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 #1
avatar+18246 
+1

lim n→∞{(1− 3/n)^3n +3*n'te√3n}

 

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) }= \ ? \)

 

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) } + \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ \hline \end{array} \)

 

\(y=\lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= \ ?\)

\(\begin{array}{|rcll|} \hline \ln(y) &=& \ln\Big( \lim \limits_{x\to 0} \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \ln \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( 3n\ln \left( 1- \frac{3}{n} \right) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{3\ln \left( 1- \frac{3}{n} \right)} {n^{-1}} \Big) \\\\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } 3\cdot \ln(1-3\cdot n^{-1}) \text{ lautet } 3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } \\\\ &=& \lim \limits_{x\to 0} \Big( \frac{3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } } {(-1)\cdot n^{-2}} \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot[(-3)(-1)\cdot n^{-2} ] } { (1-\frac{3}{n})(-1)\cdot n^{-2} } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot(-3)} { (1-\frac{3}{n}) } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ -9 } { (1-\frac{3}{n}) } \Big) \\ \ln(y) &=& \frac{ -9 } { 1-0 } \\ \ln(y) &=& -9 \\ e^{\ln(y)} = y &=& e^{-9}\\ y &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= e^{-9} \\ \hline \end{array}\)

 

 

\(\lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } = \ ?\)

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot (3n)^{\frac{1}{n}} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{\ln \left((3n)^{\frac{1}{n}} \right) } \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{ \frac{\ln(3n)} {n} } \Big) } \\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big( \frac{\ln(3n)} {n} } \Big) } \\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } \frac{\ln(3n)} {n} \text{ lautet } \frac{ \frac{3}{3n} } {1} = \frac{3}{3n} = \frac{1}{n} \\\\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big(\frac{1}{n}} \Big) } \\ &=& 3 \cdot e^0 \\ &=& 3 \cdot 1 \\ &=& 3 \\ \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } &=& 3 \\ \hline \end{array} \)

 

Somit erhalten wir insgesamt:

 

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } = e^{-9} + 3\)

 

laugh

heureka 21.07.2017
 #2
avatar+18246 
+1

Compute

\(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

 

Let \(r =\frac12\)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} \\\\ && r = \frac12 \\\\ \hline s_n &=& 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \dots + n \cdot r^n \\ rs_n &=& \qquad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1) \cdot r^n + n\cdot r^{n+1} \\ \hline s_n -r\cdot s_n &=& r+r^2+r^3+ \dots +r^n-n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ \hline && S_n = r+r^2+r^3+ \dots +r^n \\ && rS_n = \quad r^2+r^3 + \dots + r^{n+1} \\ \hline && S_n -r\cdot S_n = r - r^{n+1} \\ && S_n\cdot (1-r) = r - r^{n+1} \\ && S_n = \frac{r - r^{n+1}}{1-r} \\ \hline s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \\ s_n &=& \frac{1}{1-r} \cdot \Big( \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \Big) \\\\ && \frac{1}{1-r} = 2 \\\\ s_n &=& 2 \cdot \Big( 2\cdot(r - r^{n+1}) -n\cdot r^{n+1} \Big) \\ s_n &=& 2 \cdot ( 2\cdot r - 2\cdot r^{n+1} - n\cdot r^{n+1} ) \\ s_n &=& 2 \cdot r\cdot ( 2 - 2\cdot r^n - n\cdot r^n ) \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 \cdot r\cdot \Big( 2 - (2+n)\cdot r^n \Big)} \quad & | \quad r=\frac12 \\ \\ s_n & = & 2 \cdot \frac12 \cdot \Big( 2 - (2+n)\cdot (\frac12)^n \Big) \\ s_n & = & 2 - (2+n)\cdot \frac{1}{2^n} \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 - \frac{2+n}{2^n} } \\ \hline \end{array}\)

 

laugh

heureka 21.07.2017
 #4
avatar+18246 
+2

cos(pi/7) + (cos(pi/7))^2 - 2(cos(pi/7))^3 = ?

step by step

 

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ Substitution } } \\ x = \frac{\pi}{7} && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ \hline \end{array}\)

 

\(\mathbf{2. \cos^2(x) \text{ to } \cos(2x) \text{ and } \cos^3(x) \text{ to } \cos(3x) }\)

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \cos(2x) &=& 2\cos^2(x) - 1 \qquad \text{ or } \qquad \cos^2(x) = \frac12\Big(1+\cos(2x)\Big) \\ \cos(3x) &=& 4\cos^3(x) - 3\cos(x) \qquad \text{ or } \qquad \cos^3(x) = \frac14\Big(3\cos(x)+\cos(3x)\Big) \\ \\ && \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \cos(x) + \frac12\Big(1+\cos(2x)\Big) - \frac24\Big(3\cos(x)+\cos(3x)\Big) \\ &=& \frac12\Big( 2\cos(x) + 1 + \cos(2x)-3\cos(x) - \cos(3x) \Big) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ \hline \end{array} \)

