hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_{n \mathop \in \Z} \dbinom {r + s} n_q q^{n \left({n - 1}\right) / 2} x^n       | r = \prod_{k \mathop = 1}^{r + s} \left({1 + q^{k - 1} x}\right)       | c = Gaussian Binomial Theorem }} {{eqn | r = \prod_{k \mathop = 1}^r \left({1 + q^{k - 1} x}\right) \prod_{k \mathop = r + 1}^s \left({1 + q^{k - 1} x}\right)       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^r \left({1 + q^{k - 1} x}\right) \prod_{k \mathop = 1}^{s - r} \left({1 + q^{r + k - 1} x}\right)       | c = Translation of Index Variable of Product }} {{end-eqn}} {{ProofWanted|Not sure what direction to take this in}}
{{begin-eqn}} {{eqn | l = \dbinom n k_{q^{-1} }       | r = \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - \paren {q^{-1} }^{n - j} } {1 - \paren {q^{-1} }^{j + 1} }       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{eqn | r = \prod_{j \mathop = 0}^{k - 1} \dfrac {\frac {q^{n - j} - 1} {q^{n - j} } } {\frac {q^{j + 1} - 1} {q^{j + 1} } }       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{k - 1} \dfrac {q^{j + 1} } {q^{n - j} } \dfrac {q^{n - j} - 1} {q^{j + 1} - 1}       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{k - 1} q^{1 - n + 2 j} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \paren {q^{1 - n} }^k \prod_{j \mathop = 0}^{k - 1} q^{2 j} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = q^{k - n k} q^{k \paren {k - 1} } \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c = Closed Form for Triangular Numbers: $\displaystyle \sum_{j \mathop = 0}^{k - 1} 2 j = 2 \sum_{j \mathop = 1}^{k - 1} j = 2 \dfrac {\paren {k - 1} k} 2$ }} {{eqn | r = q^{k^2 - n k} \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \dfrac 1 {q^{k \paren {n - k} } } \dbinom n k_q       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$A_{n k} = \left({n + 1}\right) \dbinom n k - \dbinom n {k + 1} + \dbinom n k$ === Basis for the Induction === $P \left({0}\right)$ is the case $A_{0 k}$: Let $k = 0$. Then: {{begin-eqn}} {{eqn | l = A_{0 k}       | r = 1       | c =  }} {{eqn | r = \left({0 + 1}\right) \dbinom 0 0 - \dbinom 0 {0 + 1}       | c = Zero Choose n }} {{end-eqn}} Let $k > 0$. Then: {{begin-eqn}} {{eqn | l = A_{0 k}       | r = 0       | c =  }} {{eqn | r = \left({0 + 1}\right) \dbinom 0 k - \dbinom 0 {k + 1}       | c = Zero Choose n }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$A_{r k} = \left({r + 1}\right) \dbinom r k - \dbinom r {k + 1}$ from which it is to be shown that: :$A_{\left({r + 1}\right) k} = \left({r + 2}\right) \dbinom {r + 1} k - \dbinom {r + 1} {k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = A_{\left({r + 1}\right) k}       | r = A_{r k} + A_{r \left({k - 1}\right)} + \dbinom {r + 1} k       | c = by hypothesis }} {{eqn | r = \left({r + 1}\right) \dbinom r k - \dbinom r {k + 1} + \left({r + 1}\right) \dbinom r {k - 1} - \dbinom r k + \dbinom {r + 1} k       | c = Induction Hypothesis }} {{eqn | r = \left({r + 1}\right) \left({\dbinom r k + \dbinom r {k - 1} }\right) - \left({\dbinom r {k + 1} + \dbinom r k}\right) + \dbinom {r + 1} k       | c =  }} {{eqn | r = \left({r + 1}\right) \dbinom {r + 1} k - \dbinom {r + 1} {k + 1} + \dbinom {r + 1} k       | c = Pascal's Rule }} {{eqn | r = \left({r + 2}\right) \dbinom {r + 1} k - \dbinom {r + 1} {k + 1}       | c =  }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n, k \in \Z_{\ge 0}: A_{n k} = \left({n + 1}\right) \dbinom n k - \dbinom n {k + 1}$ {{qed}}
