Hello again! Mr.Rob or if anyone else happens to read this, I have the same question except only two options and they are really tough. I did test to n=1 billion and got an answer, but I am not sure about it.

a)n^{1/2}

b)2^{(logn)1/2}

So there are two things throwing me off: first is the obvious square root on the exponent. The second thing is that the logn is base 10.

Function 'a' maps 1 000 000 000 to 31622

Function 'b' maps 1 000 000 000 to 8.0

The obvious answer is function b = bigOh(functiona)

However, because the second is exponential, I am very iffy. I got a question wrong because I tested to 100,000, but afterward, I realize it took 500, 000 for a certain function to pass different one specifically because... well, see for yourself. Observe the following:

a) 2^{2logn}

b) n^{2.5}

I didn't realize that with base 2, 'a' is essentially means 2^n. So at the time, ignoring that fact,

I simply ran a test to 100,000 which returned a bigger value for function 'b' than function 'a'. Function 'a' = BigOh(function 'b')

But for a test that ran to 500,000, function 'a' far surpassed function 'b'. Function 'b' = BigOh(function 'a')

And the reason that function 'a' took so long is because function 'a' has a constant divisor in its exponent because logn is of base 10, which becomes log_{2}n/log_{2}10 (constant - log_{2}n

In this thinking, for some value n_{o}, 2^{(logn)1/2} might surpass n^{1/2}