 

3. Change minus to plus

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ \cos(x) = -\cos(\pi-x) \\\\ & \cos(\frac{\pi}{7}) &=& -\cos(\pi-\frac{\pi}{7}) \\ & &=& -\cos(\frac67 \pi) \\ & \mathbf{\cos(x)} &\mathbf{=}&\mathbf{ -\cos(6x)} \\\\ & \cos(\frac{3\pi}{7}) &=& -\cos(\pi-\frac{3\pi}{7}) \\ & &=& -\cos(\frac47 \pi) \\ & \mathbf{\cos(3x) } &\mathbf{=}& \mathbf{-\cos(4x)} \\\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline && \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+ \cos(6x)+\cos(2x)+\cos(4x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ \hline \end{array} \)

 

\(\begin{array}{|l|rcll|} \hline \mathbf{4. \text{ Substitution } } \\ y = 2x && \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( 1+\cos(y)+\cos(2y)+ \cos(3y) \Big) \\ \hline \end{array} \)

 

5. complex numbers \(1+\cos(y)+\cos(2y)+ \cos(3y) = \ ?\)

\(\begin{array}{|rcll|} \hline C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \\ S &=& \qquad \sin(y)+\sin(2y)+ \sin(3y) \\ \mathbf{T} &\mathbf{=}& \mathbf{ C+i\cdot S } \\\\ T &=& 1+\cos(y)+\cos(2y)+ \cos(3y)+i\cdot \Big( \sin(y)+\sin(2y)+ \sin(3y) \Big)\\ &=& 1+\underbrace{\cos(y)+i\cdot \sin(y)}_{=e^{iy}} +\underbrace{\cos(2y)+i\cdot \sin(2y)}_{=e^{i2y}} +\underbrace{ \cos(3y)+i\cdot \sin(3y)}_{=e^{i3y}} \\ &=& 1+ e^{iy} + e^{i2y} + e^{i3y} \\ &=& 1+ e^{iy} + (e^{iy})^2 + (e^{iy})^3 \qquad \text{geometric series } r=e^{iy} \\ T &=& 1+ r + r^2 + r^3 \\ \hline T &=& 1+ r + r^2 + r^3 \\ r\cdot T &=& \qquad r + r^2 + r^3 + r^4 \\ \hline r\cdot T - T &=& r^4 - 1 \\ T\cdot (r-1) &=& r^4 - 1 \\ T &=& \frac{r^4 - 1} {r-1} \\ &=& \frac{(e^{iy})^4 - 1} {e^{iy}-1} \\ \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ \hline \end{array}\)

 

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ e^{i\phi} - 1 = 2i\cdot \sin(\frac{\phi}{2})\cdot e^{i\frac{\phi}{2}} \\\\ & \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ & &=& \frac{ 2i\cdot \sin(2y) \cdot e^{i2y} } { 2i\cdot \sin(\frac{y}{2})\cdot e^{ i\frac{y}{2} }} \qquad | \qquad y = 2x \\ & &=& \frac{ \sin(4x) \cdot e^{i4x} } { \sin(x)\cdot e^{ ix }} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i4x-ix} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i3x} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \Big( \cos(3x) + i\cdot \sin(3x) \Big) \\ & &=& \underbrace{\frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x)}_{=C} + i\cdot \underbrace{ \frac{ \sin(4x) } { \sin(x)} \cdot \sin(3x) }_{=S} \\\\ \hline \end{array}\)

 

6. Solution

\(\begin{array}{|rcll|} \hline & C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \qquad | \qquad y = 2x \\ & &=& 1+\cos(2x)+\cos(4x)+ \cos(6x) \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \sin(x) = \sin(\pi-x) \\\\ & \cos(\frac{4\pi}{7}) &=& \sin(\pi-\frac{4\pi}{7}) \\ & &=& \sin(\frac37 \pi) \\ & \mathbf{\sin(4x)} &\mathbf{=}&\mathbf{ \sin(3x)} \\\\ & \cos(\frac{6\pi}{7}) &=& \sin(\pi-\frac{6\pi}{7}) \\ & &=& \sin(\frac17 \pi) \\ & \mathbf{\sin(6x)} &\mathbf{=}&\mathbf{ \sin(x)} \\\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \quad & | \quad x=\frac{\pi}{7} \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \mathbf{\sin(4x)= \sin(3x)}\\ &=& \frac12\Big( \frac{ \sin(3x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \sin(3x)\cos(3x)=\frac{\sin(6x)}{2} \\ &=& \frac12\Big( \frac{ \sin(6x) } { 2 \sin(x)} \Big) \quad & | \quad \mathbf{\sin(6x)=\sin(x)} \\ &=& \frac12\Big( \frac{ \sin(x) } { 2 \sin(x)} \Big) \\ &=& \frac12\Big( \frac12 \Big) \\ &=& \mathbf{\frac14} \\ \hline \end{array}\)

