- #1
mathmari
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MHB
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Hey! ![Smiling face with smiling eyes :blush: 😊](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60a.png)
We have the iteration formula $$x_{j+1}=2x_j-3x_j^2$$ and using the starting point $x_0=0.3$ we get the following approximations for $\frac{1}{3}$ :
\begin{align*}&x_1=2x_0-3x_0^2=2\cdot 0.3-3\cdot 0.3^2=0.33 \\ &x_2=2x_1-3x_1^2=2\cdot 0.33-3\cdot 0.33^2=0.3333 \\ &x_3=2x_2-3x_2^2=2\cdot 0.3333-3\cdot 0.3333^2=0.33333333\end{align*}
Now I want to show that for $j\geq 1$ the first $2^j$ digits of $x_j$ are correct for $\frac{1}{3}$.
It is $\frac{1}{3}=0.33333333333333333333333333\ldots$.
The approximation $x_1$ has the first $2=2^1$ digits correct.
The approximation $x_2$ has the first $4=2^2$ digits correct.
The approximation $x_3$ has the first $8=2^3$ digits correct.To show that it holds for $j\geq 1$ do we have to use induction? :unsure:
![Smiling face with smiling eyes :blush: 😊](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60a.png)
We have the iteration formula $$x_{j+1}=2x_j-3x_j^2$$ and using the starting point $x_0=0.3$ we get the following approximations for $\frac{1}{3}$ :
\begin{align*}&x_1=2x_0-3x_0^2=2\cdot 0.3-3\cdot 0.3^2=0.33 \\ &x_2=2x_1-3x_1^2=2\cdot 0.33-3\cdot 0.33^2=0.3333 \\ &x_3=2x_2-3x_2^2=2\cdot 0.3333-3\cdot 0.3333^2=0.33333333\end{align*}
Now I want to show that for $j\geq 1$ the first $2^j$ digits of $x_j$ are correct for $\frac{1}{3}$.
It is $\frac{1}{3}=0.33333333333333333333333333\ldots$.
The approximation $x_1$ has the first $2=2^1$ digits correct.
The approximation $x_2$ has the first $4=2^2$ digits correct.
The approximation $x_3$ has the first $8=2^3$ digits correct.To show that it holds for $j\geq 1$ do we have to use induction? :unsure:
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