Consider the recursion relation:
x_n = (x_(n-1)*x_(n-2))^k
Given x_1, x_2, and k, x_n can be found by this definition. Write a function which takes as input arguments x_1, x_2, n and k. The output should be x_n.
For example, if x_1=exp(1), x_2=pi, n=5, and k=5/9 then:
R = get_recurse(exp(1),pi,5,4/9)
R = 2.31187497992966
is there a story as to why this is "infernal"?
I hate recursion, so it is all infernal to me! Iteration is the way to go...
Test suite could helpfully include n=1 and n=2 tests. I thought that it needed extra complexity to handle these, but @bmtran's top solution shows it doesn't.
I spend a lot of time to get the correct R from the example. The issue is that you say k = 5/9 instead of 4/9 . Just a typo to correct . Thanks
it cost much too time,so .....
Fails for both n=1 and n=2?
holy moly, you posted this one mere seconds after i posted mine.
Fails for n=1?
that was a photo-finish :)
OK for both n=1 and n=2.
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