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 #6
avatar+18829 
0

Compute the sum    2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5)+...

 

see link: http://web2.0calc.com/questions/algebra_47009

 

In General:

\(\begin{array}{lcll} s = \dfrac{1}{1 \cdot 2 } + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4 } + \dfrac{1}{4 \cdot 5 } + \cdots \ + \dfrac{1}{n \cdot (n+1)} + \cdots =\ \frac{1}{1!0!}\cdot \frac{1}{1} = 1 \\\\ s =\mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \frac{1}{2!0!}\cdot \frac{1}{2} } = \frac{1}{4} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3)} + \cdots =\ \frac{1}{3!0!}\cdot \frac{1}{3} = \frac{1}{18} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3 \cdot (n+4) } + \cdots =\ \frac{1}{4!0!}\cdot \frac{1}{4} = \frac{1}{96} \\\\ \ldots \\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdots } + \dfrac{1}{2 \cdot 3 \cdot 4 \cdots } + \dfrac{1}{3 \cdot 4 \cdot 5 \cdots} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)\cdot~\cdots ~\cdot (n+m) } + \cdots =\ \frac{1}{m!0!}\cdot \frac{1}{m} = \frac{1}{m\cdot m!} = \frac{1}{(m+1)!-m!} \\\\ \end{array} \\ \)

 

 

 

 

laugh

heureka 5 hours ago
 #3
avatar+18829 
+1

A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system.

How many lattice points (points with integer coordinates) lie within or on this circle?

 

A Calculation of the Number of Lattice Points within or on the circle:

 

Let \( \lfloor x \rfloor \) be the largest integer equal to or less than x.

 

Example:
\(\lfloor 3.53553390593 \rfloor = 3\)
\(\lfloor -3.53553390593 \rfloor = -4\)

 

 

 

Noted by Gauss:

Let r  radius of the circle = 5

Let \(x = r^2\)

 

\(\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\lfloor \sqrt{x} \rfloor + 4 \lfloor \sqrt{\frac{x}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{x}{2}} \rfloor + 1 }^{\lfloor \sqrt{x} \rfloor} \lfloor \sqrt{x-y_1^2} \rfloor \qquad & | \quad x = r^2 = 5^2 \\\\ &=& 1 + 4\lfloor \sqrt{5^2} \rfloor + 4 \lfloor \sqrt{\frac{5^2}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{5^2}{2}} \rfloor + 1 }^{\lfloor \sqrt{5^2} \rfloor} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 3 + 1 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 4 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( \lfloor \sqrt{5^2-4^2} \rfloor +\lfloor \sqrt{5^2-5^2} \rfloor \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( 3 + 0 \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 24 \\\\ &=& 1 + 20 + 36 + 24 \\ &\mathbf{=} & \mathbf{81} \\ \hline \end{array}\)

 

81 lattice points (points with integer coordinates) lie within or on this circle with radius 5.

 

Example:
\(r = 0 \ldots 20\)

 

Number of lattice points in circle:

\(\begin{array}{|r|r|r|} \hline r & \text{lattice points in circle} & \text{lattice points in sphere } \\ \hline 0 & 1 & 1 \\ 1 & 5 & 7 \\ 2 & 13 & 33 \\ 3 & 29 & 123 \\ 4 & 49 & 257 \\ {\color{red}5} & {\color{red}81} & 515 \\ 6 & 113 & 925 \\ 7 & 149 & 1419 \\ 8 & 197 & 2109 \\ 9 & 253 & 3071 \\ 10 & 317 & 4169 \\ 11 & 377 & 5575 \\ 12 & 441 & 7153 \\ 13 & 529 & 9171 \\ 14 & 613 & 11513 \\ 15 & 709 & 14147 \\ 16 & 797 & 17077 \\ 17 & 901 & 20479 \\ 18 & 1009 & 24405 \\ 19 & 1129 & 28671 \\ 20 & 1257 & 33401 \\ \hline \end{array} \)

 

laugh

heureka 15-ene-2018
 #3
avatar+18829 
+1

Algrebra Help

 

b) Find all cube roots of 8i.

