How to Find the Maclaurin Series of ( dfrac{e^z}{1-z} ) Through Convolution

The Maclaurin series expansion is a fundamental tool in mathematics, particularly useful for functions that are difficult to compute directly. We aim to derive the Maclaurin series for ( dfrac{e^z}{1-z} ) by multiplying two well-known series. This approach not only simplifies the process but also provides a deeper understanding of the underlying structure of the function.

Step 1: Representing Well-Known Maclaurin Series

The Maclaurin series for ( dfrac{1}{1-z} ) is given by:

[ dfrac{1}{1-z} sum_{k0}^{infty} z^k ]

And the Maclaurin series for ( e^z ) is:

[ e^z sum_{k0}^{infty} dfrac{1}{k!} z^k ]

Step 2: Multiplying the Series

By multiplying these two series, we obtain the following:

[ dfrac{e^z}{1-z} left( sum_{k0}^{infty} dfrac{1}{k!} z^k right) left( sum_{l0}^{infty} z^l right) ]

This product can be represented as a two-dimensional sum:

[ dfrac{e^z}{1-z} sum_{l0}^{infty} sum_{k0}^{l} dfrac{1}{k!} z^{kl} ]

Step 3: Rewriting the Series

Next, we rewrite this infinite sum in a more manageable form. We define a new index ( m kl ) and transform the series:

[ dfrac{e^z}{1-z} sum_{m0}^{infty} left( sum_{k0}^{m} dfrac{1}{k!} right) z^m ]

Here, we introduce a coefficient ( a_m ) defined as:

[ a_m sum_{k0}^{m} dfrac{1}{k!} ]

Thus, the resulting series can be written as:

[ dfrac{e^z}{1-z} sum_{m0}^{infty} a_m z^m ]

Step 4: Deriving the Coefficients

We now use the coefficient definition to derive the values of ( a_m ). From the definition, we know:

[ a_0 1 ]

For ( m > 0 ), we have:

[ a_m a_{m-1} dfrac{1}{m!} ]

By simplifying the fractions, we obtain:

[ dfrac{e^z}{1-z} 1 z left( dfrac{1}{1!} dfrac{z}{1!} right) left( dfrac{1}{1!} dfrac{z^2}{2!} right) ldots ]

This is a more compact way of expressing the series, where each term is formed by the convolution of the series coefficients:

[ dfrac{e^z}{1-z} 1 z dfrac{5}{2} z^2 dfrac{8}{3} z^3 ldots ]

This simplification shows the pattern clearly, making it easier to understand and verify each term.

Step 5: Convolution and General Form

In general, a convolution of two series is given by:

[ sum_{k0}^{m} a_k b_{m-k} ]

Here, ( a_k dfrac{1}{k!} ) is part of the exponential series, and ( b_{m-k} 1 ) is part of the geometric series. Since all ( b_{m-k} ) are equal to 1, the convolution simplifies to a sum of simple terms:

[ sum_{k0}^{m} dfrac{1}{k!} ]

This process highlights the power and utility of using convolution in expanding complex functions as Maclaurin series.