Lösung 3.3:1a - Versionsgeschichte

Sekretariat1 um 11:17, 28. Aug. 2009

Sekretariat1 — Fri, 28 Aug 2009 11:17:43 GMT

← Nächstältere Version		Version vom 11:17, 28. Aug. 2009
Zeile 1:		Zeile 1:
-	In Polarform haben wir den Moivreschen Satz,	+	In Polarform haben wir den Moivreschen Satz

	{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}

-	Also bringen wir <math>1+i</math> ~~auf~~ Polarform, verwenden den Moivreschen Satz und schreiben die Potenz zuletzt wieder in der Form <math>a+ib</math>.	+	Also bringen wir <math>1+i</math> in Polarform, verwenden den Moivreschen Satz und schreiben die Potenz zuletzt wieder in der Form <math>a+ib</math>.

	<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>		<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>

Moni um 20:51, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 20:51:33 GMT

← Nächstältere Version		Version vom 20:51, 22. Aug. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}

-	Also bringen wir <math>1+i</math> auf Polarform, ~~und~~ verwenden den Moivreschen Satz, und schreiben die Potenz zuletzt wieder in der Form <math>a+ib</math>.	+	Also bringen wir <math>1+i</math> auf Polarform, verwenden den Moivreschen Satz und schreiben die Potenz zuletzt wieder in der Form <math>a+ib</math>.

	<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>		<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>

-	Durch die Rechnungen oben sehen wir dass	+	Durch die Rechnungen oben sehen wir, dass

	{{Abgesetzte Formel\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}

Martin um 14:42, 18. Mai 2009

Martin — Mon, 18 May 2009 14:42:38 GMT

← Nächstältere Version		Version vom 14:42, 18. Mai 2009
Zeile 1:		Zeile 1:
-	Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,	+	In Polarform haben wir den Moivreschen Satz,

	{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}

-	~~The equation above is called de Moivre's formula.~~	+	Also bringen wir <math>1+i</math> auf Polarform, und verwenden den Moivreschen Satz, und schreiben die Potenz zuletzt wieder in der Form <math>a+ib</math>.
-		+
-	~~The plan is therefore to rewrite~~ <math>1+i</math> ~~in polar form~~, ~~raise the expression to the power <math>12</math> using de Moivre's formula and then to write the answer~~ in ~~the form~~ <math>a+ib</math>.	+

	<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>		<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>

-	~~Using the calculations above, we see that~~	+	Durch die Rechnungen oben sehen wir dass

	{{Abgesetzte Formel\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}

-	~~De Moivre's formula now gives~~	+	Und durch den Moivreschen Satz erhalten wir

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

Tekbot: Solution 3.3:1a moved to Lösung 3.3:1a: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:42:35 GMT

Solution 3.3:1a moved to Lösung 3.3:1a: Robot: moved page

← Nächstältere Version

Version vom 10:42, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:10:44 GMT

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

← Nächstältere Version		Version vom 13:10, 10. Mär. 2009
Zeile 1:		Zeile 1:
	Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,		Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,

-	{{~~Displayed math~~\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}

	The equation above is called de Moivre's formula.		The equation above is called de Moivre's formula.
Zeile 11:		Zeile 11:
	Using the calculations above, we see that		Using the calculations above, we see that

-	{{~~Displayed math~~\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}

	De Moivre's formula now gives		De Moivre's formula now gives

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(1+i)^{12}		(1+i)^{12}
	&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos \Bigl(12\cdot\frac{\pi}{4}\Bigr) + i\sin \Bigl(12\cdot\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]		&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos \Bigl(12\cdot\frac{\pi}{4}\Bigr) + i\sin \Bigl(12\cdot\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]

Tek um 08:50, 30. Okt. 2008

Tek — Thu, 30 Oct 2008 08:50:04 GMT

← Nächstältere Version		Version vom 08:50, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form:	+	Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,
-		+
-		+
-	~~<math>\left( r\left( \cos \alpha +i\sin \alpha \right) \right)^{n}=r^{n}\left( \cos n\alpha +i\sin n\alpha \right)</math>~~	+

		+	{{Displayed math\|\|<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}

	The equation above is called de Moivre's formula.		The equation above is called de Moivre's formula.

