hcju/proofwikiqa huggingface dataset:
We have: {{begin-eqn}} {{eqn | l = F_{k_j + 1} \hat \phi + F_{k_j}       | r = \hat \phi^{k_j + 1}       | c = Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less }} {{eqn | ll= \leadsto       | l = F_{k_j + 1}       | r = \hat \phi^{k_j} - \frac {F_{k_j} } {\hat \phi}       | c =  }} {{eqn | r = \hat \phi^{k_j} + \phi F_{k_j}       | c = Reciprocal Form of One Minus Golden Mean }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1}       | r = \phi \paren {F_{k_1} + F_{k_2} + \cdots + F_{k_r} } + \paren {\hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} }       | c =  }} {{eqn | r = \phi n + \paren {\hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} }       | c =  }} {{end-eqn}} We have that: :$\hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots \le \hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} \le \hat \phi^2 + \hat \phi^4 + \hat \phi^6 + \cdots$ Then: {{begin-eqn}} {{eqn | l = \hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots       | r = \hat \phi^3 \paren {1 + \hat \phi^2 + \hat \phi^4 + \cdots}       | c =  }} {{eqn | r = \dfrac {\hat \phi^3} {1 - \hat \phi^2}       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \dfrac {\phi^2 \hat \phi^3} {\phi^2  - \phi^2 \hat \phi^2}       | c =  }} {{eqn | r = \dfrac {\paren {-1}^2 \hat \phi} {\phi^2 - \paren {-1}^2}       | c = Golden Mean by One Minus Golden Mean equals Minus 1 }} {{eqn | r = \dfrac {\hat \phi} {\phi^2 - 1}       | c =  }} {{eqn | r = \dfrac {1 - \phi} {\phi^2 - 1}       | c = {{Defof|One Minus Golden Mean}} }} {{eqn | r = \dfrac {1 - \phi} {1 + \phi - 1}       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi^{-1} - 1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \hat \phi^2 + \hat \phi^4 + \hat \phi^6 + \cdots       | r = \frac 1 {\hat \phi} \paren {\hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots}       | c =  }} {{eqn | r = \frac 1 {\hat \phi} \paren {\phi^{-1} - 1}       | c = from above }} {{eqn | r = -\phi \paren {\frac 1 \phi - 1}       | c = Reciprocal Form of One Minus Golden Mean }} {{eqn | r = -1 + \phi       | c =  }} {{eqn | r = \frac 1 {\phi}       | c = {{Defof|Golden Mean|index = 3}} }} {{eqn | r = \phi^{-1}       | c =  }} {{end-eqn}} Thus: :$\phi^{-1} - 1 \le \hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} \le \phi^{-1}$ and the result follows. {{qed}}
Follows directly from Zeckendorf Representation of Integer shifted Left, substituting $F_{k_j - 1}$ for $F_{k_j}$ throughout. {{qed}}
Let $S \left({x}\right)$ converge when $x = x_0$. Then by Existence of Interval of Convergence of Power Series, $x_0$ is in the interval of convergence of $S \left({x}\right)$ whose midpoint is $0$. The result follows by definition of interval of convergence. {{qed}}
By definition: :$G \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$ :$H \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} b_n z^n$ Let $G \left({z}\right)$ and $H \left({z}\right)$ converge to $x$ and $y$ respectively for some $z_0 \in \R_{>0}$. Then from Linear Combination of Convergent Series: :$\displaystyle \sum_{n \mathop \ge 0} \left({\alpha a_n + \beta b_n}\right) z^n = \alpha \sum_{n \mathop \ge 0} a_n z^n + \beta \displaystyle \sum_{n \mathop \ge 0} b_n z^n$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = z^m G \left({z}\right)       | r = z^m \sum_{n \mathop \ge 0} a_n z^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n + m}       | c =  }} {{eqn | r = \sum_{n + m \mathop \ge 0} a_{n - m} z^n       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{n \mathop \ge m} a_{n - m} z^n       | c =  }} {{end-eqn}} By letting $a_n = 0$ for all $n < 0$: :$z^m G \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} a_{n - m} z^n$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = z^{-m} G \left({z}\right)       | r = z^{-m} \sum_{n \mathop \ge 0} a_n z^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n - m}       | c =  }} {{eqn | r = \sum_{n + m \mathop \ge 0} a_{n + m} z^n       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{n \mathop \ge 0} a_{n + m} z^n + \sum_{k \mathop = -m}^{-1} a_{k + m} z^k       | c = splitting up and changing variable  }} {{eqn | r = \sum_{n \mathop \ge 0} a_{n + m} z^n + \sum_{k \mathop = 0}^{m - 1} a_k z^{k - m}       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{n \mathop \ge 0} a_{n + m} z^n + z^{-m} \sum_{k \mathop = 0}^{m - 1} a_k z^k       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{n \mathop \ge 1} a_n z^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop > m} a_n z^n + \sum_{n \mathop = 1}^m a_n z^n       | c =  }} {{eqn | r = \sum_{n \mathop > m} \paren {\sum_{n \mathop = 1}^m c_k a_{n-k} } z^n + \sum_{n \mathop = 1}^m b_n z^n       | c = from sequence definition }} {{eqn | r = \sum_{n \mathop = 1}^m c_k \paren {\sum_{n \mathop > m}a_{n-k} z^n} + \sum_{n \mathop = 1}^m b_n z^n       | c = Exchange of Order of Summation }} {{eqn | r = \sum_{n \mathop = 1}^m c_k \paren {z^m \map G z} + \sum_{n \mathop = 1}^m b_n z^n       | c = Generating Function by Power of Parameter }} {{eqn | r = \map G z \sum_{n \mathop = 1}^m c_k z^m + \sum_{n \mathop = 1}^m b_n z^n       | c =  }} {{eqn | ll= \leadsto       | l = \map G z \paren {1 - \sum_{n \mathop = 1}^m c_k z^m}       | r = \sum_{n \mathop = 1}^m b_n z^n       | c =  }} {{eqn | ll= \leadsto       | l = \map G z       | r = \dfrac {\sum_{n \mathop = 1}^m b_n z^n} {1 - \sum_{n \mathop = 1}^m c_k z^m}       | c =  }} {{eqn | r = \dfrac {b_1 z + b_2 z^2 + \cdots + b_m z^m} {1 - c_1 z - c_2 z^2 - \cdots - c_m z^m}       | c =  }} {{end-eqn}} {{qed}}
By definition of generating function: :$\map G z = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$ :$\map H z = \displaystyle \sum_{n \mathop \ge 0} b_n z^n$ Then: :$\begin{array}{c|ccccc} \times & b_0 & b_1 z & b_2 z^2 & b_3 z^3 & \cdots \\ \hline a_0 & a_0 b_0 & a_0 b_1 z & a_0 b_2 z^2 & a_0 b_3 z^3 & \cdots \\ a_1 z & a_1 b_0 z & a_1 b_1 z^2 & a_1 b_2 z^3 & a_1 b_3 z^4 & \cdots \\ a_2 z^2 & a_2 b_0 z^2 & a_2 b_1 z^3 & a_2 b_2 z^4 & a_2 b_3 z^5 & \cdots \\ a_3 z^3 & a_3 b_0 z^3 & a_3 b_1 z^4 & a_3 b_2 z^5 & a_3 b_3 z^6 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots  \end{array}$ The result follows by definition of generating function. {{qed}}
By definition of sequence of partial sums of $s$: {{begin-eqn}} {{eqn | l = \left\langle{c_n}\right\rangle       | r = a_0 + \left({a_0 + a_1}\right) + \left({a_0 + a_1 + a_2}\right) + \cdots       | c =  }} {{eqn | r = \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n a_k}\right)       | c =  }} {{end-eqn}} Consider the sequence $\left\langle{b_n}\right\rangle$ defined as: :$\forall n \in \Z_{\ge 0}: b_n = 1$ Let $H \left({z}\right)$ be the generating function for $\left \langle {b_n}\right \rangle$. By Generating Function for Constant Sequence: :$H \left({z}\right) = \dfrac 1 {1 - z}$ Then: {{begin-eqn}} {{eqn | l = G \left({z}\right) H \left({z}\right)       | r = \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n a_k b_{n - k} }\right) z^n       | c = Product of Generating Functions }} {{eqn | ll= \leadsto       | l = \dfrac 1 {1 - z} G \left({z}\right)       | r = \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n a_k}\right) z^n       | c = as all $b_{n - k} = 1$ }} {{eqn | r = \sum_{n \mathop \ge 0} c_n z^n       | c =  }} {{end-eqn}} Hence the result by definition of generating function. {{qed}}
