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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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up vote 2 down vote accepted

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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Does that mean we can always get $$\frac{d^kf(a)}{dx^k} = \gamma_k$$? – Gong-Yi Liao Aug 26 '10 at 14:57
Almost: replace a by 0 and divide by k!. – whuber Aug 26 '10 at 15:49
In fact, I got another form $$ (n+1)!\delta_{n+1} = \gamma_{n+1}+\sum_{k=1}^nk\gamma_k\cdot[(n-k)!\delta_{n-k}] $$ – Gong-Yi Liao Aug 30 '10 at 17:27

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