Round off to 4 decimal places

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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.

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