Let the elements of $S$ be (totally) ordered in some way, by assigning an index to each element. Thus let $S = \left\{ {a_1, a_2, a_3, \ldots, a_n}\right\}$. Thus each $k$-combination of $S$ with repetition can be expressed as: :$\left\{ {\left\{ {a_{r_1}, a_{r_1}, \ldots, a_{r_k} }\right\} }\right\}$ where: :$r_1, r_2, \ldots, r_k$ are all elements of $\left\{ {1, 2, \ldots, n}\right\}$ and such that: :$r_1 \le r_2 \le \cdots \le r_k$ Hence the problem reduces to the number of integer solutions $\left({r_1, r_2, \ldots, r_k}\right)$ such that $1 \le r_1 \le r_2 \le \cdots \le r_k \le n$. This is the same as the number of solutions to: :$0 < r_1 < r_2 + 1 < \cdots < r_k + k - 1 < n + k$ which is the same as the number of solutions to: :$0 < s_1 < s_2 < \cdots < s_k < n + k$ which is the number of ways to choose $k$ elements from $n + k - 1$. This is the same as the number of subsets as a set with $n + k - 1$ elements. From Cardinality of Set of Subsets: :$N = \dbinom {n + k - 1} k$ {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | o =        | r = \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m}       | c =  }} {{eqn | r = \sum_k \delta_{1 \paren {k + 1} } {k \brace m} \paren {-1}^{k - m}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = \sum_k \delta_{0 k} {k \brace m} \paren {-1}^{k - m}       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = {0 \brace m} \paren {-1}^{- m}       | c = All terms but where $k = 0$ vanish }} {{eqn | r = \delta_{0 m} \paren {-1}^{- m}       | c = Stirling Number of the Second Kind of 0 }} {{eqn | r = \delta_{0 m}       | c = multiplier irrelevant when $m \ne 0$ }} {{eqn | r = \sqbrk {0 \ge m} \dfrac {0!} {m!}       | c = Factorial of Zero }} {{end-eqn}} So $\map P 0$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \sqbrk {r + 1 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r \ge m} \dfrac {r!} {m!}$ from which it is to be shown that: :$\displaystyle \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m}       | c =  }} {{eqn | r = \sum_k \paren {\paren {r + 1} {r + 1 \brack k + 1} + {r + 1 \brack k} } {k \brace m} \paren {-1}^{k - m}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = \paren {r + 1} \sum_k {r + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}       | c =  }} {{eqn | r = \sqbrk {r \ge m} \dfrac {\paren {r + 1} r!} {m!} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}       | c = Induction Hypothesis }} {{eqn | r = \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k}       | c =  }} {{eqn | r = \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \delta_{m \paren {r + 1} }       | c = First Inversion Formula for Stirling Numbers }} {{eqn | r = \cases {\dfrac {\paren {r + 1}!} {m!} & : $r + 1 > m$ \cr 1 & : $r + 1 = m$ \cr 0 & : $r + 1 < m$} }} {{eqn | r = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!} }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$ {{qed}}
From Sum over $k$ of $\dbinom {m + r + s} k$ by $\dbinom {n + r - s} {n - k}$ by $\dbinom {r + k} {m + n}$: :$\displaystyle \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m - n} = \binom r m \binom s n$ Setting $\tuple {m, n, r, s, k} \gets \tuple {m + k, l - k, m + n, n + l, j}$: {{begin-eqn}} {{eqn | o =        | r = \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k}       | c =  }} {{eqn | r = \sum_{j, k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {l + k} j \dbinom {m + k} {l - k - j} \dbinom {m + n + j} {m + l}       | c =  }} {{eqn | r = \sum_{j, k \mathop \in \Z} \paren {-1}^k \dbinom {2 l + 2 j} {l - j + k} \dfrac {\paren {m + n + j}!} {\paren {2 l - 2 j}! \, j! \, \paren {m - l + j}! \, \paren {n + j - l}!}       | c =  }} {{end-eqn}} Unless $j = l$, the sum on $k$ vanishes. {{qed}} {{Namedfor|Alfred Cardew Dixon|cat = Dixon, A.C.}}
