Example

Given the recursive relationship [latex]{{z}_{n+1}}={{z}_{n}}+2,\quad{{z}_{0}}=4[/latex], generate several terms of the recursive sequence. We are given the starting value, [latex]{{z}_{0}}=4[/latex]. The recursive formula holds for any value of n, so if [latex]n = 0[/latex], then [latex]{{z}_{n+1}}={{z}_{n}}+2[/latex] would tell us [latex]{{z}_{0+1}}={{z}_{0}}+2[/latex], or more simply, [latex]{{z}_{1}}={{z}_{0}}+2[/latex]. Notice this defines [latex]{{z}_{1}}[/latex] in terms of the known [latex]{{z}_{0}}[/latex], so we can compute the value:

[latex]{{z}_{1}}={{z}_{0}}+2=4+2=6[/latex].

Now letting [latex]n = 1[/latex], the formula tells us [latex]{{z}_{1+1}}={{z}_{1}}+2[/latex], or [latex]{{z}_{2}}={{z}_{1}}+2[/latex]. Again, the formula gives the next value in the sequence in terms of the previous value.

[latex]{{z}_{2}}={{z}_{1}}+2=6+2=8[/latex]

Continuing,

[latex]{{z}_{3}}={{z}_{2}}+2=8+2=10\\{{z}_{4}}={{z}_{3}}+2=10+2=12[/latex]

Example

Given the recursive relationship [latex]{{z}_{n+1}}={{z}_{n}}\cdot{i}+(1-i),\quad{{z}_{0}}=4[/latex], generate several terms of the recursive sequence. We are given [latex]{{z}_{0}}=4[/latex]. Using the recursive formula:

[latex]{{z}_{1}}={{z}_{0}}\cdot{i}+(1-i)=4\cdot{i}+(1-i)=1+3i\\{{z}_{2}}={{z}_{1}}\cdot{i}+(1-i)=(1+3i)\cdot{i}+(1-i)=i+3{{i}^{2}}+(1-i)=i-3+(1-i)=-2\\{{z}_{3}}={{z}_{2}}\cdot{i}+(1-i)=(-2)\cdot{i}+(1-i)=-2i+(1-i)=1-3i\\{{z}_{4}}={{z}_{3}}\cdot{i}+(1-i)=(1-3i)\cdot{i}+(1-i)=i-3{{i}^{2}}+(1-i)=i+3+(1-i)=4\\{{z}_{5}}={{z}_{4}}\cdot{i}+(1-i)=4\cdot{i}+(1-i)=1+3i[/latex]

Notice this sequence is exhibiting an interesting pattern—it began to repeat itself.

Example

Determine if [latex]c=1+i[/latex] is part of the Mandelbrot set. We start with [latex]{{z}_{0}}=0[/latex]. We continue, omitting some detail of the calculations

[latex]{{z}_{1}}={{z}_{0}}^{2}+1+i=0+1+i=1+i\\{{z}_{2}}={{z}_{1}}^{2}+1+i={{(1+i)}^{2}}+1+i=1+3i\\{{z}_{3}}={{z}_{2}}^{2}+1+i={{(1+3i)}^{2}}+1+i=-7+7i\\{{z}_{4}}={{z}_{3}}^{2}+1+i={{(-7+7i)}^{2}}+1+i=1-97i[/latex]

We can already see that these values are getting quite large. It does not appear that [latex]c=1+i[/latex] is part of the Mandelbrot set.

Example

Determine if [latex]c=0.5i[/latex] is part of the Mandelbrot set. We start with [latex]{{z}_{0}}=0[/latex]. We continue, omitting some detail of the calculations

[latex]{{z}_{1}}={{z}_{0}}^{2}+0.5i=0+0.5i=0.5i\\{{z}_{2}}={{z}_{1}}^{2}+0.5i={{(0.5i)}^{2}}+0.5i=-0.25+0.5i\\{{z}_{3}}={{z}_{2}}^{2}+0.5i={{(-0.25+0.5i)}^{2}}+0.5i=-0.1875+0.25i\\{{z}_{4}}={{z}_{3}}^{2}+0.5i={{(-0.1875+0.25i)}^{2}}+0.5i=-0.02734+0.40625i[/latex]

While not definitive with this few iterations, it does appear that this value is remaining small, suggesting that 0.5i is part of the Mandelbrot set.