heureka

Lösung:

Subtrahiere den Quotienten aus -(3:8) und ein viertel von -(3:2)

\(\begin{array}{lcl} -\dfrac32 - \dfrac{ -\dfrac38 }{ \dfrac14 } \\\\ &=& -\dfrac32 - \left( -\dfrac38 \cdot \dfrac41 \right) \\\\ &=& -\dfrac32 + \left( \dfrac38 \cdot \dfrac41 \right) \\\\ &=& \left( \dfrac38 \cdot \dfrac41 \right) -\dfrac32 \\\\ &=& \left( \dfrac32 \cdot \dfrac11 \right) -\dfrac32 \\\\ &=& \dfrac32 -\dfrac32 \\\\ &=& 0 \\\\ \mathbf{ -\dfrac32 - \dfrac{ -\dfrac38 }{ \dfrac14 } } & \mathbf{=} & \mathbf{0} \end{array}\)

( + )*(-x)=-x-1

\(\left( 1+\dfrac{1}{x} \right) \cdot (-x) = -x-1\)

All Solutions for x:

\(\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i (\pi \pm 2k\pi) } + 1 \\ e^{2x} &=& e^{i(\pi \pm 2k\pi)}\\ 2x &=& i(\pi \pm 2k\pi) \\ x &=& \dfrac{ i(\pi \pm 2k\pi) } {2}\\ \mathbf{x} & \mathbf{=}& \mathbf{ \dfrac{ i\pi }{2} \pm k\pi i } \qquad k \in N \qquad | \qquad \dfrac{ i\pi }{2} - \pi i = -\dfrac{ i\pi }{2} \\ \mathbf{x} &\mathbf{=}& \mathbf{ -\dfrac{ i\pi }{2} \pm k\pi i \qquad k \in N } \\ \end{array}\)

what values of x give cosh (x) =0 ?

\(\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i\pi} + 1 \\ e^{2x} &=& e^{i\pi}\\ 2x &=& i\pi \\ x &=& \dfrac{ i\pi } {2} \end{array}\)

35% of 80

\(\begin{array}{rcl} 80 \cdot 35\ \% &=& 80 \cdot \frac{35}{100} \\ &=& 35 \cdot \frac{80}{100} \\ &=& 35 \cdot \frac{8}{10} \\ &=& \frac{35\cdot 8}{10} \\ &=& \frac{7\cdot 8}{2} \\ &=& 7\cdot 4 \\ &=& 28 \end{array}\)

was bedeutet sinh?

Was der sin(x) im Einheitskreis \((x^2+y^2)=1 \) ist, das ist der sinh(x) in der Hyperbel \((x^2-y^2)=1\)

Was der cos(x) im Einheitskreis \((x^2+y^2)=1 \) ist, das ist der cosh(x) in der Hyperbel \((x^2-y^2)=1\)

Einheitskreis:        \((x^2+y^2)=1 \)   x = cos(z) und y = sin(z)       \(\cos^2{(z)} + \sin^2{(z)} = 1\)

Hyperbelfunktion:  \((x^2-y^2)=1\)  x = cosh(z) und y = sinh(z)   \(\cosh^2{(z)}- \sinh^2{(z)} = 1\)

Der Name Hyperbelfunktionen stammt daher, dass sinh und cosh zur Parametrisierung der Hyperbel verwendet werden können:

ganz in Analogie zum Kreis , der durch Sinus und Kosinus parametrisiert werden kann:

Box A contained 579 paper clips. Box B contained 756 paper clips. After Kevin moved some paper clips from Box B to Box A, Box A contained twice as many paper clips as Box B.

b) How many paper clips did Kevin move from Box B to Box A?

Box A = 579 paper clips

Box B = 756 paper clips

Kevin moved some paper clips from Box B to Box A:

\(\begin{array}{rcl} A+x &=& 2\cdot ( B-x ) \\ A+x &=& 2B-2x \\ 3x &=& 2B - A \\ x &=& \dfrac{2B-A}{3}\\ \hline \\ x &=& \dfrac{2\cdot 756- 579}{3} \\\\ x &=& \dfrac{933}{3} \\\\ x &=& 311 \\ \end{array}\)

Kevin moved from Box B to Box A 311 paper clips

a) How many paper clips did Box B contain in the end?

