heureka

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 #1
 #7
avatar+26364 
+3

find the sum of n terms of the series:

 

A) 1+ 2x + 3x^2 + 4x^3 +.....

like Euler:

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1+ 2x + 3x^2 + 4x^3 +\ldots +n x^{n-1}} \quad & | \quad \cdot dx \\ s_n dx &=& dx+ 2xdx + 3x^2dx + 4x^3dx +\ldots +nx^{n-1}dx \quad & | \quad \int{} \\ \int{s_n dx} &=& x+ 2\frac{x^2}{2} + 3\frac{x^3}{3} + 4\frac{x^4}{4} +\ldots +n\frac{x^n}{n} \\ \int{s_n dx} &=& x+ x^2 + x^3 + x^4 +\ldots + x^n = \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx} &=& \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx}&=& \frac{x-x^{n+1}} {1-x} = [x-x^{n+1}][(1-x)^{-1}] \quad & | \quad \text{derivate} \\ s_n &=& [1-(n+1)x^n](1-x)^{-1}+ (x-x^{n+1})(-1)(1-x)^{-2}(-1) \\ s_n &=& \frac{1-(n+1)x^n} {1-x} + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-(n+1)x^n} {1-x} \left(\frac{1-x}{1-x}\right) + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{[1-(n+1)x^n](1-x)+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-x-(n+1)x^n+(n+1)x^{n+1}+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+(n+1)x^{n+1} -x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+[-1+(n+1)]x^{n+1} } {(1-x)^2} \\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{1 -(n+1)x^n+n x^{n+1} } {(1-x)^2} } \\ \hline \end{array}\)

 

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1\cdot 2^2 + 2\cdot3^2 + 3\cdot4^2 +\ldots +n \cdot (n+1)^2 } \\ s_n &=& \sum \limits_{k=1}^{n} k \cdot (k+1)^2 \\ &=& \sum \limits_{k=1}^{n} k \cdot (k^2+2k+1) \\ &=& \sum \limits_{k=1}^{n} (k^3+2k^2+k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + \sum \limits_{k=1}^{n} (2k^2) + \sum \limits_{k=1}^{n} (k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + 2\sum \limits_{k=1}^{n} (k^2) + \sum \limits_{k=1}^{n} (k) \\ && \begin{array}{|rcll|} \hline \sum \limits_{k=1}^{n} (k^3) &=& 1^3+2^3+3^3+ \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\ \sum \limits_{k=1}^{n} (k^2) &=& 1^2+2^2+3^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum \limits_{k=1}^{n} (k) &=& 1+2+3+ \ldots + n = \frac{n(n+1)}{2} \\ \hline \end{array} \\ &=& \left(\frac{n(n+1)}{2}\right)^2 + 2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ &=& \left(\frac{n(n+1)}{2}\right)\left( \frac{n(n+1)}{2} + \frac23\cdot(2n+1) +1\right) \\ &=& \left(\frac{n(n+1)}{2}\right)\left(\frac{3n(n+1)+4(2n+1)+6}{6}\right) \\ &=& \left(\frac{n(n+1)}{12}\right) \Big( 3n(n+1)+4(2n+1)+6 \Big) \\ &=& \frac{n(n+1)[3n(n+1)+4(2n+1)+6]}{12} \\ &=& \frac{n(n+1)(3n^2+3n+8n+4+6)}{12} \\ &=& \frac{n(n+1)(3n^2+11n+10)}{12} \\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{n(n+1)(n+2)(3n+5)}{12}} \\ \hline \end{array}\)

 

laugh

13.11.2017
 #7
avatar+26364 
+3

With gamma and the beta function

 

Formula: \(\displaystyle \int \limits_{0}^{\pi/2} \cos^{2u-1}(x)\cdot \sin^{2v-1}(x) \;dx = \frac12 \cdot B(u,v), \qquad \text{Re } u>0, \quad \text{Re } v>0\)
Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}, \qquad \text{Re } u>0, \quad \text{Re } v>0\)

 

4)\(\displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\;dx\)

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 15 & 2v-1 = 0 \\ 2u = 16 & 2v = 1\\ u = 8 & v = \frac12 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx &=& \frac12 \cdot B(8,\frac12) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \Gamma(\frac12)}{\Gamma(8+\frac12)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \Gamma(8) = 7! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \displaystyle \Gamma(8+\frac12) = \frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! 8! 4^8}{16!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 4^8}{ 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14\cdot 15\cdot 16} \\\\ &=& \displaystyle \dfrac{165150720 }{518918400 } \\\\ &=& \displaystyle \dfrac{2048\cdot 80640 }{6435\cdot 80640 } \\\\ &=& \displaystyle \dfrac{2048 }{6435 } \\ \end{array}\)

 

 

5) \(\displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx\)

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 0 & 2v-1 = 13 \\ 2u = 1 & 2v = 14 \\ u = \frac12 & v = 7 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx &=& \frac12 \cdot B(\frac12,7) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(\frac12) \cdot \Gamma(7)}{\Gamma(\frac12+7)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot \Gamma(7)}{\Gamma(7+\frac12)} \quad & | \quad \Gamma(7) = 6! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\Gamma(7+\frac12)} \quad & | \quad \displaystyle \Gamma(7+\frac12) = \frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{6! 7! 4^7}{14!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \cdot 4^7}{ 8\cdot 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14} \\\\ &=& \displaystyle \dfrac{5898240 }{17297280 } \\\\ &=& \displaystyle \dfrac{1024\cdot 5760 }{3003\cdot 5760 } \\\\ &=& \displaystyle \dfrac{1024 }{3003 } \\ \end{array}\)

 

laugh

09.11.2017
 #3
avatar+26364 
+2

What is the scale factor used to create the dilation?
The smaller triangle is a pre-image of the bigger triangle. The center of dilation is (2, -1).
What is the scale factor used to create the dilation?

 

Let \(\vec{A} = \binom{-1}{-1} \) before dilation
Let \(\vec{A'} = \binom{8}{-1}\) after dilation
Let \(\vec{C} = \binom{2}{-1}\) the center of dilation
Let \(\lambda \) is the scale factor used to create the dilation

 

Formula for dilation with vector A:
 \(\lambda = -2\)

\(\begin{array}{|rcll|} \hline \vec{A'} &=& (\vec{A}-\vec{C})\cdot \lambda + \vec{C} \quad & | \quad \lambda = -2 \\ \binom{8}{-1} &\overset{?}{=}& \Big(\binom{-1}{-1}-\binom{2}{-1} \Big)\cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-1-2}{-1-(-1)} \cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3}{0} \cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3\cdot (-2)}{0\cdot (-2)} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{6}{0} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{6+2}{0-1} \\ &\overset{!}{=}& \binom{8}{-1}~ \checkmark \\ \hline \end{array}\)

 

\( \lambda = 2\)
\(\begin{array}{|rcll|} \hline \vec{A'} &=& (\vec{A}-\vec{C})\cdot \lambda + \vec{C} \quad & | \quad \lambda = 2 \\ \binom{8}{-1} &\overset{?}{=}& \Big(\binom{-1}{-1}-\binom{2}{-1} \Big)\cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-1-2}{-1-(-1)} \cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3}{0} \cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3\cdot 2}{0\cdot 2} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-6}{0} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-6+2}{0-1} \\ & \ne & \binom{-4}{-1} \\ \hline \end{array}\)

 

 

laugh

07.11.2017