Let $G \left({z}\right) H \left({z}\right)$ be the generating function for the sequence $\left\langle{c'_n}\right\rangle$. By definition of generating function: {{begin-eqn}} {{eqn | l = G \left({z}\right) H \left({z}\right)       | r = \sum_{k \mathop \ge 0} \dfrac {a_k} {k!} z^k \sum_{k \mathop \ge 0} \dfrac {b_k} {k!} z^k       | c =  }} {{eqn | r = \left({\dfrac {a_0} {0!} + \dfrac {a_1} {1!} z + \dfrac {a_2} {2!} z^2 + \cdots}\right) \left({\dfrac {b_0} {0!} + \dfrac {b_1} {1!} z + \dfrac {b_2} {2!} z^2 + \cdots}\right)       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = c'_n       | r = \sum_{k \mathop = 0}^n \dfrac {a_k} {k!} \dfrac {b_{n - k} } {\left({n - k}\right)!}       | c = Product of Generating Functions }} {{eqn | r = \dfrac 1 {n!} \left({\sum_{k \mathop = 0}^n \dfrac {n!} {k! \left({n - k}\right)!} a_k b_{n - k} }\right)       | c =  }} {{eqn | r = \dfrac 1 {n!} \left({\sum_{k \mathop = 0}^n \dbinom n k a_k b_{n - k} }\right)       | c = {{Defof|Binomial Coefficient/Integers|index = 1|disp = Binomial Coefficient}} }} {{eqn | r = \dfrac {c_n} {n!}        | c = where $c_n = \displaystyle \sum_{k \mathop = 0}^n \dbinom n k a_k b_{n - k}$ }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = G \left({c z}\right)       | r = \sum_{n \mathop \ge 0} a_n \left({c z}\right)^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop \ge 0} \left({a_n c^n}\right) z^n       | c =  }} {{end-eqn}} Hence the result by definition of generating function. {{qed}}
Consider the sequence $\sequence {b_n}$ defined as: : $\forall n \in \Z_{\ge 0}: b_n = 1$ Let $\map H z$ be the generating function for $\sequence {b_n}$. Then: {{begin-eqn}} {{eqn | l = \map H z       | r = \dfrac 1 {1 - z}       | c = Generating Function for Constant Sequence }} {{eqn | ll= \leadsto       | l =\map H {c z}       | r = \dfrac 1 {1 - c z}       | c =  }} {{eqn | r = \sum_{n \mathop \ge 0} \paren {b_n c^n} z^n       | c = Generating Function of Multiple of Parameter }} {{eqn | r = \sum_{n \mathop \ge 0} c^n z^n       | c = as $b_n = 1$ throughout }} {{end-eqn}} The result follows from the definition of a generating function. {{qed}}
{{begin-eqn}} {{eqn | l = G \left({z}\right)       | r = \sum_{n \mathop \ge 0} a_n z^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}       | c = separating out odd and even integers }} {{eqn | l = G \left({-z}\right)       | r = \sum_{n \mathop \ge 0} a_n \left({-z}\right)^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop \ge 0} \left({-1}\right)^n a_n z^n       | c =  }} {{eqn | r = \sum_{r \mathop \ge 0} \left({-1}\right)^{2 r} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} \left({-1}\right)^{2 r + 1} a_{2 r + 1} z^{2 r + 1}       | c = separating out odd and even integers }} {{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} - \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}       | c =  }} {{eqn | ll= \leadsto       | l = G \left({z}\right) + G \left({-z}\right)       | r = 2 \sum_{r \mathop \ge 0} a_{2 r} z^{2 r}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{n \mathop \ge 0} a_n z^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}       | c = separating out odd and even integers }} {{eqn | l = \map G {-z}       | r = \sum_{n \mathop \ge 0} a_n \paren {-z}^n       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{n \mathop \ge 0} \paren {-1}^n a_n z^n       | c =  }} {{eqn | r = \sum_{r \mathop \ge 0} \paren {-1}^{2 r} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} \paren {-1}^{2 r + 1} a_{2 r + 1} z^{2 r + 1}       | c = separating out odd and even integers }} {{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} - \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \map G z - \map G {-z}       | r = 2 \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \omega^{-k r} G \left({\omega^k z}\right)       | r = \sum_{j \mathop \ge 0} a_j \omega^{k \left({j - r}\right)} z^j       | c = {{Defof|Generating Function}} }} {{eqn | ll= \leadsto       | l = \dfrac 1 m \, \sum_{0 \mathop \le k \mathop < m} \omega^{-k r} G \left({\omega^k z}\right)       | r = \dfrac 1 m \, \sum_{0 \mathop \le k \mathop < m} \,\, \sum_{j \mathop \ge 0} a_j \omega^{k \left({j - r}\right)} z^j       | c =  }} {{eqn | r = \dfrac 1 m \sum_{j \mathop \ge 0} a_j z^j \,\, \sum_{0 \mathop \le k \mathop < m} \omega^{k \left({j - r}\right)}       | c =  }} {{end-eqn}} By definition of $\omega$: :$\omega = \exp \dfrac {2 \pi i} m \implies \omega^m = 1$ and so: {{begin-eqn}} {{eqn | l = \sum_{0 \mathop \le k \mathop < m} \omega^{k \left({j - r}\right)}       | r = \begin{cases} \dfrac {1 - \omega^{m \left({j - r}\right)} } {1 - \omega^{j - r} } & : j - r \not \equiv 0 \pmod m \\ m & : j \equiv r \pmod m \end{cases}       | c = Sum of Geometric Progression }} {{eqn | ll= \leadsto       | l = \dfrac 1 m \sum_{0 \mathop \le k \mathop < m} \omega^{-k r} G \left({\omega^k z}\right)       | r = \dfrac 1 m \sum_{j \mathop \ge 0} m a_j \left[{j \equiv r \pmod m}\right] z^j       | c = where $\left[{\cdots}\right]$ is Iverson's convention }} {{eqn | r = \sum_{n \bmod m \mathop = r} a_n z^n       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac \d {\d z} \map G z       | r = \frac \d {\d z} \paren {\sum_{k \mathop \ge 0} a_k z^k}       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{k \mathop \ge 0} \paren {\frac \d {\d z} a_k z^k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} k a_k z^{k - 1}       | c = Power Rule for Derivatives }} {{eqn | r = \sum_{k \mathop \ge 1} k a_k z^{k - 1}       | c = as the zeroth term vanishes when $k = 0$ }} {{eqn | r = \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^k       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = G' \left({z}\right)       | r = \sum_{k \mathop \ge 0} \left({k + 1}\right) a_{k + 1} z^k       | c = Derivative of Generating Function }} {{eqn | ll= \leadsto       | l = z G' \left({z}\right)       | r = \sum_{k \mathop \ge 0} \left({k + 1}\right) a_{k + 1} z^{k + 1}       | c =  }} {{eqn | r = \sum_{k + 1 \mathop \ge 0} k a_k z^k       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{k \mathop \ge 0} k a_k z^k       | c = as the term vanishes when $k = 0$ }} {{end-eqn}} The result follows by definition of generating function. {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^z G \left({t}\right) \rd t       | r = \int_0^z \left({\sum_{k \mathop \ge 0} a_k t^k}\right) \rd t       | c = {{Defof|Generating Function}} }} {{eqn | r = \sum_{k \mathop \ge 0} \left({\int_0^z a_k t^k \rd t}\right)       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \left({\left[{a_k \dfrac {t^{k + 1} } {k + 1} }\right]_0^z}\right)       | c = Primitive of Power }} {{eqn | r = \sum_{k \mathop \ge 0} \left({a_k \dfrac {z^{k + 1} } {k + 1} }\right)       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{qed}}