\(\begin{array}{rcl} B -x &=&756 - 311 = 445 \end{array}\)

Box B did contain in the end 445 paper clips

Wäre der koeff. von x^4, wenn man (x+2)^7 auslöst 420 ?  NEIN

\(\begin{array}{rcl} (x+2)^7 \\ &=& \binom70 x^7 \cdot 2^0 \\ &=& \binom71 x^6 \cdot 2^1 \\ &=& \binom72 x^5 \cdot 2^2 \\ &=& \color{red}{\mathbf{ \binom73 x^4 \cdot 2^3 }}\\ &=& \binom74 x^3 \cdot 2^4 \\ &=& \binom75 x^2 \cdot 2^5 \\ &=& \binom76 x^1 \cdot 2^6 \\ &=& \binom77 x^0 \cdot 2^7 \\ \end{array}\\ \begin{array}{rcl} \text{Der Koeffizient von } x^4 \text{ in } (x+2)^7 \text{ ist } 2^3 \cdot \binom73 \qquad &|& \qquad \binom73 = \frac{7}{3}\cdot\frac{6}{2}\cdot\frac{5}{1} = 35\\ 2^3 \cdot \binom73 &= & 8 \cdot 35 = 280 \end{array}\)

Der Koeffizient von \(x^4\) ist 280

\((x+2)^7 = 128+448 x+672 x^2+560 x^3+\mathbf{280} x^4+84 x^5+14 x^6+x^7\)

3,5 1/2, 10 1/2, 20 1/2,...

find the difference

write a recursive rule and a explict rule

\(\begin{array}{rrrr} 3,& 5 \dfrac12,& 10 \dfrac12,& 20 \dfrac12,~\cdots\\ \end{array}\)

find the difference:

\(\small{ \begin{array}{rrrrr} n & a(n) & 1.~ \text{difference} & 2.~ \text{difference} & 3.~ \text{difference} \\ \\ \hline \\ 0 & a_0 = 3= \frac62 &\\ & &d_1= \frac52 \\ 1 & a_1 = \frac{11}2 & &d_2= \frac52 \\ & & \frac{10}2 & &d_3= \frac52 \\\\ 2 & a_2 = \frac{21}2 & & \frac{10}2 \\ & & \frac{20}2 \\ 3 & a_3 = \frac{41}2 & &\\ ~\cdots\\ \end{array} }\)

explict rule:

\(\small{ \begin{array}{rcl} a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot d_1 + \binom{n}{2}\cdot d_2 + \binom{n}{3}\cdot d_3\\\\ a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad \binom{n}{0} = 1 \\\\ a_n &=& a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad a_0 = 3 \\\\ a_n &=& 3 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52\\ a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\ \end{array} }\)

recursive rule:

\(\small{ \begin{array}{rcl} a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1} &=& 3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right]\\ \hline a_{n+1}-a_n &=&3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - 3 - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1}-\binom{n}{1} + \binom{n+1}{2}-\binom{n}{2} + \binom{n+1}{3}- \binom{n}{3} \right]\\\\ \binom{n+1}{1}-\binom{n}{1} &=& \binom{n}{0} \\ \binom{n+1}{2}-\binom{n}{2} &=& \binom{n}{1} \\ \binom{n+1}{3}-\binom{n}{3} &=& \binom{n}{2} \\\\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right]\\\\ a_{n+1} &=&a_n + \frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\\\\\ \end{array} }\)

Wie bekommt man die 0-stellen aus f(x)= (-1)x^4+4x³-4x²+1 ?

1. Nullstelle \( x_1 = 1 \) durch raten:       ( x = 1 einsetzen in \(-x^4+4x^3-4x^2+1\) )

   \(\begin{array}{rcl} (-1)\cdot 1^4+4\cdot 1^3-4\cdot 1^2 + 1 &=& 0\\ -1+4-4+1&=&0 \end{array}\)

weitere Nullen durch Polygondivision:   \(-x^4+4x^3-4x^2+1 : ( x-1)= -x^3+3x^2-x-1\)

Wir erhalten: \(( x-1)(-x^3+3x^2-x-1)=0\)

2. Nullstelle \(x_2 = 1\) durch raten:   ( x = 1 einsetzen in \(-x^3+3x^2-x-1\))

   \(\begin{array}{rcl} -1\cdot 1^3+ 3\cdot 1^2 -1 - 1 &=& 0\\ -1+3-1+1&=&0 \end{array}\)

weitere Nullen durch Polygondivision:  \(-x^3+3x^2-x-1 : ( x-1)= -x^2+2x+1\)

Wir erhalten: \((x-1)(x-1)(-x^2+2x+1)=0\)

3. und 4. Nullstelle:

\(\begin{array}{rcl} (-x^2+2x+1)&=&0 \qquad | \qquad \cdot(-1)\\ x^2-2x-1&=&0 \\ x_{3,4} &=& \dfrac{2 \pm \sqrt{2^2-4(-1)} }{2} \\ x_{3,4} &=& \dfrac{2 \pm \sqrt{8} }{2} \\ x_{3,4} &=& \dfrac{2 \pm 2\cdot\sqrt{2} }{2} \\ x_{3,4} &=& 1 \pm \sqrt{2}\\ x_3 &=& 1 + \sqrt{2}\\ x_4 &=& 1 - \sqrt{2} \end{array} \)

Probe:

\(-(x-1)(x-1)(x-1-\sqrt{2})(x-1+\sqrt{2}) = 0\)

\(x_1 = 1 \\ x_2 = 1 \\ x_3 = 1+\sqrt{2}\\ x_4 = 1-\sqrt{2}\)