Mitglied: heureka

$$\small{\text{
$
E=m*c^2 = 2.4*10^{-3} \ * \ (2*10^{-6})^2 =2.4*10^{-3} \ * \ 2^2*(10^{-6})^2
$
}}\\
\small{\text{
$
=2.4*10^{-3} \ * \ 2^2*10^{-12}=2.4*10^{-3} \ * \ 4*10^{-12} = 2.4*4*10^{-3} 10^{-12} = 9.6 * 10^{-3} 10^{-12} $
}}\\
\small{\text{
$
= 9.6 * 10^{-15}
$
}}$$

heureka26.01.2015

+26387

Let

$f(x) = \frac{x^2}{x^2 - 1}.$

Find the largest $n$ so that

$$$ f(2)\cdot f(3)\cdot f(4)\cdots f(n-1)\cdot f(n) < 1.98. $$$

$$\boxed{ \small{\text{ $ \prod \limits_{i=2}^{n} \frac{i^2}{i^2-1} < 1.98 $ }} } \qquad \dfrac{i^2}{i^2-1} = \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right)
$\\ \\ \\$
\small{\text{
$ \prod \limits_{i=2}^{n} \dfrac{i^2}{i^2-1} =
\prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \prod\limits_{i=2}^{n} \left(\dfrac{i}{i+1} \right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{2-1} \cdot
\frac{3}{3-1} \cdot \frac{4}{4-1} \cdot \frac{5}{5-1} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{2+1} \cdot
\frac{3}{3+1} \cdot \frac{4}{4+1} \cdot \frac{5}{5+1} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{1} \cdot
\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{3} \cdot
\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{3}} }{ \textcolor[rgb]{0,0,1}{\not{2}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{4}} }{ \textcolor[rgb]{0,0,1}{\not{3}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{5}} }{ \textcolor[rgb]{0,0,1}{\not{4}} } \dots
\frac{\textcolor[rgb]{1,0,0}{\not{n}} }{ \textcolor[rgb]{0,0,1}{\not{n-1}} } \right)
\cdot
\left(
\frac{ \textcolor[rgb]{0,0,1}{\not{2}} }{\textcolor[rgb]{1,0,0}{\not{3}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{3}} }{\textcolor[rgb]{1,0,0}{\not{4}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{4}} }{\textcolor[rgb]{1,0,0}{\not{5}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{5}} }{\textcolor[rgb]{1,0,0}{\not{6}}} \dots
\frac{ \textcolor[rgb]{0,0,1}{\not{n-1}} }{\textcolor[rgb]{1,0,0}{\not{n}} }
\cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right)
$
}} $\\\\$
\boxed{
\small{\text{
$
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right) < 1.98
$
}}
} \\\\
\small{\text{$ \dfrac{n}{n+1} < \dfrac{1.98}{2} $ }$ \\\\$
\small{\text{ $ \dfrac{n}{n+1} < 0.99 $ } $\\\\$
\small{\text{ $ n < 0.99 *(n+1) $ }} $\\$
\small{\text{ $ n < 0.99 *n+ 0.99 $ }} $\\$
\small{\text{ $ n - 0.99 *n < 0.99 $ }} $\\$
\small{\text{ $ n *(1 - 0.99) < 0.99 $ }} $\\$$$

$$\\
\small{\text{ $ n * 0.01 < 0.99 $ }} $\\$
\small{\text{ $ n < \frac{ 0.99 } { 0.01 } $ }} $\\$
\small{\text{ $ n < 99 $ }} \\
\boxed{ \small{\text{ $ n = 98 $ }} }$$

heureka26.01.2015

+26387

+10

Alpha writes the infinite arithmetic sequence $10, 8, 6, 4, 2, 0 ,\ldots.$

Beta writes the infinite geometric sequence $9, 6, 4, \frac{8}{3}, \frac{16}{9}, \ldots.$

Gamma makes a sequence whose $n^{\text{th}}$ term is the product of the $n^{\text{th}}$ term of Alpha's sequence and the $n^{\text{th}}$ term of Beta's sequence: $10\cdot 9 \quad,\quad 8\cdot 6\quad ,\quad 6\cdot 4\quad,\quad 4\cdot \frac83\quad,\quad 2\cdot \frac{16}{9}\quad,\quad \ldo...$

What is the sum of Gamma's entire sequence ?

