups of units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ On page 230 of Dummit and Foote's Abstract Algebra, they say: the units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ are determined by the integers $a,b$ with $a^2+ab+b^2=\pm1$ i.e. with $(2a+b)^2+3b^2=4$, from which is is easy to see the group of units is a group of order $6$ given by $\{\pm1,\pm\rho,\pm\rho^2\}$ where $\rho=\frac{-1+\sqrt{-3}}{2}$.
First, why change the characterization of unit from integers solutions of $a^2+ab+b^2=\pm1$ to integers solutions of $(2a+b)^2+3b^2=4$? How did they arrive at their answer?
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The reason you may want to change it from $a^2+ab+b^2=\pm 1$ to $(2a+b)^2+3b^2 = \pm 4$ is because the latter is a sum of squares, so this immediately cuts down on the possibilities: for one thing, you can tell that the answer must be $4$ and not $-4$ (sum of squares), that you must have $|2a+b|\leq 2$ and $3b^2\leq 4$; so that $b$ must be either $0$, $1$, or $-1$, etc. Whereas in $a^2+ab+b^2 = \pm 1$, $ab$ may be negative, which makes searching for solutions somewhat more complicated. (Not much, but things are not so quickly evident).
To go from $a^2+ab+b^2 = \pm 1$ to $(2a+b)^2 + 3b^2=\pm 4$, multiply by $4$ and complete the square:
$$\begin{align*}
a^2 + ab + b^2 &= \pm 1\\
4a^2 + 4ab + 4b^2 &= \pm 4\\
(2a)^2 + 2(2a)b + b^2 +3b^2 &=\pm 4\\
(2a+b)^2 + 3b^2 &= \pm 4
\end{align*}$$
Alternatively, start by completing the square and then clear denominators:
$$\begin{align*}
a^2 + ab + b^2 &= \pm1\\
\left(a+\frac{1}{2}b\right)^2 + \frac{3}{4}b^2 &=\pm 1\\
\left(2a+b\right)^2 + 3b^2 &=\pm 4.
\end{align*}$$
|
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|
Rigorous proof of an infinite product. I'll give a proof of the following expansion:
$$\frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i}$$
$${\sin x} = 2 \cos \frac{x}{2}\sin \frac{x}{2}$$
$${\sin x} = 2^2 \cos \frac{x}{2}\cos \frac{x}{4}\sin \frac{x}{4}$$
$$ {\sin x} = 2^3 \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \sin\frac{x}{8} $$
$$ {\sin x} = 2^k \cos \frac{x}{2} \cos \frac{x}{4} \cdots\cos \frac{x}{2^k} \sin\frac{x}{2^k} $$
$${\sin x} = 2^k \sin\frac{x}{2^k} \prod_{i=1}^{k} \cos \frac{x}{2^i} $$
$$ \frac{\sin x}{x} = \frac{\sin \displaystyle \frac{x}{2^k}}{\displaystyle \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i} $$
Let $k \to \infty$, then $\displaystyle \frac{x}{2^k} \to 0$
$$ \frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i} $$
Is there any extra observation needed to make the proof complete?
I guess since $$\cos \frac{x}{2^i} \to 1$$
the convergence is not at stake.
|
Your expression
$$ \frac{\sin x}{x} = \frac{\sin \frac{x}{2^k}}{ \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i} $$
is correct. Maybe we should separate out the very special case $x=0$, and from then on assume that $x\ne 0$. For $x=0$, $\frac{\sin x}{x}$ is formally undefined, but it is natural to set it equal to $1$. Then the formula works. If we are feeling in a very formal mood, we should prove the correctness of your expression by induction on $k$. However, I think that would be overkill.
We want to find the limit of the expression on the right as $k\to \infty$. As you observed, $\frac{x}{2^k}\to 0$ as $k\to \infty$. That certainly does not need proof. However, it is necessary to observe that since
$$\lim_{t=0}\frac{\sin t}{t}=1,$$
we have
$$\lim_{k\to\infty} \frac{\sin \frac{x}{2^k}}{ \frac{x}{2^k}}=1.$$
There will unfortunately be some special cases that require special treatment. We deal first with the "general" case when $x$ is not an integer multiple of $\pi$.
