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Consider the following exponential expansion form: $$ \exp\left[\sum_{k=1}^\infty \gamma_k x^k\right] = \sum_{j=0}^\infty\delta_j x^j $$ where $\gamma_k$'s are known, and $\delta_0=1$, $$\delta_{j+1} = \frac{1}{j+1}\sum_{i=1}^{j+1}i\gamma_i\delta_{j+1-i}$$ Does anyone know how to prove it? |
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A brute-force approach would use a Taylor series expansion of $\exp(\cdot)$ at 0 and group terms appropriately to demonstrate the relationship. I have not attempted to do so myself but an inspection of the first few terms does indicate that the proposed relationship holds. Perhaps, there is a more elegant approach? $$\exp\left[\gamma_1 x^1\right] = 1 + \frac{\gamma_1 x^1}{1!} + \frac{\left(\gamma_1 x^1\right)^2}{2!} + \ldots$$ $$\exp\left[\gamma_2 x^2\right] = 1 + \frac{\gamma_2 x^2}{1!} + \frac{\left(\gamma_2 x^2\right)^2}{2!} + \ldots$$ $$\exp\left[\gamma_3 x^3\right] = 1 + \frac{\gamma_3 x^3}{1!} + \frac{\left(\gamma_3 x^3\right)^2}{2!} + \ldots$$ |
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Srikant asks for a "more elegant approach." Perhaps the following will respond to this challenge. Let the argument of the exponential be $f(x)$ (so its power series coefficients are the gammas) and let the right hand side be $g(x)$ (with deltas as its coefficients), so that by definition $$g(x) = \exp(f(x)).$$ Differentiating both sides and replacing $\exp(f)$ with $g$ yields $$g' = f' * g.$$ Writing this out as power series gives the desired result: the delta comes from $g'$ while the convolution of the gammas and deltas comes from $f' * g$. You don't have to worry about convergence (and the whole machinery of Taylor series), by the way: all these calculations can be performed in the ring of formal power series. |
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