$$\\\small{\text{
Sequence alpha:
$ a_n = a_1 + (n-1)*d = (a_1-d) + n*d
\quad a_1=10$ and $d=8-10=a_{n+1}-a_n=-2$
}}\\
\small{\text{
Sequence beta:
$ b_n = b_1 *r^{n-1} \quad b_1=9$ and $r=\frac{6}{9}=\frac{b_{n+1}}{b_n}=\frac{2}{3}$
}}$\\\\$
\small{\text{
Sequence gamma :
$ g_n = a_n*b_n = b_1 *r^{n-1} [(a_1-d) + n*d]
$
}}\\
\small{\text{
$ \boxed{ g_n = \underbrace{ \left[ b_1(a_1-d) \right] *r^{n-1} }_{sum\ = \dfrac{b_1(a_1-d)}{1-r} } \ +\ b_1d* n r^{n-1} }
$ The sequence of Gamma has two parts.
}}$\\\\$
\small{\text{
The sum of the first part $ \left[ b_1(a_1-d) \right] *r^{n-1} $ is the sum of a geometric sequence $
= \frac{ b_1(a_1-d)}{1-r}
$
}}$\\\\$
\small{\text{
The sum s of the second part $ b_1d* n r^{n-1} $ is:
}} \\
\begin{array}{rcrrrrr}
s & = & (b_1d) * 1 * r^0 +&
(b_1d)* 2 * r^1 \ + &(b_1d) * 3 * r^2 \ +&(b_1d) * 4 * r^3 \ + &(b_1d) * 5 * r^4 \ + \dots \\
r*s & = & & (b_1d)* 1 * r^1 \ + &(b_1d) * 2 * r^2 \ +&(b_1d) * 3 * r^3 \ + &(b_1d) * 4 * r^4 \ + \dots \\
\hline
s-r*s & = & (b_1*d) \ + & (b_1d)*r^1 \ + &(b_1d)*r^2 \ +&(b_1d)*r^3 \ +&(b_1d)*r^4 \ + \dots \\
\end{array}\\
\small{\text{
$
s-r*s = \underbrace{ (b_1*d) \ + (b_1d)*r^1 \ + (b_1d)*r^2 \ +(b_1d)*r^3 \ +(b_1d)*r^4 \ + \dots }_{\text{sum of a geometric sequence }\ = \frac{b_1d}{1-r} }
$
} $\\$
\small{\text{
$
s-r*s = \frac{b_1d}{1-r}
$
}}$\\$
\small{\text{
$
s(1-r) = \frac{b_1d}{1-r}
$
}}$\\\\$
\small{\text{
$
s = \dfrac{b_1d}{(1-r)^2}
$
}}$\\\\$
\small{\text{
The sum of Gamma's sequence
$ = \dfrac{b_1(a_1-d)}{1-r} \ + \dfrac{b_1d}{(1-r)^2}
$
}}$\\\\$
\small{\text{
$ = \left( \dfrac{b_1}{1-r} \right) * \left[ (a_1-d)+\dfrac{d}{(1-r)} \right]
$
}}$$

$$\small{\text{
The sum of Gamma's sequence
$ = \left(
\dfrac{ 9 }{ 1 - \frac{2}{3} }
\right) *
\left[ (10-(-2))+ \dfrac{ (-2) } { 1 - \frac{2}{3} }
\right] $
}}\\
\small{\text{
$ = 9*3 *(12-2*3) = 27*6 = 162
$
}}$$

heureka26.01.2015

+26387

P.S.

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\
= 1+ \frac{1}{3} + \frac{2}{3}
+ \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}
+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}
+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}
+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}
+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}
+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}
+\dots \ \\\\\\
= 1+ 1
+ \frac{2}{3}}
+ \frac{2^2}{3^2}}
+ \frac{2^3}{3^3}}
+\dots \ \\\\\\
s=
1+1*( \frac{2}{3} ) ^0
+1*(\frac{2}{3})^1
+1*(\frac{2}{3})^2
+1*(\frac{2}{3})^3
+1*(\frac{2}{3})^4+
\dots \$$

heureka26.01.2015

+26387

Was ergibt wenn man 2+4+6+8+10 bis 100 rechnet ?