Take the sequence: :$S_n = 1, 1, 1, \ldots$ From Generating Function for Constant Sequence, this has the generating function: :$\displaystyle \map G z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$ By Integral of Generating Function: {{begin-eqn}} {{eqn | l = \int_0^z \map G t \rd t       | r = \sum_{k \mathop \ge 1} \dfrac {z^k} k       | c =  }} {{eqn | r = z + \dfrac {z^2} 2 + \dfrac {z^3} 3 + \dfrac {z^4} 4 + \cdots       | c =  }} {{end-eqn}} which is the power series whose coefficients are $\sequence {a_n}$. But: :$\map G z = \dfrac 1 {1 - z}$ and so by Primitive of Reciprocal and the Integration by Substitution: {{begin-eqn}} {{eqn | l = \int_0^z \map G t \rd t       | r = \int_0^z \dfrac 1 {1 - t} \rd t       | c =  }} {{eqn | r = \map \ln {\dfrac 1 {1 - z} }       | c =  }} {{end-eqn}} The result follows from the definition of generating function. {{qed}}
Take the sequence: :$S_n = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$ From Generating Function for Sequence of Reciprocals of Natural Numbers, this has the generating function: :$\map H z = \map \ln {\dfrac 1 {1 - z} }$ By definition, $\sequence {a_n}$ is the sequence of partial sums of $\sequence {a_n}$. The result follows from Generating Function for Sequence of Partial Sums of Series. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 {\paren {1 - z}^{n + 1} }       | r = \paren {1 + \paren {-z} }^{- n - 1}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \binom {- n - 1} k \paren {-z}^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop \ge 0} \dbinom {n + 1 + k - 1} k \paren {-1}^k \paren {-z}^k       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \sum_{k \mathop \ge 0} \dbinom {n + k} k z^k       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sqbrk {z^n} \map G z       | r = \dfrac 1 {n!} \map {G^{\paren n} } 0       | c = Derivative of Generating Function/General Result/Corollary }} {{eqn | r = \dfrac 1 {n!} \paren {\dfrac {n!} {2 \pi i} \int_{\partial D} \frac {\map G z} {z^{n + 1} } \d z}       | c = Cauchy's Integral Formula/General Result }} {{eqn | r = \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }       | c =  }} {{end-eqn}} {{qed}}
=== Lemma 1 === {{:Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1}} === Lemma 2 === {{:Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2}} Then: {{begin-eqn}} {{eqn | l = e^{\map \ln {\map G z} }       | r = \map \exp {\sum_{k \mathop \ge 1} \dfrac {S_k z^k} k}       | c = Lemma 2 }} {{eqn | ll= \leadsto       | l = \map G z       | r = \prod_{k \mathop \ge 1} e^{S_k z^k / k}       | c = Exponential of Sum }} {{eqn | r = \paren {1 + S_1 z + \dfrac { {S_1}^2 z^2} {2!} + \cdots} \paren {1 + \dfrac {S_2 z^2} 2 + \dfrac { {S_2}^2 z^4} {2^2 \times 2!} + \cdots} \cdots       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}} }} {{eqn | r = \sum_{m \mathop \ge 0} \paren {\sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !} } z^m       | c =  }} {{end-eqn}} {{finish|It still needs to be demonstrated that $\paren {1 + S_1 z + \dfrac { {S_1}^2 z^2} {2!} + \cdots} \paren {1 + \dfrac {S_2 z^2} 2 + \dfrac { {S_2}^2 z^4} {2^2 \times 2!} + \cdots} \cdots {{=}} \sum_{m \mathop \ge 0} \paren {\sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop {{=}} m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !} } z^m$}} Thus, by definition of generating function: :$h_m = \displaystyle \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$