Follows directly from Dixon's Identity/General Case: :$\displaystyle \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$ setting $l = m = n$. {{qed}} {{Namedfor|Alfred Cardew Dixon|cat = Dixon, A.C.}}
We have: {{begin-eqn}} {{eqn | l = \dbinom y {n + 1}       | o = \le       | r = \dbinom x {n + 1}       | c = Ordering of Binomial Coefficients }} {{eqn | o = \le       | r = \dbinom {y + 1} {n + 1}       | c =  }} {{eqn | r = \dbinom y {n + 1} + \dbinom y n       | c = Pascal's Rule }} {{end-eqn}} {{finish}}
If $z \ge n$, then from Ordering of Binomial Coefficients: :$\dbinom z {n + 1} \le \dbinom y {n + 1}$ Otherwise $n - 1 \le z \le n$, and: :$\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$ In either case: :$(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$ Therefore: {{begin-eqn}} {{eqn | l = \dbinom {z + 1} {n + 1}       | r = \dbinom z {n + 1} + \dbinom z n       | c = Pascal's Rule }} {{eqn | o = \le       | r = \dbinom y {n + 1} + \dbinom z n       | c =  }} {{eqn | r = \dbinom x {n + 1}       | c = by hypothesis }} {{end-eqn}} and so $x \ge z + 1$. Now we are to show that every term of the summation: :$\displaystyle \binom x {n + 1} - \binom y {n + 1} = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - k} t_k$ where: :$t_k = \dbinom {x - z - 1 + k} {k + 1} - \dbinom {y - z - 1 + k} {k + 1}$ is negative. Because $z \ge n - 1$, the binomial coefficient $\dbinom {z - k} {n - k}$ is non-negative. Because $x \ge z + 1$, the binomial coefficient $\dbinom {x - z - 1 + k} {k + 1}$ is also non-negative. Therefore: :$z \le y \le x$ implies that: :$\dbinom {y - z - 1 + k} {k + 1} \le \dbinom {x - z - 1 + k} {k + 1}$ When $x = y$ and $z = n - 1$ the result becomes: :$\dbinom x n \le \dbinom x n + \dbinom {n - 1} {n - 1}$ which reduces to: :$\dbinom x n \le \dbinom x n + 1$ which is true. Otherwise: {{begin-eqn}} {{eqn | l = \dbinom x n - \dbinom y n - \dbinom z {n - 1}       | r = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - 1 - k} \left({t_k - \delta_{k 0} }\right)       | c = where $\delta_{k 0}$ is the Kronecker delta }} {{eqn | r = \sum_{k \mathop \ge 0} \dfrac {n - k} {z - n + 1} \dbinom {z - k} {n - k} \left({t_k - \delta_{k 0} }\right)       | c = Factors of Binomial Coefficient }} {{end-eqn}} This is less than or equal to: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \dfrac {n - 1} {z - n + 1} \dbinom {z - k} {n - k} \left({t_k - \delta_{k 0} }\right)       | r = \dfrac {n - 1} {z - n + 1} \left({\dbinom x {n + 1} - \dbinom y {n + 1} - \dbinom z n}\right)       | c =  }} {{eqn | r = 0       | c = because $t_0 - 1 = x - y - 1 \le 0$ }} {{end-eqn}} Hence the result. {{qed}}
From Lower and Upper Bound of Factorial, we have that: :$\dfrac {k^k} {e^{k - 1} } \le k!$ so that: :$(1): \quad \dfrac 1 {k!} \le \dfrac {e^{k - 1} } {k^k}$ Then: {{begin-eqn}} {{eqn | l = \dbinom n k       | r = \dfrac {n^\underline k} {k!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | o = \le       | r = \dfrac {n^k} {k!}       | c =  }} {{eqn | o = \le       | r = \dfrac {n^k e^{k - 1} } {k^k}       | c = from $(1)$ }} {{eqn | o = \le       | r = \dfrac {n^k e^k} {k^k}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Lower and Upper Bound of Factorial, we have that: :$k! \le \dfrac {k^{k + 1} } {e^{k - 1} }$ so that: :$(1): \quad \dfrac 1 {k!} \ge \dfrac {e^{k - 1} } {k^{k + 1} }$ Then: {{begin-eqn}} {{eqn | l = \dbinom n k       | r = \dfrac {n^\underline k} {k!