Then your expression can be rewritten as
$$\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}=\prod_{i=1}^{k} \cos \frac{x}{2^i}.$$
Since
$$\lim_{k\to\infty}\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}$$
exists and is equal to $\frac{\sin x}{x}$, we conclude that
$$\lim_{k\to\infty} \prod_{i=1}^{k} \cos \frac{x}{2^i}$$
also exists and is equal to $\frac{\sin x}{x}$.
If $x\ne 0$ is an integer multiple of $\pi$, we have to choose $k$ large enough so that $\sin\frac{x}{2^k}\ne 0$, else when we rewrite your expression, we might be dividing by $0$. Minor point. For such $x\ne 0$, $\frac{\sin x}{x}=0$, and one of the cosines is $0$. So, in that case also, apart from a technicality discussed below, the formula looks correct.
Technical remark: In the formal definition of an infinite product, we say that
$$\prod_{i=1}^\infty a_i$$
converges if
$$\lim_{k\to\infty}\prod_{i=1}^k a_i$$
exists and is not equal to $0$. So technically when $x$ is an integer multiple of $\pi$, the infinite product does not converge!
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|
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
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$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0 \Rightarrow 12 \cdot 2^{2x} - 35 \cdot 2^x \cdot 3^x + 18 \cdot 3^{2x} = 0 \Rightarrow$
$\Rightarrow 12 \cdot \left(\frac{2}{3}\right)^{2x}-35 \cdot \left(\frac{2}{3}\right)^{x}+18=0 $
Now make substitution : $\left(\frac{2}{3}\right)^{x} = t$ , and solve quadratic equation .
|
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|
Derive a formula to find the number of trailing zeroes in $n!$
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
I know that I have to find the number of factors of $5$'s, $25$'s, $125$'s etc. in order to do this. But how can you derive such a formula for any number $n$?
|
The number of trailing zeroes in $n!$ is the exponent of $5$ in the prime factorization of $n!$, which by the de Polignac's formula$^1$ is given by
$$e_5(n!)=\sum_{i= 1}^{\left\lfloor \log n/\log 5\right\rfloor} \left \lfloor \frac{n}{5^i} \right \rfloor.\tag{1}$$
Added. By the same de Polignac's formula the exponent of $2$ in the prime factorization of $n!$ is
$$e_2(n!)=\sum_{i= 1}^{\left\lfloor\log n/\log 2\right\rfloor} \left \lfloor \frac{n}{2^i} \right \rfloor.\tag{2}$$
Edited. For every $n$ there exists $m$ such that $n!=2^{e_2(n!)}\cdot 5^{e_5(n!)}m=(2^{e_2(n!)-e_5(n!)}m)10^{e_5(n!)}$, which proves that the number of trailing zeroes in $n!$ is the exponent of $5$ in the prime factorization of $n!$ .
Example: $n=50$. The exponent of $2$ in the prime factorization of $50!$ is
$$\begin{align}
e_2(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{2^{i}}\right\rfloor &= \left\lfloor \dfrac{50}{2}\right\rfloor+\left\lfloor \dfrac{50}{2^{2}} \right\rfloor + \left\lfloor \dfrac{50}{2^{3}} \right\rfloor + \left\lfloor\dfrac{50}{2^{4}}\right\rfloor +\left\lfloor \dfrac{50}{2^{5}}\right\rfloor
\\
&=25+12+6+3+1
\\
&=47,
\end{align}$$
and the exponent of $5$ is $$e_5(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{5^{i}}\right\rfloor=\left\lfloor \dfrac{50}{5}\right\rfloor +\left\lfloor \dfrac{50}{5^{2}}\right\rfloor =10+2=12.$$
So, the number of trailing zeroes in $50!=2^{47}5^{12}m=(2^{35}m)10^{12}$ is $12$.
$^1$ For every integer $n$ the exponent of the prime $p$ in the prime
factorization of $n!$ equals $$\displaystyle\sum_{i= 1}^{\left\lfloor \log n/\log p\right\rfloor}\left\lfloor \frac{n}{p^{i}}
\right\rfloor .$$
This exponent is obtained by adding to the numbers between $1$ and $n$ which are
divisible by $p$ the number of those divisible by $p^{2}$, then the number
of those divisible by $p^{3}$, and so on. The process terminates at the
greatest power $p^{i}\leq n$.
|
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Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$ How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find:
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$
For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$
Thus, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$
|
+1 for your handwriting :-)
*
*$\frac{f(x)}{g(x)} \ge 0$ is not so different from $f(x)\times g(x) \ge 0$ (except for zeros of $g$.)