{{begin-eqn}} {{eqn | l = \map {G'} z       | r = \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} } }       | c =  }} {{eqn | r = \dfrac 1 {1 - z} \map {\dfrac \d {\d z} } {\map \ln {\dfrac 1 {1 - z} } } + \map \ln {\dfrac 1 {1 - z} } \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} }       | c = Product Rule for Derivatives }} {{eqn | r = \dfrac 1 {1 - z} \paren {\paren {-1} \dfrac {-1} {\paren {1 - z}^2} \paren {\dfrac 1 {1 - z} }^{-1} } + \map \ln {\dfrac 1 {1 - z} } \paren {-1} \paren {\dfrac {-1} {\paren {1 - z}^2} }       | c = Derivative of Logarithm Function, Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = \dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } + \dfrac 1 {\paren {1 - z}^2}       | c = simplifying }} {{end-eqn}} {{qed}}
Let $R$ be the radius of convergence of the power series: :$\displaystyle \map f x = \sum_{n \mathop = 0}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} x^n$ Then: {{begin-eqn}} {{eqn | l = \frac 1 R       | r = \lim_{n \mathop \to \infty} \frac {\size {\alpha \paren {\alpha - 1} \dotsm \paren {\alpha - n} } } {\paren {n + 1}!} \frac {n!} {\size {\alpha \paren {\alpha - 1} \dotsm \paren {\alpha - n + 1} } }       | c = Radius of Convergence from Limit of Sequence }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {\size {\alpha - n} } {n + 1}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Thus for $\size x < 1$, Power Series is Differentiable on Interval of Convergence applies: :$\displaystyle D_x \map f x = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} n x^{n - 1}$ This leads to: {{begin-eqn}} {{eqn | l = \paren {1 + x} D_x \map f x       | r = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} x^{n - 1} + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} x^n       | c =  }} {{eqn | r = \alpha + \sum_{n \mathop = 1}^\infty \paren {\frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {n!} + \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} } x^n       | c =  }} {{eqn | r = \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {\paren {n - 1}!} \paren {\frac 1 n + \frac 1 {\alpha - n} } x^n       | c =  }} {{eqn | r = \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {\paren {n - 1}!} \, \frac \alpha {n \paren {\alpha - n} } x^n       | c =  }} {{eqn | r = \alpha \paren {1 + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} x^n}       | c =  }} {{eqn | r = \alpha \, \map f x       | c =  }} {{end-eqn}} Gathering up: :$\paren {1 + x} D_x \map f x = \alpha \, \map f x$ Thus: :$\map {D_x} {\dfrac {\map f x} {\paren {1 + x}^\alpha} } = -\alpha \paren {1 + x}^{-\alpha - 1} \map f x + \paren {1 + x}^{-\alpha} D_x \map f x = 0$ So $\map f x = c \paren {1 + x}^\alpha$ when $\size x < 1$ for some constant $c$. But $\map f 0 = 1$ and hence $c = 1$. {{qed}}
From Sum over k of r-kt choose k by r over r-kt by z^k: :$\displaystyle \sum_n \dbinom {\alpha - n t} k \dfrac \alpha {\alpha - n t} z^n = x^\alpha$ where: : $z = x^{t + 1} - x^t$ : $x = 1$ for $z = 0$. Setting $t = 0$: {{begin-eqn}} {{eqn | l = \sum_k \dbinom {\alpha - n \times 0} n \dfrac \alpha {\alpha - n \times 0} z^n       | r = x^\alpha       | c =  }} {{eqn | ll= \leadsto       | l = \sum_n \dbinom \alpha n \dfrac \alpha \alpha z^n       | r = \left({1 + z}\right)^\alpha       | c =  }} {{eqn | ll= \leadsto       | l = \sum_n \dbinom \alpha n z^n       | r = \left({1 + z}\right)^\alpha       | c =  }} {{end-eqn}} {{qed}}