FIXED POINT ITERATION
We are not given a starting value h₀, so let h₀=0, then:
f(h₀)=0.605946 approx when we substitute for V and r.
Now substitute this value for h into the square root expression to get h₁:
h₁=0.664328
h₂=0.681836
h₃=0.687628
h₄=0.689600
h₅=0.690278
h₆=0.690511
h₇=0.690592
h₈=0.690620
We have achieved 4 decimal place accuracy with this iteration so the solution is h=0.6906.
If h₀=1 as possibly indicated in the table, we get f(h₀)=0.836961
and h₁=0.750068
h₂=0.712624
h₃=0.698428
h₄=0.693351
h₅=0.691575
h₆=0.690960
h₇=0.690747
h₈=0.690673
h₉=0.690648
h₁₀=0.690639
h₁₁=0.690636
So, again, h=0.6906 is the solution. Note that we needed more iterations using h₀=1 than we did for h₀=0.