Lösung 3.1:1f - Versionsgeschichte

Moni um 11:50, 22. Aug. 2009

2009-08-22T11:50:42Z

← Nächstältere Version		Version vom 11:50, 22. Aug. 2009
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	\end{align}</math>}}		\end{align}</math>}}

-	Wir sehen dass ~~nachdem~~ <math>i^4=1</math>, können wir <math>i^{11}</math> und <math>i^{20}</math> in ~~Termen~~ von <math>i^4</math> zerlegen,	+	Wir sehen, dass <math>i^4=1</math>, deshalb können wir <math>i^{11}</math> und <math>i^{20}</math> in Terme von <math>i^4</math> zerlegen,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

2009-05-12T08:45:14Z

← Nächstältere Version		Version vom 08:45, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~Let's begin by calculating some powers of~~ ''i'',	+	Wir berechnen einige Potenzen von ''i'', um zu sehen was passiert,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Now, we observe that because~~ <math>i^4=1</math>, ~~we can try to factorize~~ <math>i^{11}</math> ~~and~~ <math>i^{20}</math> in ~~terms of~~ <math>i^4</math>,	+	Wir sehen dass nachdem <math>i^4=1</math>, können wir <math>i^{11}</math> und <math>i^{20}</math> in Termen von <math>i^4</math> zerlegen,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 14:		Zeile 14:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The answer becomes~~	+	Wir erhalten also

	{{Abgesetzte Formel\|\|<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}

2009-03-11T10:30:15Z

← Nächstältere Version

Version vom 10:30, 11. Mär. 2009

2009-03-10T13:05:34Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 13:05, 10. Mär. 2009
Zeile 1:		Zeile 1:
	Let's begin by calculating some powers of ''i'',		Let's begin by calculating some powers of ''i'',

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	i^2 &= i\cdot i = -1\,,\\[5pt]		i^2 &= i\cdot i = -1\,,\\[5pt]
	i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt]		i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt]
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	Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>,		Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt]		i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt]
	i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.}		i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.}
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	The answer becomes		The answer becomes

-	{{~~Displayed math~~\|\|<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}

2008-10-29T15:07:12Z

← Nächstältere Version		Version vom 15:07, 29. Okt. 2008
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-	~~{{NAVCONTENT_START}}~~	+	Let's begin by calculating some powers of ''i'',
-	Let's begin by calculating some powers of i:	+

-	<math>\begin{align}i^2&=i\cdot i=-1,\\	+	{{Displayed math\|\|<math>\begin{align}
-	i^3&=i^2\cdot i = (-1)\cdot i = -i,\\	+	i^2 &= i\cdot i = -1\,,\\[5pt]
-	i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\end{align}</math>	+	i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt]
		+	i^4 &= i^2\cdot i^2 = (-1)\cdot (-1) = 1\,\textrm{.}
		+	\end{align}</math>}}

-	Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>,	+	Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>,

-	<math>\begin{align}~~i^2&=i\cdot i=-1,\\~~	+	{{Displayed math\|\|<math>\begin{align}
-	i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\	+	i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt]
-	i^{20}&=i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1=1\end{align}</math>	+	i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.}
		+	\end{align}</math>}}

	The answer becomes		The answer becomes

-	<math>i^{20}+i^{11}=1-i</math>	+	{{Displayed math\|\|<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}
-		+
-	~~{{NAVCONTENT_STOP~~}}	+

2008-09-18T12:05:19Z

← Nächstältere Version		Version vom 12:05, 18. Sep. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_1f~~.~~gif]]~~ </~~center~~>	+	Let's begin by calculating some powers of i:
		+
		+	<math>\begin{align}i^2&=i\cdot i=-1,\\
		+	i^3&=i^2\cdot i = (-1)\cdot i = -i,\\
		+	i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\end{align}</math>
		+
		+	Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>,
		+
		+	<math>\begin{align}i^2&=i\cdot i=-1,\\
		+	i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\
		+	i^{20}&=i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1=1\end{align}</math>
		+
		+	The answer becomes
		+
		+	<math>i^{20}+i^{11}=1-i</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-09-16T14:50:41Z

← Nächstältere Version

Version vom 14:50, 16. Sep. 2008

2008-08-21T08:05:04Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-04-07T10:42:48Z

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