 

 

laugh

heureka 18.07.2017
 #3
avatar+18246 
+3

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

 

another approach

 

 

Let \(BC = 8\)
Let \(AB = 6\)

\(\text{Let } A = \text{area}_{ABC} = \frac{AB\cdot BC}{2} = \frac{6\cdot 8}{2} = 24\)

 

\(\text{Let } A_1 = \text{area}_{BPN} = \text{area}_{NPC} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_2 = area_{BCP} = area_{CAP} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_3 = area_{APC} = \ ? \\ \text{Let area}_{BAN} = \frac{ AB\cdot \frac{BC}{2} } {2} = \frac{ AB\cdot BC } {4}\\ \text{Let area}_{BMC} = \frac{ BC\cdot \frac{AB}{2} } {2} = \frac{ AB\cdot BC } {4}\)

 

\(\begin{array}{|rcll|} \hline \text{area}_{BAN} &=& \text{area}_{BMC} = \frac{ AB\cdot BC } {4} \quad & | \quad \text{area}_{BAN} = A_1 +2A_2 \qquad \text{area}_{BMC} = A_2 +2A_1 \\ A_1 +2A_2 &=& A_2 +2A_1 \\ 2A_2-A_2 &=& 2A_1-A_1 \\ A_2 &=& A_1 \\\\ A_1 +2A_2 &=& \frac{ AB\cdot BC }{4} \quad & | \quad A_2 = A_1 \\ 3A_1 &=& \frac{ AB\cdot BC }{4} \\ A_1 &=& \frac{ AB\cdot BC }{12} \quad & | \quad AB = 6 \qquad BC = 8 \\ A_1 &=& \frac{ 6\cdot 8 }{12} \\ \mathbf{A_1} & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A &=& 2A_1+2A_2+A_3 \quad & | \quad A_2 = A_1 \\ A &=& 2A_1+2A_1+A_3 \\ A &=& 4A_1+A_3 \\ A_3 &=& A-4A_1 \quad & | \quad A = 24 \qquad A_1 = 4\\ A_3 &=& 24-4\cdot 4 \\ A_3 &=& 24-16 \\ \mathbf{A_3 } & \mathbf{=} & \mathbf{8} \\ \hline \end{array} \)

 

The area of triangle APC is 8 square units

 

laugh

heureka 14.07.2017
 #2
avatar+18246 
+3

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

 

 

Centroid of a Triangle: The point where the three medians of the triangle intersect.

So P is the centroid of the triangle ABC.

 

\(\begin{array}{lcll} \text{Let } AB = 6 \\ \text{Let } BC = 8 \\\\ \text{Let } \vec{B} = \binom {0}{0} \\ \text{Let } \vec{A} = \binom {0}{AB} \\ \text{Let } \vec{C} = \binom {BC}{0} \\ \end{array} \)

 

\(\vec{P} \text{ the centroid of the triangle ABC}\ =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 (\vec{A}+\vec{B}+\vec{C}) \\ \vec{P} &=& \frac13 \Big( \binom {0}{AB}+\binom {0}{0}+\binom {BC}{0} \Big) \\ \vec{P} &=& \frac13 \binom {BC}{AB} \quad & | \quad AB = 6 \qquad BC = 8 \\ \vec{P} &=& \frac13 \binom {8}{6} \\ \vec{P} &=& \binom {\frac{8}{3}}{2} \\ \hline \end{array} \)

 

 

\(\text{Area}_{\text{APC}} = \ ? \)

\(\begin{array}{|rcll|} \hline \text{Area}_{\text{APC}} &=& \frac12 | (\vec{C}-\vec{P}) \times (\vec{A}-\vec{P}) | \\ &=& \frac12 | \Big(\binom {BC}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{AB}- \binom {\frac{8}{3}}{2} \Big) | \quad & | \quad AB = 6 \qquad BC = 8 \\ &=& \frac12 | \Big(\binom {8}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{6}- \binom {\frac{8}{3}}{2} \Big) | \\ &=& \frac12 | \binom {8-\frac{8}{3}}{0-2} \times \binom {0-\frac{8}{3}}{6-2} | \\ &=& \frac12 | \binom {\frac{16}{3}}{-2} \times \binom {-\frac{8}{3}}{4} | \\ &=& \frac12 \cdot \Big( \frac{16}{3} \cdot 4 - (-2)\cdot (-\frac{8}{3}) \Big) \\ &=& \frac12 \cdot \Big( \frac{64}{3} - \frac{16}{3} \Big) \\ &=& \frac12 \cdot \frac{48}{3} \\ &=& \frac{48}{6} \\ \text{Area}_{\text{APC}} &=& 8 \text{ square units } \\ \hline \end{array}\)

 

laugh

heureka 14.07.2017