heureka

5 is by definition a integer

2 is by definition a integer

so:  5 \ 2   means integer divided by integer is by definition a integer in programming.    = [2].5

see also floor-function from gauss.

example: 5. \ 2 means real number divided by integer is by definiton a real number in programming. = 2.5

atan2(4,3)

$$\small{ \begin{array}{lcl} \tan{(\alpha)} = \frac{a}{b} = \frac{\Delta y}{\Delta x}\\\\ \alpha = \mathbf{atan}{(\frac{a}{b}) }\\\\ \alpha =\mathbf{atan2} { ( a,b) } _{\text{in right quadrant}} =\mathbf{atan2}{ ( \text{numerator},\text{denominator} ) } =\mathbf{atan2}{ ( \Delta y,\Delta x ) }\\ \end{array} }$$

if c is the midpoint of KN , what is KC?

$$\dfrac{\vec{N}-\vec{K}}{2} = \vec{KC}\\\\ \overline{KC} = |\vec{KC}|$$

What is X= 5\2 = [5\2] = 2

atan2(4,3) ?

$$\mathbf{atan2}{(4,3)}=53.130102354156\ensurement{^{\circ}}\\ \Delta y = 4 \text{ and } \Delta x = 3$$

arctan with two parameters. see web2.0calculator. You can input like above.

$$\small{\text{ $ \frac{\pi}{4}=8\cdot \arctan{(\frac{1}{10})}-($rational fraction$)$. What is the rational fraction }} \\\\ \small{\text{ We start with:\qquad $\tan{(\alpha)} = \frac{1}{10}$ }}\\\\ \small{\text{ Using the formula for double angles three times we get $\tan{(8\alpha)}$ : }}\\\\ \small{\text{$ \begin{array}{lrcl} \qquad \qquad &\tan{(2\alpha)} &=& \frac{2\cdot \tan{(\alpha)}} {1-\tan^2{(\alpha)}} = \frac{2\cdot \frac{1}{10} } {1- (\frac{1}{10})^2 } = \dfrac{20}{99} \\ \end{array} $ }}\\\\ \small{\text{ Second Double angle formula, we get : }}\\\\ \small{\text{$ \begin{array}{lrcl} \qquad \qquad &\tan{(4\alpha)} &=& \frac{2\cdot \tan{(2\alpha)}} {1-\tan^2{(2\alpha)}} = \frac{2\cdot \frac{20}{99} } {1- (\frac{20}{99})^2 } = \dfrac{3960}{9401} \\ \end{array} $ }}\\\\ \small{\text{ Third Double angle formula, we get : }}\\\\ \small{\text{$ \begin{array}{lrcl} \qquad \qquad &\tan{(8\alpha)} &=& \frac{2\cdot \tan{(4\alpha)}} {1-\tan^2{(4\alpha)}} = \frac{2\cdot \frac{3960}{9401} } {1- (\frac{3960}{9401})^2 } = \dfrac{74455920}{72697201} \\ \end{array} $ }}\\\\ \small{\text{ $8\alpha$ differs from $\frac{\pi}{4}$, and $\tan{( \frac{\pi}{4} )}=1 $ we have : }}\\\\ \small{\text{$ \begin{array}{lrcl} \qquad \qquad &\tan{ (8\alpha-\frac{\pi}{4}) } &=& \frac{ \tan{(8\alpha)} - \tan{(\frac{\pi}{4})} } { \tan{(8\alpha)} + \tan{(\frac{\pi}{4})} } = \dfrac{ \frac{74455920}{72697201} - 1 } { \frac{74455920}{72697201} + 1 } = \dfrac{1758719}{147 153 121} \\ \end{array} $ }}\\\\$$

$$\small{\text{ Taking the arctan of both sides, we have : }}\\\\ \small{\text{$ \begin{array}{lrcl} \qquad \qquad & 8\alpha-\frac{\pi}{4}&=& \arctan{ (\frac{1758719}{147 153 121}) }\\\\ \qquad \qquad &\frac{\pi}{4}&=& 8\alpha - \arctan{ (\frac{1758719}{147 153 121}) } \\\\ \qquad \qquad &\mathbf{ \frac{\pi}{4} }& \mathbf{=} & \mathbf{ 8 \arctan{(\frac{1}{10})} - \arctan{ (\frac{1758719}{147 153 121}) } } \end{array} $ }}$$

Ein summand ist um 230 größer als der andere die summe ist 730

zahl1 + zahl2 = 730

zahl1 - zahl2 = 230

zahl1 = 730 - zahl2

zahl1 = 230 + zahl2

zahl1 = 730 - zahl2 = 230 + zahl2

730 - zahl2 = 230 + zahl2

2*zahl2 = 730 - 230

2*zahl2 = 500

zahl2 = 250

zahl1 = 230 +  zahl2

zahl1 = 230 + 250

zahl1 = 480

Probe: 

480 + 250 = 730 (okay)

480 - 250 = 230 (okay)

Die 1. Zahl lautet 480 und die 2. Zahl lautet 250.