*With $(x^2-1) = (x-1)(x+1)$ etc, your problem reduces to the form of $(x-a)(x-b)(x-c)(x-d)...(x-z) \ge 0$
Edit: oops I only read the hand-written part! Anyways thanks to the monotonicity of $x^5-1$ etc, you can still use similar argument.
Edit 2: Plot it in google.
|
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|
Solving LP from tableau $$\begin{array}{cccccc}
& x1 & x2 & x3 & x4& x5 \\
-4& 2 & 0& -2 & 0& 3\\
3 & 1 & 0 & -1& 1 & 3\\
2 &0& 1 & 0& 0.5 & 2\\
\end{array}$$
When I learned about solving LP represented by the tableau in class, I thought I need to select the column that has negative value at the top. According to that, x3 is the only column that has -2 value. However, it's subsequent values -1 in row 1 and 0 in row 2 are neither positive, which I need it to pivot the row.
Am I missing something here?
The top row with z =-4, x1=2, x3=-2 is the cost function
|
$$\begin{array}{cccccc}
& x1 & x2 & x3 & x4& x5 \\
-4& 2 & 0& -2 & 0& 3\\
3 & 1 & 0 & -1& 1 & 3\\
2 &0& 1 & 0& 0.5 & 2\\
\end{array}$$
To begin with a bfs
$$\begin{array}{cccccc}
& x1 & x2 & x3 & x4 & x5 \\
-10& 0 & 0 & 0 & -2 & -3 \\
3 & 1 & 0 & -1 & 1 & 3 \\
2 & 0 & 1 & 0 &0.5 & 2 \\
\end{array}$$
Suppose we pick column4, then row 2 is most profitable
$$\begin{array}{cccccc}
& x1 & x2 & x3 & x4 & x5 \\
-2 & 0 & 4 & 0 & 0 & 5 \\
-1 & 1 & -2 & -1 & 0 & -1\\
4 & 0 & 2 & 0 &1 & 4 \\
\end{array}$$
Now all cost are positive and we are done. Optimal value of z is -2.
=====
Did you miss your objective functions? And you need to find a basis(basic feasible solution) first.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
|
Here is a hint: Use the identity $(a^3-b^3)=(a-b)\cdot(a^2+ab+b^2)$ with $a$, $b$ being suitable cube roots. Otherwise, the method is similar to the one you tried.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$
Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something?
|
Here is an alternate approach.
Using de Moivre's Formula, we get for $\theta=\frac{\pi}{5}$
$$
0=(\cos(\theta)+i\sin(\theta))^5+1\tag{1}
$$
Looking at the real part of $(1)$ yields
$$
\begin{align}
0
&=\color{red}{\cos^5(\theta)}\color{green}{-10\cos^3(\theta)\sin^2(\theta)}\color{blue}{+5\cos(\theta)\sin^4(\theta)}+1\\
&=\color{red}{\cos^5(\theta)}\color{green}{-10\cos^3(\theta)+10\cos^5(\theta)}\color{blue}{+5\cos(\theta)-10\cos^3(\theta)+5\cos^5(\theta)}+1\\
&=16\cos^5(\theta)-20\cos^3(\theta)+5\cos(\theta)+1\\
&=(4\cos^2(\theta)-2\cos(\theta)-1)^2(\cos(\theta)+1)\tag{2}
\end{align}
$$
Since $\cos(\theta)=-1$ only when $\theta$ is an odd multiple of $\pi$, we are left with
$$
4\cos^2\left(\frac{\pi}{5}\right)-2\cos\left(\frac{\pi}{5}\right)-1=0\tag{3}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/113466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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|
A problem about parametric integral How to solve the following integral.
$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|<1$
|
Note that the integrand always changes sign at $x=\frac\pi2$ for $\theta\ne0$.
In fact, this is an even, nonpositive function,
since $\cos(\pi-x)=-\cos x$ and since, for $r=\theta\cos x$,
$|r|<1$ and $\ln(1+r)+$ $\ln(1-r)=$ $\ln(1-r^2)<0$ $\implies$
$$
\eqalign{
\int_0^\pi~\ln\big(1+\theta\,\cos\,x\big)\;dx &=
\int_0^\frac\pi2~\ln\big(1+\theta\cos x\big)\;dx +
\int_\frac\pi2^\pi~\ln\big(1+\theta\cos x\big)\;dx \\ &=
\int_0^\frac\pi2~\ln\big(1-\theta^2\cos^2 x\big)\;dx
\le 0
\,,
}
$$
with equality iff $\theta=0$.