\(\begin{array}{|rcll|} \hline && \sqrt[3]{8i} \\ &=& \sqrt[3]{8}\cdot \sqrt[3]{i} \\ &=& \sqrt[3]{2^3}\cdot \sqrt[3]{i} \\ &=& 2\cdot \sqrt[3]{i} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} z= a+bi &=& i \\ |z| &=& \sqrt{0^2+1^2} \\ &=& \sqrt{1} \\ &=& 1 \\ \sin(\varphi) &=& \frac{b}{|z|} \\ &=& \frac{1}{1} \\ &=& 1 \\ \varphi &=& \arcsin(1) +2k\pi \\ \varphi &=& \frac{\pi}{2} +2k\pi \\\\ z &=& |z|e^{i\varphi} \\ & = & 1\cdot e^{i\cdot ( \frac{\pi}{2}+2k\pi) } \\ \mathbf{z} & \mathbf{=} & \mathbf{ e^{i\cdot ( \frac{\pi}{2}+2k\pi) }} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \sqrt[3]{i} \\ &=& (i)^{\frac13} \\ &=& \left( e^{i\cdot ( \frac{\pi}{2}+2k\pi)}\right)^{\frac13} \\ &=& e^{i\cdot \left( \frac{\frac{\pi}{2}+2k\pi}{3} \right)} \\ &=& e^{i\cdot \left( \frac{\pi}{6}+\frac{2}{3}\cdot k\pi \right) } \quad & | \quad k =(0,1,2) \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6} \right)} \quad & | \quad k = 0 \\ &=& \cos( \frac{\pi}{6} ) + i \cdot \sin(\frac{\pi}{6} ) \quad & | \quad \frac{\pi}{6} = 30^{\circ}\\ &=& \cos( 30^{\circ} ) + i \cdot \sin(30^{\circ}) \\ &=& \frac{\sqrt{3}}{2} + i\cdot \frac12 \\ 2\cdot \sqrt[3]{i} &=& 2\cdot \left( \frac{\sqrt{3}}{2} + i\cdot \frac12 \right) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{\sqrt{3} + i} \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6}+\frac23 \pi \right)} \quad & | \quad k = 1 \\ &=& e^{i\cdot \left( \frac{5\pi}{6} \right)} \\ &=& \cos(\frac{5\pi}{6}) + i \cdot \sin(\frac{5\pi}{6}) \quad & | \quad \frac{5\pi}{6} = 150^{\circ}\\ &=& \cos( 150^{\circ} ) + i \cdot \sin(150^{\circ}) \\ &=& -\cos( 30^{\circ} ) + i \cdot \sin(30^{\circ}) \\ &=& -\frac{\sqrt{3}}{2} + i\cdot \frac12 \\ 2\cdot \sqrt[3]{i} &=& 2\cdot \left( -\frac{\sqrt{3}}{2} + i\cdot \frac12 \right) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{-\sqrt{3} + i} \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6}+2\cdot \frac23 \pi \right)} \quad & | \quad k = 2 \\ &=& e^{i\cdot \left( \frac{3\pi}{2} \right)} \\ &=& \cos(\frac{3\pi}{2}) + i \cdot \sin(\frac{3\pi}{2}) \quad & | \quad \frac{3\pi}{2} = 270^{\circ}\\ &=& \cos( 270^{\circ} ) + i \cdot \sin(270^{\circ}) \\ &=& 0 + i \cdot \sin(90^{\circ}) \\ &=& -i \\ 2\cdot \sqrt[3]{i} &=& 2\cdot( -i ) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{-2i} \\\\ \hline \end{array}\)


All 3rd roots of \(8 i\) :
\(\sqrt{3} + i \\ -\sqrt{3} + i \\ -2i\)

 

laugh

heureka 12-dic-2017