Correct the returned series when factorizing, with tests

[lbenet]

This fixes #58 and other similar cases.

[lbenet]

I'll add a summary here: Currently, in master, we get the following:

julia> t = Taylor1(5)
 1.0 t + 𝒪(t⁶)

julia> 1/(1+t)
 1.0 - 1.0 t + 1.0 t² - 1.0 t³ + 1.0 t⁴ - 1.0 t⁵ + 𝒪(t⁶)

julia> t/(t+t^2) # should be identical to the result above
 1.0 - 1.0 t + 1.0 t² - 1.0 t³ + 1.0 t⁴ + 𝒪(t⁶)

julia> t*sqrt(1+t)
 1.0 t + 0.5 t² - 0.125 t³ + 0.0625 t⁴ - 0.0390625 t⁵ + 𝒪(t⁶)

julia> sqrt(t^2+t^3) # should be identical to the result above
 1.0 t + 0.5 t² - 0.125 t³ + 𝒪(t⁶)

This PR corrects this.

[lbenet]

Tests should pass now...

[coveralls]

Coverage increased (+0.01%) to 97.185% when pulling f9d1f0f on lb/truncation into 34f7a48 on master.

[lbenet]

@dpsanders I think this is ready!

[dpsanders]

I don't follow the calculations, but if the new tests work, LGTM!

[lbenet]

The basic idea is to mimic an "extension" of the order of the independent variable, by using zeros appropriately; this is used to compute the coefficients (which before were simply set to zero) to the correct order.

Considering the example of sqrt(t^2+t^3) with t=Taylor1(5), the idea is to treat the independent variable of the expression as being of order 7 (instead of order 5), because t^2 can be factorized (which are the two orders lost in the returned expression). So the idea is to mimic the behavior of sqrt( Taylor1(7)^2+Taylor1(7)^3 ) (using t and zeros for order 6 and 7) and return the result up to order 5.

I hope this short explanation makes some sense.

EDIT: Corrected the markdown...