Out of perverse curiosity, let us define, slightly more generally
(substituting $a$ for 1, $b$ for $\theta$ and $\theta$ for $x$)
$$
I(a,b)=\int_0^\pi\ln\big(a+b\cos\theta\big)\;d\theta.
$$
Then
$$
\frac{\partial I}{\partial b}
=\int_0^\pi \frac{\cos\theta\,d\theta}{a+b\cos\theta}
=\frac{\pi}{b}-\frac{a}b\int_0^\pi\frac{d\theta}{a+b\cos\theta}
\qquad
\text{since}
\qquad
\frac{b\cos\theta}{a+b\cos\theta}
=1-\frac{a }{a+b\cos\theta}
.
$$
But (thanks to Ragib Zaman):
$$
\eqalign{\frac\pi{a} - \frac{b}{a} \, \frac{\partial I}{\partial b}
& = \int_0^\pi \frac{d\theta}{a+b\cos{\theta}}
\\& = \int_0^\pi \frac{d\theta}
{a\left(\sin^2\frac\theta2+\cos^2\frac\theta2\right)
+b\left(\cos^2\frac\theta2-\sin^2\frac\theta2\right)}
\\& = \int_0^\pi \frac{d\theta}
{(a-b)\sin^2\frac\theta2+(a+b)\cos^2\frac\theta2}
\\& = \int_0^\pi \frac{\sec^2\frac\theta2\;d\theta}
{(a+b)+(a-b)\tan^2\frac\theta2}
\\& = 2\int_0^\infty \frac{dt}{(a+b)+(a-b)t^2}
\\& = 2\int_0^\infty \frac{dt}{(\sqrt{a+b})^2+(\sqrt{a-b})^2t^2}
\\& = \frac{2}{{\sqrt{a^2-b^2}}}~
\left.\tan^{-1}
\left(
\frac{\sqrt{a-b}}{\sqrt{a+b}}~t
\right)
\right|_{0}^{\infty} \\& = \frac{\pi}{\sqrt{a^2-b^2}}
}
$$
so that
$$
\frac{\partial I}{\partial b} =
\frac\pi{b} \left( 1-\frac{a}{\sqrt{a^2-b^2}} \right)
$$
or
$$
I(a,b)
= \pi \int \frac{db}{b}
- \pi a \int \frac{db}{b \sqrt{a^2-b^2}}
\,.
$$
For $a>|b|>0$, we can use the substitution
$b=a\sin\phi,~db=a\cos\phi\,d\phi$ to continue thus:
$$
\eqalign{
I(a,b)&=\pi\ln|b| - \pi a \int \frac{d\phi}{\sin\phi}
\\ &=\pi\ln|b| - \pi a \int \csc\phi ~d\phi
\\ &=\pi\ln|b| + \pi a ~\ln\, \big| \csc\phi + \cot\phi \big| + c
\\ &=\pi\ln|b| + \pi a ~\ln\, \left| \frac{1+\cos\phi}{\sin\phi} \right| + c
\\ &=\pi\ln|b| + \pi a ~\ln\, \left|
\frac{a}{b} +\sqrt{\left(
\frac{a}{b} \right)^2-1}
\right| + c
\,.
}
$$
Using $I(a,0)=\pi\,\ln a$,
we find that $c$ depends on a limit
which exists and is $\ln2$ iff $a=1$,
$$
\frac{\pi\ln a-c}{a}
=\lim_{b\rightarrow0}\,\ln
\left|
\frac{
\frac{a}{b} +\sqrt{\left(
\frac{a}{b} \right)^2-1}
}{|b|^{-1/a}}
\right|
=\lim_{b\rightarrow0}\,\ln
\left|
b^{1/a-1}
\left(
a+\sqrt{a^2-b^2}
\right)
\right|
$$
in which case $c=-\pi\ln2$.
So for our original problem,
$$
\int_0^\pi~\ln\big(1+\theta\,\cos\,x\big)\;dx=I(1,\theta)
= \pi \ln\frac{1+\sqrt{1-\theta^2}}{2}
\,.
$$
As already noted, this exists and
is nonpositive for $|\theta|<1$.
On the other hand we see from the RHS above
that the integral is bounded below by
$-\pi\ln2\approx-2.17758609030360$.