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | o = \ge       | r = \dfrac {\left({n - k}\right)^k} {k!}       | c =  }} {{eqn | o = \ge       | r = \dfrac {\left({n - k}\right)^k e^{k - 1} } {k^{k + 1} }       | c = from $(1)$ }} {{eqn | r = \dfrac {\left({n - k}\right)^k e^k} {k^k} \dfrac 1 {e k}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = H_{2^m}^{\paren r}       | r = H_{2^{m - 1} }^{\paren r} + \dfrac 1 {\paren {2^{m - 1} + 1}^r} + \dfrac 1 {\paren {2^{m - 1} + 2}^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} + 2^m}^r}       | c =  }} {{eqn | o = \le       | r = H_{2^{m - 1} }^{\paren r} + \dfrac 1 {\paren {2^{m - 1} }^r} + \dfrac 1 {\paren {2^{m - 1} }^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} }^r}       | c =  }} {{eqn | r = H_{2^{m - 1} }^{\paren r} + \dfrac {2^m} {2^{\paren {m - 1} r} }       | c =  }} {{eqn | r = H_{2^{m - 1} }^{\paren r} + \dfrac 1 {2^{m \paren {r - 1} } }       | c =  }} {{eqn | ll= \leadsto       | l = H_{2^{m + 1} }^{\paren r}       | o = \le       | r = \sum_{0 \mathop \le k \mathop < m} \dfrac {2^k} {2^{k r} }       | c =  }} {{eqn | r = \dfrac {2^{r - 1} } {2^{r - 1} - 1}       | c = Sum of Geometric Sequence }} {{end-eqn}} {{qed}} {{mistake|I wish I wasn't so tired.}}
=== Step 1 === We will show that $\map \sigma \EE \subseteq \EE_\Omega$. Define:  :$\OO = \set {o \in \Omega: \EE_o \in \map \sigma \EE}$ By the definition of a $\sigma$-algebra: :$\EE_1 \subseteq \map \sigma \EE$. From $\sigma$-Algebra of Countable Sets, if $\beta$ immediately precedes $\alpha$, then $\EE_\alpha$ is a $\sigma$-algebra containing $\EE_\beta$. If $\beta$ strictly precedes $\alpha$ but is not an immediate predecessor of $\alpha$, then $\EE_\alpha$ is a countable union of measurable sets. By the definition of a $\sigma$-algebra and of union, $\EE_\alpha$ is a $\sigma$-algebra containing $\EE_\beta$. Thus the hypotheses of well-ordered induction are satisfied, and $\OO = \Omega$. Thus $\EE_\alpha \subseteq \map \sigma \EE$ for all $\alpha \in \Omega$. By Set Union Preserves Subsets:General Result: :$\ds \bigcup_{\alpha \mathop \in \Omega} \EE_\alpha \subseteq \map \sigma \EE$ Thus $\EE_{\Omega} \subseteq \map \sigma \EE$. {{qed|lemma}} === Step 2 === We identify numbers in $\N$ with the finite ordinals, using the definition of ordinals as initial segments. This identification is justified from Minimal Infinite Successor Set forms Peano Structure. Thus every $j \in \N$ can be treated as a finite initial segment of $\Omega$. By the definition of a $\sigma$-algebra generated by $\EE$, the reverse inclusion: :$\EE_\Omega \supseteq \map \sigma \EE$ will follow if $\EE_\Omega$ is a $\sigma$-algebra. First note that: :$\EE = \EE_1 \in \EE_\Omega$ so $\EE_\Omega$ has a unit. Let:  :$\family {E_j}_{j \mathop \in \N}$ be an arbitrary countable indexed family of sets in $\EE_\Omega$. By the inductive construction of $\EE_\Omega$, for all $j \in \N$: :$E_j, E_j^\complement \in \EE_{\alpha_j}$ for some initial segment $\alpha_j$,  Thus $\EE_\Omega$ is closed under complement. The set: :$\family {\alpha_i}_{i \mathop \in \N} = \set {\alpha_1, \alpha_2, \alpha_3, \ldots}$ is countable, because it is indexed by $\N$. By Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound, $\family {\alpha_i}$ has an upper bound. Call this bound $\gamma$. Then $\EE_\gamma$ is an initial segment. Also, $E_j \in \EE_\gamma$ for all $j \in \N$, by the definition of upper bound. For any $\delta$ strictly succeeding $\gamma$: :$\ds \bigcup_{j \mathop \in \N} E_j \in \EE_\delta$ Thus by the construction of $\EE_\Omega$: :$\EE_\delta \subseteq \EE_\Omega$ This means that $\EE_\Omega$ is closed under countable union. We have that: :$\EE_\Omega$ has a unit :$\EE_\Omega$  is closed under complement. :$\EE_\Omega$  is closed under countable union. By the definition of $\sigma$-algebra, $\EE_\Omega$ is indeed a $\sigma$-algebra. By the definition of a generated $\sigma$-algebra: :$\map \sigma \EE \subseteq \EE_\Omega$ The result follows from Part 1 and Part 2 of the proof combined, by the definition of set equality. {{qed}} {{AoC|Countable Union of Countable Sets is Countable}}
By definition of the generating function for Bernoulli numbers: {{begin-eqn}} {{eqn | l = \frac t {e^t - 1}       | r = \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k  }} {{end-eqn}} By Power Series Expansion for Exponential Function: {{begin-eqn}} {{eqn | l = e^{t x}       | r = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac {t e^{t x} } {e^t - 1}       | r = \paren {\sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k} \paren {\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k} }} {{end-eqn}} Combining like powers of $t$ we obtain:  {{begin-eqn}} {{eqn | l = \frac {t e^{t x} } {e^t - 1}       | r = \sum_{k \mathop = 0}^\infty t^k \sum_{m \mathop = 0}^k \frac {B_{k - m} } {\paren {k - m}!} \frac {x^m} {m!} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \frac {k!} {\paren {k - m}! m!} B_{k - m} x^m }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \binom k m B_{k - m} x^m        | c = {{Defof|Binomial Coefficient/Integers|index = 1|Binomial Coefficient}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k       | c = {{Defof|Bernoulli Polynomial}} }} {{end-eqn}} {{qed}} Category:Generating Functions Category:Bernoulli Polynomials 2ek5dmioawdr8d7for4w05jtm1vk19i
Let $\map G {t, x}$ denote the Generating Function of Bernoulli Polynomials: :$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$ Then: {{begin-eqn}} {{eqn | l = \map G {t, 1 - x}       | r = \frac {t e^{t \paren {1 - x} } } {e^t - 1} }} {{eqn | r = \frac {t e^{t - t x} } {e^t - 1} }} {{eqn | r = \frac {t e^{-t x} } {1 - e^{-t} } }} {{eqn | r = \frac {-t e^{-t x} } {e^{-t} - 1} }} {{eqn | r = \map G {-t, x} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \map G {t, 1 - x}       | r = \map G {-t, x} }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} {1 - x} } {k!} t^k        | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \map {B_k} x} {k!} t^k  }} {{eqn | ll= \leadsto       | l = \map {B_k} {1 - x}       | r = \paren {-1}^k \map {B_k} x }} {{end-eqn}} {{qed}} Category:Bernoulli Polynomials oxeh5ryiuvxvq35scu191wx0yy2329r
{{begin-eqn}} {{eqn | l = B_{2 n + 1}\left({1 - x}\right)       | r = \left({-1}\right)^{2 n + 1} B_{2 n + 1} \left({x}\right)       | c = Symmetry of Bernoulli Polynomial }} {{eqn | r = \left({-1}\right) B_{2 n + 1} \left({x}\right) }} {{eqn | ll= \leadsto       | l = B_{2 n + 1} \left({\frac 1 2}\right)       | r = \left({-1}\right) B_{2 n + 1} \left({\frac 1 2}\right) }} {{eqn | ll= \leadsto       | l = 2 B_{2 n + 1} \left({\frac 1 2}\right)       | r = 0 }} {{eqn | l = B_{2 n + 1} \left({\frac 1 2}\right)       | r = 0 }} {{end-eqn}} {{qed}} Category:Bernoulli Polynomials a1elo598r9iaz6g4rj5asn5bv1kbh9o
From Approximate Size of Sum of Harmonic Series: :$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$ where $0 < \epsilon < \dfrac 1 {252 n^6}$. We have that: :$\dfrac 1 {2 n} + \dfrac 1 {120 n^4} > \dfrac 1 {12 n^2} + \dfrac 1 {252 n^6}$ and the result follows. {{qed}}
