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I am trying to use double interpolation(linear and forward quartic) but the interpolator is the zero coupon so at each time i need to convert the instantaneous forward rate to zc my problem is at the junction suppose i would like to ensure continuity in the instantaneous forward. Let s formulate the problem and how i think it could be solved and if someone can help i will appreciate. Suppose i have (t1,…,tn) with (zc1,…,zcn) and i want to have linear interpolation at the first pillars (t1,…,ti) and (ti,…,tn) forward quartic in such a way i ensure that fwd(ti+,linear) =fwd(ti-,quartic). I think a solution is to introduce a pillar before ti and use it but i don’t know how to do it properly. How can we solve this problem please. Thanks

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If the problem is continuity, most types of interpolation (incl. linear and quartic splines) ensure that the interpolation is a continuous function -- switching types of interpolation shouldn't cause a discontinuity as long as the linear interpolation goes between $t_0$ and $t_i$ and the quartic interpolation goes between $t_i$ and $t_n$ -- both linear and quartic should result in continuity at $t_i$ (as well as all other points).

If you're concerned about smoothness, the way splines remain visually smooth is by ensuring that the derivatives match at the pillars, and in general there needs to be some decision made about what the derivatives must equal at the end points (in your case, $t_i$ and $t_n$. So depending on how exactly you plan to fit this spline, there might be a place where you can input your choices for the derivatives at those points, or you might have to code it yourself. In either case, if I've understood you correctly, matching the derivatives at $t_i$ between the linear section and the spline section should do what you're looking for.

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