# Taylor Series expansion for an implicitely defined family of functions

Can we find a Taylor Series expansion for $y(x)$ implicitly defined by: $$\sum _{i=1}^nA_ie^{a_ix+b_iy} = 1 ?$$

In financial mathematics, the two-additive-factors Model G2++ is commonly used for interest rates forecasting. G2++ Model parameters are calibrated on market Caps and Swaptions prices. On the basis of these calibrated parameters complex financial product prices are computed using Monte-Carlo methods.

For further details on G2++ models refer for example to Brigo & Mercurio

Numerical computational challenges arise due to the necessity to integrate an implicitly defined function of the form:

$$\sum _{i=1}^nA_ie^{a_ix+b_iy} = 1 ,$$

which implicitly defines $y(x)$. All other parameters $A_i$, $a_i$ and $b_i$ are constant positive real numbers.

Now comes my question: What is the Taylor of $y(x)$ around $x=0$?

I started with a highly simplified version of the equation:

$$e^{x+y}+e^{x+by} = 1,$$

And found explicit closed form solutions for $y(x)$ when $b \in \{0,1,2,3,4\}$. This corresponds to the cases when the equation becomes Galois solvable. But what if $b$ is irrational, or arbitrarily large?

At least for this simplified single parametric version, wouldn't there be a strategy to obtain Taylor Series expansions?

Looking at the case $b=2$, it seems that $y(x)$ is really close to its linear asymptotes $x+y=0$ and $x+by=0$. maybe a clue? see here.
• I can't give you a reference. May be somone else will have a suggestion. What I had in mind is more or less in "Elementary Theory of Analytic Functions of One or Several Complex Variables" by H. Cartan. In Chapter 1, section 9 he goes over the inverse function theorem. The recurssion in (9.1) involves the polynomial $P_n$ that has coefficients in $\mathbb N_0$, so $|P_n(a_1,..b_{n-1})| \le P_n( |a_1|,...,|b_{n-1}|)$, which leads to the majorization in (9.2), etc. – VictorZurkowski Aug 22 '16 at 8:41
• Thanks a lot that reference really helped! Even if we are not dealing with inverse functions here it is totally applicable to the situation. It is sufficient to know $a_0$ (by numerically solving $e^{a_0}+e^{b a_0}=1$), to recursively deduct the following coefficients. $$a_1 = -1/{(e^{a_0}*(1-b)+b)},$$ $$a2 = -((1-e^{a_0} (a_1 b+1)^2+e^{a_0}*(a_1+1)^2)/(2*(b*(1-e^{a_0})+e^{a_0})), etc...$$ However, the convergence appears to be rather slow and there is no benefit gained from the fact that $y(x)+x$ tends towards $0$ at $+\infty$. Still a great step forward. Thanks a lot!! – handelskai Aug 22 '16 at 12:21