[M.B.E.]

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Great job ! Very good explanation !

But a small error here:

For the case of a=1,b=2, n = 2 (heads up) , this is:
c = 0.4

c=0.5

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No, that's still wrong. If there's only $2 in the pot, you don't want to be betting $2 half the time, because then your opponent could cream you by tightening up a little bit. So that when you both call, your opponent is the favourite to win $3 of your money, but the times your opponent folds and you call you'll only win $1 of his.

For example, suppose you're playing heads up against me with a=1 and b=2. You bet when you're dealt anything greater than 1/2, but I play tighter and only bet with something greater than 5/8. Let's calculate my EV.

The probability of a push (we both fold) is (1/2)(5/8) = 5/16.
The probability that you bet and I fold is (1/2)(5/8) = 5/16: I lose $1.
The probability that you fold and I bet is (1/2)(3/8) = 3/16: I win $1.
The probability we both bet is (1/2)(3/8).
The probability we both bet and I win is (1/2)(3/8)(5/8) = 15/128: I win $3.
The probability we both bet and I lose is (1/2)(3/8)(3/8) = 9/128: I lose $3.

So my total EV on each hand is

(5/16)(-$1) + (3/16)($1) + (15/128)($3) + (9/128)(-$3) = $(1/64).

But if you're playing optimally, how can I have positive EV of about 1.5 cents per hand?

Actually the correct answer is c = 2/3.