Here is a plot of the solution using sage,
with the factor of $\pi$ removed (in blue),
and a comparable function (in red)
from an earlier erroneous draft:
var('t')
#assume(t != 0)
G = plot(log( (1 + sqrt(1-t^2)) / 2 ), (t, -1, 1), color='blue')
G+= text('log((1 + sqrt(1-t^2)) / 2)', (-.6,-.65), color='blue')
G+= plot( 1 - arcsin(t)/t , (t, -1, 1), color='red')
G+= text( '1 - arcsin(t)/t' , (.85,-.06), color='red')
G#.show(aspect_ratio=1)
In particular, the graph has a minimum
of $-\ln2\approx-0.693147180559945$
at $\theta=\pm1$.
To check the endpoint, I computed an approximate numerical integral, which agrees with the above (the tuple gives the integral and error estimates):
numerical_integral(2*log(sin(x)), 0, pi/2)
$\left(-2.17758608788, 1.09713268507 \times 10^{-06}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/114401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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|
How were trigonometrical functions of $\dfrac{2\pi}{17}$ calculated? I know they were calculated by Gauss, but how? Is there a method for calculating them?
|
Let $\omega = e^{2 \pi i/17}$. Since $3$ is a primitive root mod $17$, i.e. a generator of the multiplicative group of nonzero integers mod $17$, write $R_j = \omega^{3^j}$ for $j = 0, 1, \ldots, 15$. These and $1$ are the $17$'th roots of unity.
For $2^j \le i < 2^{j+1}$ let $x_i = \sum_{k \equiv i \mod 2^j} R_k$. Then for $2^j \le i < 2^j + 2^{j-1}$, $x_i + x_{i+2^{j-1}}$ and $x_i x_{i+2^{j-1}}$ can be expressed in terms of the previous $x_k$'s, which allows $x_i$ and $x_{i+2^{j-1}}$ to be obtained by solving a quadratic equation: if $x + y = c$ and $xy = d \ne 0$, then $x$ and $y$ are the roots of $z^2 - c z + d$.
Thus:
$$x_1 = \sum_{j=0}^{15} R_j = -1$$
$x_2 = R_0 + R_2 + \ldots + R_{14}$ and $x_3 = R_1 + R_3 + \ldots R_{15}$ satisfy
$x_2 + x_3 = x_1 = -1$ and $x_2 x_3 = 4 x_1 = -4$.
$x_4 = R_0 + R_4 + R_8 + R_{12}$ and $x_6 = R_2 + R_6 + R_{10} + R_{14}$ satisfy
$x_4 + x_6 = x_2$, $x_4 x_6 = -1$
$x_5 = R_5 + R_9 + R_{13} + R_1$ and $x_7 = R_7 + R_{11} + R_{15} + R_3$ satisfy
$x_5 + x_7 = x_3$, $x_5 x_7 = -1$
$x_8 = R_0 + R_8$ and $x_{12} = R_4 + R_{12}$ satisfy
$x_8 + x_{12} = x_4$, $x_8 x_{12} = x_5$
$x_{16} = R_0 = \omega$ and $x_{24} = R_8$ satisfy $x_{16} + x_{24} = x_8$, $x_{16} x_{24} = 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/115023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$. Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$.
Show that $$ \frac{x^2+y^2+1}{xy}= 3 \;.$$
I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks
|
$x$ divides $x^2 + y^2 + 1$ implies $y^2 = ax - 1$.
Then $y$ divides $x(x+a)$
Case 1 - $y$ divides $x$, so $x = by$.
$$1/b + b + 1/(by^2) = k$$.
$$b=x=y=1$$
$$k=3$$
Or, $b=x=2$, $y=1$, $k=3$
Case 2 - $y$ divides $x+a$, so $y = -a \text{ mod } x$, $y^2 = 1 \text{ mod } x$.
$$x^2 + y^2 +1 = 2 \text{ mod } x$$
$x$ is $1$ or $2$. (Otherwise, $x$ does not divide the equation)
If $x = 1$, $x^2 + y^2 +1 = 2 \text{ mod } y$, $y$ is $1$ or $2$
If $x = 2$, $y$ is $1$ or $5$.
Solutions: $(1,1),(2,1),(5,2)$, $k$ is always $3$
(Tough to type on the phone)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/115272",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "19",
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|
How do I integrate this? How do I integrate $\displaystyle\int (x^2 + 2)\sqrt{1-x} \; dx$ ?