(3x^3-5x^2+10x-3)/(3x-4)

7.6855172413793103 in fractions is?

1. Continued fraction of 7.6855172413793103 :

$$\small{\text{$ \begin{array}{rrrrrrrrrr} [x_0;&x_1,&x_2,&x_3,&x_4,&x_5,&x_6,&x_7,&x_8,&x_9]\\ \[[7;&1,&2,&5,&1,&1,&3,&1,&1,&2] \end{array} $ }}$$  

$$\small{\text{$ \begin{array}{lclcl} \textcolor[rgb]{1,0,0}{x_0} &=& \textcolor[rgb]{1,0,0}{7}.6855172413793103 \\ && 1/0.6855172413793103= \\ \textcolor[rgb]{1,0,0}{x_1}&=&\textcolor[rgb]{1,0,0}{1}.4587525150905433549486051115546459473433926066921394\\ && 1/0.4587525150905433549486051115546459473433926066921394=\\ \textcolor[rgb]{1,0,0}{x_2}&=&\textcolor[rgb]{1,0,0}{2}.1798245614035083186653585718683317691923088345579715\\ && 1/0.1798245614035083186653585718683317691923088345579715=\\ \textcolor[rgb]{1,0,0}{x_3}&=&\textcolor[rgb]{1,0,0}{5}.5609756097561115779298036883141182477038929421104669\\ && 1/0.5609756097561115779298036883141182477038929421104669=\\ \textcolor[rgb]{1,0,0}{x_4}&=&\textcolor[rgb]{1,0,0}{1}.7826086956521293714555765605533410043560222722903369\\ && 1/0.7826086956521293714555765605533410043560222722903369=\\ \textcolor[rgb]{1,0,0}{x_5}&=&\textcolor[rgb]{1,0,0}{1}.2777777777778505015432098790380390517833503327285810\\ && 1/0.2777777777778505015432098790380390517833503327285810=\\ \textcolor[rgb]{1,0,0}{x_6}&=&\textcolor[rgb]{1,0,0}{3}.5999999999990575000000002144187499999512197343750242\\ && 1/0.5999999999990575000000002144187499999512197343750242=\\ \textcolor[rgb]{1,0,0}{x_7}&=&\textcolor[rgb]{1,0,0}{1}.6666666666692847222222257391435185232429161265495199\\ && 1/0.6666666666692847222222257391435185232429161265495199=\\ \textcolor[rgb]{1,0,0}{x_8}&=&\textcolor[rgb]{1,0,0}{1}.4999999999941093750000152199023437106755773194376239\\ && 1/0.4999999999941093750000152199023437106755773194376239=\\ \textcolor[rgb]{1,0,0}{x_9}&=&\textcolor[rgb]{1,0,0}{2}.0000000000\ldots \end{array} $}}$$