The proof proceeds by induction over $n$. For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^1 \dfrac {x^k} k       | r = \dfrac {x^1} 1       | c =  }} {{eqn | r = x       | c =  }} {{eqn | r = 1 + \left({x - 1}\right)       | c =  }} {{eqn | r = 1 + \dbinom 1 1 \left({x - 1}\right)       | c = Binomial Coefficient with Self, for example }} {{eqn | r = H_1 + \sum_{k \mathop = 1}^1 \dbinom 1 k \dfrac {\left({x - 1}\right)^k} k       | c = Harmonic Number H1 }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{k \mathop = 1}^r \dfrac {x^k} k = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop = 1}^{r + 1} \dfrac {x^k} k = H_{r + 1} + \displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom {r + 1} k \dfrac {\left({x - 1}\right)^k} k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^{r + 1} \dfrac {x^k} k       | r = \sum_{k \mathop = 1}^r \dfrac {x^k} k + \dfrac {x^{r + 1} } {r + 1}       | c =  }} {{eqn | r = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k + \dfrac {x^{r + 1} } {r + 1}       | c = Induction Hypothesis }} {{eqn | r = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k + \dfrac {\left({x^{r + 1} - 1}\right) + 1} {r + 1}       | c =  }} {{eqn | r = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k + \dfrac {\left({x^{r + 1} - 1}\right)} {r + 1} + \dfrac 1 {r + 1}       | c =  }} {{eqn | r = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k + \left({x - 1}\right) \sum_{k \mathop = 0}^r \dfrac {x^k} {r + 1} + \dfrac 1 {r + 1}       | c =  }} {{eqn-intertext|further work needed}} {{eqn | r = H_{r + 1} + \displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom r k \dfrac {\left({x - 1}\right)^k} k       | c =  }} {{end-eqn}} {{finish}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 1}: \displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$
Differentiating {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \paren {\sum_{k \mathop = 1}^n \dfrac {x^k} k}       | r = \sum_{k \mathop = 1}^n k \dfrac {x^{k - 1} } k       | c = Power Rule for Derivatives }} {{eqn | r = \sum_{k \mathop = 1}^n x^{k - 1}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} x^k       | c = Translation of Index Variable of Summation }} {{eqn | r = \dfrac {x^n - 1} {x - 1}       | c = Sum of Geometric Sequence }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \dfrac {x^k} k       | r = \int \dfrac {x^n - 1} {x - 1} \rd x       | c =  }} {{eqn | r = \int \dfrac {\paren {z + 1}^n - 1} z \rd z       | c = Integration by Substitution }} {{eqn | r = \int \paren {\dfrac {\paren {z + 1}^n} z - \dfrac 1 z} \rd z       | c =  }} {{eqn | r = \int \paren {\dfrac 1 z \sum_{k \mathop = 0}^n \dbinom n k z^k - \dfrac 1 z} \rd z       | c = Binomial Theorem }} {{eqn | r = \int \paren {\sum_{k \mathop = 1}^n \dbinom n k z^{k - 1} + \dfrac 1 z - \dfrac 1 z} \rd z       | c =  }} {{eqn | r = \int \paren {\sum_{k \mathop = 1}^n \dbinom n k z^{k - 1} } \rd z       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \dbinom n k \dfrac {z^k} k + C       | c = Primitive of Power }} {{eqn | r = \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k + C       | c = Primitive of Power }} {{end-eqn}} When $x = 1$ we have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {1 - 1}^k} k + C       | r = \sum_{k \mathop = 1}^n \dfrac {1^k} k       | c =  }} {{eqn | ll= \leadsto       | l = 0 + C       | r = \sum_{k \mathop = 1}^n \dfrac 1 k       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = H_n       | c = {{Defof|Harmonic Number}} }} {{end-eqn}} The result follows.