I have feeling substitution might be used, but I just can't put my finger on it...
Thank you.
|
Note that $$x^2+2=(1-x)^2-2(1-x)+3,$$ which implies that
$$(x^2+2)\sqrt{1-x}=(1-x)^\frac{5}{2}-2(1-x)^{\frac{3}{2}}+3(1-x)^{\frac{1}{2}}.$$
Therefore, let $u=1-x$, we have $dx=-du$, which implies that
$$\int (x^2+2)\sqrt{1-x}dx=-\int u^\frac{5}{2}-2u^{\frac{3}{2}}+3u^{\frac{1}{2}}du$$
$$=-\frac{2}{7}u^{\frac{7}{2}}+\frac{4}{5}u^{\frac{5}{2}}+2u^{\frac{3}{2}}+C$$
$$=-\frac{2}{7}(1-x)^{\frac{7}{2}}+\frac{4}{5}(1-x)^{\frac{5}{2}}+2(1-x)^{\frac{3}{2}}+C.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/116627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Divisibility of integers Let $n > 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?
In general, for a fixed positive integer $a > 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?
|
As @AQP said, if $n$ is even then $3\mid 2^n-1$ so $2^n-1\nmid 3^n-1$.
If $n=2k-1$
then
$2^n-1 \equiv 1 \pmod{3}$ so $2^n-1$ is a quadratic residue mod 3.
$3(3^n-1)=3^{2k}-3$ so $2^n-1 \mid 3^n-1$ would require that $3^{2k}\equiv 3 \pmod{2^n-1}
$,
i.e. that 3 is a quadratic residue mod $2^n-1$.
But $2^n-1\equiv 3 \pmod{4}$ so is divisible by an odd number of primes $p\equiv 3\pmod{4}$. By quadratic reciprocity 3 cannot be a quadratic residue mod $2^n-1$, hence $2^n-1\nmid 3^n-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/116978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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|
powers of $\frac{1+\sqrt a}2$ For any a which is not a perfect square, let $x=\frac{1+\sqrt a}2$.
$x^n$ can be written uniquely as $b_nx+c_n$, where b and c are rational.
Apart from $a=0, a=1, a= 1 \pm 2^m$ for $m>2$, are there any other values of $a$ for which $b$ or $c$ is an integer for infinitely many $n$? If not, are there any upper bounds on the values of n for which $b$ or $c$ is an integer?
e.g for $a=7$
$\\b \ c\\
0 \ 1\\
1 \ 0\\
1 \ \frac{3}2\\
\frac{5}2 \ \frac{3}2\\
4 \ \frac{15}2\\
\frac{23}2 \ 6\\
\frac{35}2 \ \frac{69}4$
$b_n=b_{n-1}+c_{n-1}$ and $c_n=\frac{a-1}4b_{n-1}$
|
If $a \equiv 5$ ($\bmod 8$) then this happens infinitely many times. This follows from the relation $x^2 = x + \tfrac{a-1}{4}$ and $\tfrac{a-1}{4}$ is odd. Suppose $x^n = b_nx + c_n$ for integers $b_n,c_n$ then
$$
x^{n+1} = (b_n+c_n)x + b_n\frac{a-1}{4}
$$
So $b_n \equiv 1, 1, 0, 1, 1, 0, \dotsc$ ($\bmod 2$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/119981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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|
How I can find the value of $abc$ using the given equations? If I have been given the value of
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried by squaring the equations.
But could not get it.
Thanks in advance.
|
If $a,b,c$ solve the equation $x^3+mx^2+nx+p=0$ then you know that $S_3+mS_2+nS_1+3p=0$, where $S_i=a^i+b^i+c^i$. From the sum you find who $m$ is. The expression of $n$ is $ab+bc+ca$. You can find $p$ substituting all the values in the equation. Then you can find the product, which is $-p$ and eventually solve the equation.
Alternatively, you can find the product using the formula
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/120536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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|
The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\ 2x-3 = -(2x-3)$ when $2x-3$ $\geqslant$ $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$ $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.
|
$$\left | 2x-3 \right | = -(2x-3)$$
$let$, $t= 2x-3$
$$\left | t \right | = -t$$
$$t=<0$$
$$2x-3=<0$$
$$x \in \left(-\infty, \frac{3}{2}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121240",
"timestamp": "2023-03-29T00:00:00",
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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.