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Consistent choice of words (start → starting), adjust a sentence.
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Jendrik Stelzner
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You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$, then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$, then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two startstarting terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$$$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,, $$ This resultsresulting in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$, then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$, then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two start terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$ This results in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$, then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$, then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two starting terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,, $$ resulting in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

Add missing commas.
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Jendrik Stelzner
  • 16.6k
  • 4
  • 29
  • 70

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$, then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$, then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two start terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$ This results in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$ then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$ then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two start terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$ This results in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$, then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$, then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two start terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$ This results in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$

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Jendrik Stelzner
  • 16.6k
  • 4
  • 29
  • 70

You get two different generating functions by using two different definitions of the Fibonacci sequence $(a_n)_n$: if we choose the starting terms as $a_0 = 1$ and $a_1 = 1$ then we will get the generating function $1 / (1 - z - z^2)$; if we choose $a_0 = 0$ and $a_1 = 1$ then we get the generating function $z / (1 - z - z^2)$ instead.

If more generally $(a_n)_n$ is any sequence defined via the recursive relation $a_{n + 2} = a_{n + 1} + a_n$ and two start terms $a_0$ and $a_1$, then the generating functon $F = \sum_{n = 0}^∞ a_n z^n$ will satisfy $$ F - a_1 z - a_0 = z^2 F + z F - a_0 z \,. $$ This results in the explicit formula $$ F = \frac{a_0 + (a_1 - a_0) z}{1 - z - z^2} \,. $$