Aus Online Mathematik Brückenkurs 2

Version vom 15:12, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.4:2 moved to Solution 3.4:2: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 09:25, 28. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	If the equation has the root
-	<center> [[Image:3_4_2.gif]] </center>	+	<math>z=\text{1}</math>, this means, according to the factor rule, that the equation mustcontain the
-	{{NAVCONTENT_STOP}}	+	<math>z=\text{1}</math>, i.e. the polynomial on the left-hand side can be written as
		+
		+
		+	<math>z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right)</math>
		+
		+
		+	for some constants
		+	<math>A</math>
		+	and
		+	<math>B</math>. We can determine the other unknown factor using polynomial division:
		+
		+
		+	<math>\begin{align}
		+	& z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right) \\
		+	& z^{2}+Az+B=\frac{z^{3}-3z^{2}+4z-2}{z-1} \\
		+	& =\frac{z^{3}-z^{2}+z^{2}-3z^{2}+4z-2}{z-1} \\
		+	& =\frac{z^{2}\left( z-1 \right)-2z^{2}+4z-2}{z-1} \\
		+	& =z^{2}+\frac{-2z^{2}+4z-2}{z-1} \\
		+	& =z^{2}+\frac{-2z^{2}+2z-2z+4z-2}{z-1} \\
		+	& =z^{2}+\frac{-2z\left( z-1 \right)+2z-2}{z-1} \\
		+	& =z^{2}-2z+\frac{2z-2}{z-1} \\
		+	& =z^{2}-2z+\frac{2\left( z-1 \right)}{z-1} \\
		+	& =z^{2}-2z+2 \\
		+	\end{align}</math>
		+
		+
		+	Thus, the equation can be written as
		+
		+
		+	<math>\left( z-1 \right)\left( z^{2}-2z+2 \right)=0</math>
		+
		+
		+	The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor
		+	<math>z^{2}-2z+2</math>. This is because the left-hand side is zero only when at least one of the factors
		+	<math>z-\text{1}</math>
		+	or
		+	<math>z^{2}-2z+2</math>
		+	is zero, and we see directly that
		+	<math>z-\text{1}</math>
		+	is zero only when
		+	<math>z=\text{1}</math>.
		+
		+	Hence, we determine the roots by solving the equation
		+
		+
		+	<math>z^{2}-2z+2=0</math>
		+
		+
		+	Completing the square gives
		+
		+
		+	<math>\begin{align}
		+	& \left( z-\text{1} \right)^{2}-1^{2}+2=0 \\
		+	& \left( z-\text{1} \right)^{2}=-1 \\
		+	\end{align}</math>
		+
		+
		+	and taking the root gives that
		+	<math>z-\text{1}=\pm i</math>
		+	i.e.
		+	<math>z=1-i</math>
		+	and
		+	<math>z=1+i</math>.
		+
		+	The equation's other roots are
		+	<math>z=1-i</math>
		+	and
		+	<math>z=1+i</math>.
		+
		+	As an extra check, we investigate whether
		+	<math>z-\text{1}=\pm i</math>
		+	really are roots of the equation.
		+
		+
		+	<math>\begin{align}
		+	& z=1+i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\
		+	& =\left( \left( 1+i-3 \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\
		+	& =\left( \left( -2+i \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\
		+	& =\left( -2+i-2i-1+4 \right)\left( 1+i \right)-2 \\
		+	& =\left( 1-i \right)\left( 1+i \right)-2 \\
		+	& =1^{2}-i^{2}-2=1+1-2=0 \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& z=1-i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\
		+	& =\left( \left( 1-i-3 \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\
		+	& =\left( \left( -2-i \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\
		+	& =\left( -2-i+2i-1+4 \right)\left( 1-i \right)-2 \\
		+	& =\left( 1+i \right)\left( 1-i \right)-2 \\
		+	& =1^{2}-i^{2}-2=1+1-2=0 \\
		+	\end{align}</math>
		+
		+
		+	NOTE: Writing
		+
		+
		+	<math>z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2</math>
		+
		+
		+	is known as the Horner scheme and is used to reduce the amount of the arithmetical work.

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Version vom 09:25, 28. Okt. 2008

If the equation has the root \displaystyle z=\text{1}, this means, according to the factor rule, that the equation mustcontain the \displaystyle z=\text{1}, i.e. the polynomial on the left-hand side can be written as

\displaystyle z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right)

for some constants \displaystyle A and \displaystyle B. We can determine the other unknown factor using polynomial division:

\displaystyle \begin{align} & z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right) \\ & z^{2}+Az+B=\frac{z^{3}-3z^{2}+4z-2}{z-1} \\ & =\frac{z^{3}-z^{2}+z^{2}-3z^{2}+4z-2}{z-1} \\ & =\frac{z^{2}\left( z-1 \right)-2z^{2}+4z-2}{z-1} \\ & =z^{2}+\frac{-2z^{2}+4z-2}{z-1} \\ & =z^{2}+\frac{-2z^{2}+2z-2z+4z-2}{z-1} \\ & =z^{2}+\frac{-2z\left( z-1 \right)+2z-2}{z-1} \\ & =z^{2}-2z+\frac{2z-2}{z-1} \\ & =z^{2}-2z+\frac{2\left( z-1 \right)}{z-1} \\ & =z^{2}-2z+2 \\ \end{align}

Thus, the equation can be written as

\displaystyle \left( z-1 \right)\left( z^{2}-2z+2 \right)=0

The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor \displaystyle z^{2}-2z+2. This is because the left-hand side is zero only when at least one of the factors \displaystyle z-\text{1} or \displaystyle z^{2}-2z+2 is zero, and we see directly that \displaystyle z-\text{1} is zero only when \displaystyle z=\text{1}.

Hence, we determine the roots by solving the equation

\displaystyle z^{2}-2z+2=0

Completing the square gives

\displaystyle \begin{align} & \left( z-\text{1} \right)^{2}-1^{2}+2=0 \\ & \left( z-\text{1} \right)^{2}=-1 \\ \end{align}

and taking the root gives that \displaystyle z-\text{1}=\pm i i.e. \displaystyle z=1-i and \displaystyle z=1+i.

The equation's other roots are \displaystyle z=1-i and \displaystyle z=1+i.

As an extra check, we investigate whether \displaystyle z-\text{1}=\pm i really are roots of the equation.

\displaystyle \begin{align} & z=1+i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ & =\left( \left( 1+i-3 \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ & =\left( \left( -2+i \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ & =\left( -2+i-2i-1+4 \right)\left( 1+i \right)-2 \\ & =\left( 1-i \right)\left( 1+i \right)-2 \\ & =1^{2}-i^{2}-2=1+1-2=0 \\ \end{align}

\displaystyle \begin{align} & z=1-i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ & =\left( \left( 1-i-3 \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ & =\left( \left( -2-i \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ & =\left( -2-i+2i-1+4 \right)\left( 1-i \right)-2 \\ & =\left( 1+i \right)\left( 1-i \right)-2 \\ & =1^{2}-i^{2}-2=1+1-2=0 \\ \end{align}

NOTE: Writing

\displaystyle z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2

is known as the Horner scheme and is used to reduce the amount of the arithmetical work.