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Using [tex]$x_{n+1}=1-\frac{3}{x_n^2}$[/tex] with [tex]$x_0=-2.5$[/tex], find the values of [tex][tex]$x_1, x_2$[/tex][/tex], and [tex]$x_3$[/tex].

[tex]\[
\begin{array}{l}
x_1 = \square \\
x_2 = \square \\
x_3 = \square
\end{array}
\][/tex]


Sagot :

Sure! Let's find the values of [tex]\( x_1 \)[/tex], [tex]\( x_2 \)[/tex], and [tex]\( x_3 \)[/tex] using the recursive formula [tex]\( x_{n+1} = 1 - \frac{3}{x_n^2} \)[/tex] starting with [tex]\( x_0 = -2.5 \)[/tex].

1. Finding [tex]\( x_1 \)[/tex]:
Using the initial value [tex]\( x_0 = -2.5 \)[/tex],
[tex]\[ x_1 = 1 - \frac{3}{x_0^2} \][/tex]
[tex]\[ x_1 = 1 - \frac{3}{(-2.5)^2} \][/tex]
[tex]\[ x_1 = 1 - \frac{3}{6.25} \][/tex]
[tex]\[ x_1 = 1 - 0.48 \][/tex]
[tex]\[ x_1 = 0.52 \][/tex]

2. Finding [tex]\( x_2 \)[/tex]:
Using the value [tex]\( x_1 = 0.52 \)[/tex],
[tex]\[ x_2 = 1 - \frac{3}{x_1^2} \][/tex]
[tex]\[ x_2 = 1 - \frac{3}{(0.52)^2} \][/tex]
[tex]\[ x_2 = 1 - \frac{3}{0.2704} \][/tex]
[tex]\[ x_2 = 1 - 11.094674556213016 \][/tex]
[tex]\[ x_2 = -10.094674556213016 \][/tex]

3. Finding [tex]\( x_3 \)[/tex]:
Using the value [tex]\( x_2 = -10.094674556213016 \)[/tex],
[tex]\[ x_3 = 1 - \frac{3}{x_2^2} \][/tex]
[tex]\[ x_3 = 1 - \frac{3}{(-10.094674556213016)^2} \][/tex]
[tex]\[ x_3 = 1 - \frac{3}{102.90465879219869} \][/tex]
[tex]\[ x_3 = 1 - 0.02943991896746737 \][/tex]
[tex]\[ x_3 = 0.9705600810325326 \][/tex]

So, the values are:
[tex]\[ \begin{array}{l} x_1 = 0.52 \\ x_2 = -10.094674556213016 \\ x_3 = 0.9705600810325326 \end{array} \][/tex]
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