-	The plan is therefore to rewrite	+	The plan is therefore to rewrite <math>1+i</math> in polar form, raise the expression to the power <math>12</math> using de Moivre's formula and then to write the answer in the form <math>a+ib</math>.
-	<math>~~\text{~~1}+i</math>	+
-	in polar form, ~~raised~~ the expression to the power	+
-	<math>~~\text{~~12}</math>	+
-	using de Moivre's formula and then to write the answer in the form	+
-	<math>a+ib</math>.	+
-		+
-		+
-		+
-	~~[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]~~	+

		+	<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>

	Using the calculations above, we see that		Using the calculations above, we see that

		+	{{Displayed math\|\|<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}

-	~~<math>\text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)</math>~~	+	De Moivre's formula now gives
-		+
-		+
-	de Moivre's formula now gives	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	~~& \left~~( ~~\text{~~1}+i ~~\right~~)^{12}=\~~left~~( \sqrt{2} \~~right~~)^{12}\~~left~~( \cos 12\~~centerdot~~ \frac{\pi }{4}+i\sin 12\~~centerdot~~ \frac{\pi }{4} \~~right~~) \\	+	(1+i)^{12}
-	& =2^{~~\frac{~~1}{2}\~~centerdot~~ 12}\~~left~~( \cos 3\pi +i\sin 3\pi \~~right~~) \\	+	&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos \Bigl(12\cdot\frac{\pi}{4}\Bigr) + i\sin \Bigl(12\cdot\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]
-	& =2^{6~~}\left~~( -1+i\~~centerdot~~ 0 ~~\right~~)=64\~~centerdot \left~~( -1 ~~\right~~)~~=-64~~ \\	+	&= 2^{(1/2)\cdot 12}\Bigl(\cos 3\pi + i\sin 3\pi\Bigr)\\[5pt]
-	& \\	+	&= 2^6(-1+i\cdot 0)\\[5pt]
-	\end{align}</math>	+	&= 64\cdot (-1)\\[5pt]
		+	&= -64\,\textrm{.}
		+	\end{align}</math>}}

Ian um 07:06, 24. Okt. 2008

Ian — Fri, 24 Oct 2008 07:06:23 GMT

← Nächstältere Version		Version vom 07:06, 24. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form:
-	<~~center~~> ~~[[Image:3_3_1a-1~~(2)~~.gif]]~~ </~~center~~>	+
-	{~~{NAVCONTENT_STOP}~~}	+
-	{~~{NAVCONTENT_START}~~}	+	<math>\left( r\left( \cos \alpha +i\sin \alpha \right) \right)^{n}=r^{n}\left( \cos n\alpha +i\sin n\alpha \right)</math>
-	<~~center~~> ~~[[Image:3_3_1a-2(2).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+
		+	The equation above is called de Moivre's formula.
		+
		+	The plan is therefore to rewrite
		+	<math>\text{1}+i</math>
		+	in polar form, raised the expression to the power
		+	<math>\text{12}</math>
		+	using de Moivre's formula and then to write the answer in the form
		+	<math>a+ib</math>.
		+
		+

	[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]		[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]
		+
		+
		+	Using the calculations above, we see that
		+
		+
		+	<math>\text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)</math>
		+
		+
		+	de Moivre's formula now gives
		+
		+
		+	<math>\begin{align}
		+	& \left( \text{1}+i \right)^{12}=\left( \sqrt{2} \right)^{12}\left( \cos 12\centerdot \frac{\pi }{4}+i\sin 12\centerdot \frac{\pi }{4} \right) \\
		+	& =2^{\frac{1}{2}\centerdot 12}\left( \cos 3\pi +i\sin 3\pi \right) \\
		+	& =2^{6}\left( -1+i\centerdot 0 \right)=64\centerdot \left( -1 \right)=-64 \\
		+	& \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:1a moved to Solution 3.3:1a: Robot: moved page

Tekbot — Tue, 16 Sep 2008 15:03:01 GMT

Lösning 3.3:1a moved to Solution 3.3:1a: Robot: moved page

← Nächstältere Version

Version vom 15:03, 16. Sep. 2008

Madde um 12:39, 21. Aug. 2008

Madde — Thu, 21 Aug 2008 12:39:00 GMT

← Nächstältere Version		Version vom 12:39, 21. Aug. 2008
Zeile 5:		Zeile 5:
	<center> [[Image:3_3_1a-2(2).gif]] </center>		<center> [[Image:3_3_1a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]

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

Tekbot — Thu, 21 Aug 2008 08:11:14 GMT

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

← Nächstältere Version		Version vom 08:11, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_1a-1(2).gif]] </center>	+	<center> [[Image:3_3_1a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_1a-2(2).gif]] </center>	+	<center> [[Image:3_3_1a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}