Wir suchen sie Summe von s = 2 + 4 + 6 + 8 + 10 + ... + 96 + 98 + 100

Wir können auch schreiben s = 2*1 + 2*2 + 2*3 + 2*4 + ... + 2*48 + 2*49 + 2*50

oder s = 2 * ( 1 + 2 + 3 + 4 + ... + 48 + 49 + 50 )

Die Summe der ganzen Zahlen von 1 bis n ist $$s=\frac{n(n+1)}{2}$$

Wir haben n = 50, also haben wir ( 1 + 2 + 3 + 4 + ... + 48 + 49 + 50 ) = $$\small{\text{
$
\frac{50*(50+1)}{2}=\frac{50*51}{2}=\frac{ 2550 }{2}=1275
$
}}$$

Für unsere Summe ergibt = s = 2 * 1275 = 2550

heureka26.01.2015

+26387

$$\small{\text{
sum
$
s=
1+1*( \frac{2}{3} ) ^0+
+1*(\frac{2}{3})^1+
+1*(\frac{2}{3})^2+
+1*(\frac{2}{3})^3+
+1*(\frac{2}{3})^4+
\dots
$
}}\\
a= 1 \\
r = \frac{2}{3} \\
s = 1 + \frac{a}{1-r} = 1 + \frac{1}{1-\frac{2}{3}} = 1 + \frac{1} {\frac{1}{3}} = 1 + 3 = 4 \\
\boxed{s= 4}$$

heureka25.01.2015

+26387

+10

sqrt(8-sqrt(55))+sqrt(8+sqrt(55))

$$\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \quad | \quad \sqrt{ x^2 } \\ \\
= \sqrt{ \left( \sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \right)^2 } \\ \\
= \sqrt{ (8-\sqrt{55}) +(8+ \sqrt{55}) +2*\left( \sqrt{8-\sqrt{55}} \right)* \left(\sqrt{8+\sqrt{55}} \right)} \\ \\
= \sqrt{ 16+2*\left( \sqrt{8-\sqrt{55}} \right) * \left(\sqrt{8+\sqrt{55}} \right) } \\ \\
= \sqrt{ 16+2* \sqrt{8^2- 55 } }\\ \\
= \sqrt{ 16+2* \sqrt{64-55} } \\ \\
= \sqrt{ 16+2* \sqrt{9} } \\ \\
= \sqrt{ 16+2* 3 }\\ \\
= \sqrt{ 22 }\\ \\
= 4.69041576$$

heureka25.01.2015

+26387

Die Seiten a und b sind gleich. Der Umfang U= a+b+c.

Also U=2a+c. Die Seite a kann mit dem Pythagorassatz berechnet werden.

a^2=(c/2)^2+hc^2 (c/2)=1.125 dm

a^2=1.125^2+11.5^2

a^2=133.5156

a=11.55 dm

U=2*11.55+2.25

U=25.35 dm

heureka25.01.2015

+26387

+11

Wurzel-Iteration mit der Newton-Methode:

$$\small {
\boxed{\ x_{i+1}=x_i-\frac{f(x)}{f'(x)}\ } \qquad
f(x) = 3x^2-1-\sin{(x)} \qquad f'(x) = 6x-\cos{ (x) }
}$$

1. Nullstelle: Startwert: $$\dfrac{\sqrt{3}}{3}$$ ( 1. Nullstelle von $$3x^2-1 = 0$$ )

Iteration 1. Nullstelle: x = 0.577350269190 - ( -0.207831748217 ) = 0.785182017406

Iteration 1. Nullstelle: x = 0.785182017406 - ( 0.035610499427 ) = 0.749571517979

Iteration 1. Nullstelle: x = 0.749571517979 - ( 0.001127940021 ) = 0.748443577958
Iteration 1. Nullstelle: x = 0.748443577958 - ( 0.000001130941 ) = 0.748442447017
Iteration 1. Nullstelle x = 0.748442447017 - ( 0.000000000001 ) = 0.748442447016

 1. Nullstelle Lösung: x = 0.748442447016 
 3x^2 -1 - sin(x) = 0.000000000000 okay!

2. Nullstelle: Startwert: $$-\dfrac{\sqrt{3}}{3}$$ ( 2. Nullstelle von $$3x^2-1 = 0$$ )

Iteration 2. Nullstelle: x = -0.577350269190 - ( -0.126872131370) = -0.450478137820  
Iteration 2. Nullstelle: x = -0.450478137820 - ( -0.012263768379) = -0.438214369441  
Iteration 2. Nullstelle: x = -0.438214369441 - ( -0.000118460950) = -0.438095908491  
Iteration 2. Nullstelle: x = -0.438095908491 - ( -0.000000011070) = -0.438095897421  
Iteration 2. Nullstelle: x = -0.438095897421 - ( -0.000000000000) = -0.438095897421

 2. Nullstelle Lösung: x = -0.438095897421 
 3x^2 -1 - sin(x) = 0.000000000000 okay!

heureka25.01.2015

+26387

+13

$$\small{\text{ The fraction is
$\dfrac{79}{90}
$}}$$

heureka24.01.2015

Benutzername	heureka
Punkte	26387
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