2. The successive convergents with numerators p and denominators q are given by the

$$\small{ \text{ Formula } \boxed{ \begin{array}{lcl} \dfrac{p_{n+1}}{q_{n+1}} = \dfrac{x_{n+1}\cdot p_n + p_{n-1} }{x_{n+1}\cdot q_n + q_{n-1} } \qquad \dfrac{p_0} {q_0} = \dfrac{x_0}{1}=\dfrac{ \textcolor[rgb]{1,0,0}{7} }{1} \qquad \dfrac{p_{-1}} {q_{-1} } = \dfrac{1}{0} \\\\ $ For example $ n=0:~~ \dfrac{p_{1}}{q_{1}} = \dfrac{x_1\cdot p_0 + p_{-1} }{x_1\cdot q_0 + q_{-1} } = \dfrac{x_1\cdot x_0+1}{x_1\cdot 1+ 0 } = \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 7 + 1 }{ \textcolor[rgb]{1,0,0}{1}\cdot 1 + 0} = \dfrac{8}{1} = 8\\\\ $ For example $ n=1:~~ \dfrac{p_{2}}{q_{2}} = \dfrac{x_2\cdot p_1 + p_0 }{x_2\cdot q_1 + q_0 } = \dfrac{x_2\cdot 8 + 7}{x_2\cdot 1+ 1 } = \dfrac{\textcolor[rgb]{1,0,0}{2}\cdot 8 + 7 }{ \textcolor[rgb]{1,0,0}{2}\cdot 1+ 1} = \dfrac{23}{3} = 7.6\overline{6}\\\\ $ For example $ n=2:~~ \dfrac{p_{3}}{q_{3}} = \dfrac{x_3\cdot p_2 + p_1 }{x_3\cdot q_2 + q_1 } = \dfrac{x_3\cdot 23 + 8}{x_3\cdot 3+ 1 } = \dfrac{\textcolor[rgb]{1,0,0}{5}\cdot 23 + 8 }{ \textcolor[rgb]{1,0,0}{5}\cdot 3+ 1} = \dfrac{123}{16} = 7.6875\\\\ $ For example $ n=3:~~ \dfrac{p_{4}}{q_{4}} = \dfrac{x_4\cdot p_3 + p_2 }{x_4\cdot q_3 + q_2 } = \dfrac{x_4\cdot 123 + 23}{x_4\cdot 16 + 3 } = \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 123 + 23 }{ \textcolor[rgb]{1,0,0}{1}\cdot 16 + 3 } = \dfrac{146}{19} = 7.68421052631578947\ldots \\\\ $ For example $ n=4:~~ \dfrac{p_{5}}{q_{5}} = \dfrac{x_5\cdot p_4 + p_3 }{x_5\cdot q_4 + q_3 } = \dfrac{x_5\cdot 146 +123}{x_5\cdot 19+16 } = \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 146 +123 }{ \textcolor[rgb]{1,0,0}{1}\cdot 19+16 } = \dfrac{269}{35} = 7.68571428571428571\ldots \\\\ $ For example $ n=5:~~ \dfrac{p_{6}}{q_{6}} = \dfrac{x_6\cdot p_5 + p_4 }{x_6\cdot q_5 + q_4 } = \dfrac{x_6\cdot 269+ 146}{x_6\cdot 35+19 } = \dfrac{\textcolor[rgb]{1,0,0}{3}\cdot 269+ 146 }{ \textcolor[rgb]{1,0,0}{3}\cdot 35+19 } = \dfrac{953}{124} = 7.68548387096774194\ldots \\\\ $ For example $ n=6:~~ \dfrac{p_{7}}{q_{7}} = \dfrac{x_7\cdot p_6 + p_5 }{x_7\cdot q_6 + q_5 } = \dfrac{x_7\cdot 953+ 269 }{x_7\cdot 124 + 35 } = \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 953+ 269 }{ \textcolor[rgb]{1,0,0}{1}\cdot 124 + 35 } = \dfrac{1222}{159} = 7.68553459119496855\ldots \\\\ $ For example $ n=7:~~ \dfrac{p_{8}}{q_{8}} = \dfrac{x_8\cdot p_7 + p_6 }{x_8\cdot q_7 + q_6 } = \dfrac{x_8\cdot 1222 + 953 }{x_8\cdot 159 + 124 } = \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 1222 + 953 }{ \textcolor[rgb]{1,0,0}{1}\cdot 159 + 124 } = \dfrac{2175}{283} = 7.68551236749116607\ldots \\\\ $ For example $ n=8:~~ \dfrac{p_{9}}{q_{9}} = \dfrac{x_9\cdot p_8 + p_7 }{x_9\cdot q_8 + q_7 } = \dfrac{x_9\cdot 2175 + 1222 }{x_9\cdot 283 + 159 } = \dfrac{\textcolor[rgb]{1,0,0}{2}\cdot 2175 + 1222 }{ \textcolor[rgb]{1,0,0}{2}\cdot 283 + 159 } = \dfrac{5572}{725} = 7.68551724137931034\ldots \\\\ \end{array} } }$$

$$7.6855172413793103 \approx \dfrac{5572}{725} = 7.68551724137931034\ldots$$

$$\small{\text{$ 2.5! = \Gamma{(2.5+1)} $ }}$$

$$\small{ \text{ Formula } \boxed{ \begin{array}{lcl} \Gamma{(x+1)} = x\cdot\Gamma{(x)} \\ \Gamma{(\frac12)} = \sqrt{\pi} \end{array} } }$$

$$\small{ \begin{array}{lcl} 2.5!&=& \Gamma{(2.5+1)}\\ &=& 2.5\cdot\Gamma{(2.5)}\\ &=& 2.5\cdot \Gamma{(1.5+1)}\\ &=& 2.5\cdot 1.5\cdot \Gamma{(1.5)}\\ &=& 2.5\cdot 1.5\cdot \Gamma{(\frac12+1)}\\\\ &=& 2.5\cdot 1.5\cdot \frac12 \cdot \Gamma{(\frac12)}\\\\ &=& 2.5\cdot 1.5\cdot \frac12 \cdot \sqrt{\pi}\\\\ &=& 1.875 \cdot \sqrt{\pi} \\ &\approx& 1.875 \cdot 1.77245385091 \\ \mathbf{2.5!} &\mathbf{\approx}& \mathbf{3.32335097045} \end{array} }$$