Faster exp calculation

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Use exp(x) = exp(x.round(k)) * exp(x - x.round(k)).
exp(x.round(k)) is fast because calculation of x**n is fast.
exp(x-x.round(k) is fast because x**n/n! converges fast.

BigMath.exp(1, 10000);
# 0.062810s → 0.062706s (no difference)

BigMath.exp(BigDecimal(3).div(7, 1000), 1000)
# 0.01463742s → 0.00326753s

BigMath.exp(BigDecimal(3).div(7, 10000), 10000)
# 8.571512s → 0.552467s

Why this optimization?

Currently, exp(x) uses exp(x/10**cnt)**(10**cnt) to make |x|<1.
We can make |x| even smaller, but the computation cost of y**10 is large currently O(prec**2).
exp(x) = exp(x.round(k)) * exp(x - x.round(k)) might not be optimal for the long run, but considering the simplicity of the change, improvement is significant and worth enough.

- y = _exp_taylor(x, prec2)
+ x_small_prec = x.round(Integer.sqrt(prec2))
+ y = _exp_taylor(x_small_prec, prec2).mult(_exp_taylor(x - x_small_prec, prec2), prec2)