avewright/proofwiki-math huggingface dataset:
Theorem: Lagrange's Trigonometric Identities/Sine/Sine Form

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \sin k x
      | r = \sin 0 + \sin x + \sin 2 x + \sin 3 x + \cdots + \sin n x
      | c = 
}}
{{eqn | r = \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}
      | c = 
}}
{{end-eqn}}
where $x$ is not an integer multiple of $2 \pi$.

Proof:
Let $x$ be a real number that is not a integer multiple of $2 \pi$.
Let $k$ be a non-negative integer.
We have, from Euler's Formula:
:$\map \exp {i k x} = i \sin k x + \cos k x$
Summing from $k = 0$ to $k = n$, we have: 
:$\ds \sum_{k \mathop = 0}^n \map \exp {i k x} = i \sum_{k \mathop = 0}^n \sin k x + \sum_{k \mathop = 0}^n \cos k x$
As $\sin k x$ and $\cos k x$ are both real for real $k, x$, we have: 
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \sin k x
      | r = \map \Im {\sum_{k \mathop = 0}^n \map \exp {i k x} }
}}
{{eqn | r = \map \Im {\paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2} }
      | c = Sum of $\map \exp {i k x}$ 
}}
{{eqn | r = \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Infinite Series of Product of Power and Cosine

Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty r^k \cos k x
      | r = 1 + r \cos x + r^2 \cos 2 x + r^3 \cos 3 x + \cdots
      | c = 
}}
{{eqn| r = \dfrac {1 - r \cos x} {1 - 2 r \cos x + r^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty r^k \cos k x
      | r = \map \Re {\sum_{k \mathop = 0}^\infty r^k e^{i k x} }
      | c = 
}}
{{eqn | r = \map \Re {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}
      | c = 
}}
{{eqn | r = \map \Re {\frac 1 {1 - r e^{i x} } }
      | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
      | c = 
}}
{{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
      | c = 
}}
{{eqn | r = \map \Re {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + r^2} }
      | c = {{Corollary|Euler's Formula}}
}}
{{eqn | r = \dfrac {1 - r \cos x} {1 - 2 r \cos x + r^2} 
      | c = after simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Series of Product of Power and Cosine

Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n r^k \map \cos {k x}
      | r = 1 + r \cos x + r^2 \cos 2 x + r^3 \cos 3 x + \cdots + r^n \cos n x
      | c = 
}}
{{eqn | r = \dfrac {r^{n + 2} \cos n x - r^{n + 1} \map \cos {n + 1} x - r \cos x + 1} {1 - 2 r \cos x + r^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n r^k \map \cos {k x}
      | r = \map \Re {\sum_{k \mathop = 0}^n r^k e^{i k x} }
      | c = 
}}
{{eqn | r = \map \Re {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^k}
      | c = 
}}
{{eqn | r = \map \Re {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } }
      | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Re {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } } 
      | c = 
}}
{{eqn | r = \map \Re {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} } 
      | c = 
}}
{{eqn | r = \map \Re {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} }
      | c = Euler's Formula and {{Corollary|Euler's Formula}}
}}
{{eqn | r = \dfrac {r^{n + 2} \cos n x - r^{n + 1} \map \cos {n + 1} x - r \cos x + 1} {1 - 2 r \cos x + r^2}
      | c = after simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Series of Product of Power and Sine

Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n r^k \map \sin {k x}
      | r = r \sin x + r^2 \sin 2 x + r^3 \sin 3 x + \cdots + r^n \sin n x
      | c = 
}}
{{eqn | r = \dfrac {r \sin x - r^{n + 1} \map \sin {n + 1} x + r^{n + 2} \sin n x} {1 - 2 r \cos x + r^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n r^k \map \sin {k x}
      | r = \map \Im {\sum_{k \mathop = 1}^n r^k e^{i k x} }
      | c = 
}}
{{eqn | r = \map \Im {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^n}
      | c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$
}}
{{eqn | r = \map \Im {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } } 
      | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Im {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } } 
      | c = 
}}
{{eqn | r = \map \Im {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
      | c = 
}}
{{eqn | r = \map \Im {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} } 
      | c = Euler's Formula and {{Corollary|Euler's Formula}}
}}
{{eqn | r = \dfrac {r \sin x - r^{n + 1} \map \sin {n + 1} x + r^{n + 2} \sin n x} {1 - 2 r \cos x + r^2}
      | c = after simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {1 + x}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = 
}}
{{eqn | r = 1 - x + x^2 - x^3 + x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {1 + x}
      | r = \frac 1 {1 - \paren {-x} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-x}^k
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {1 + x}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = 
}}
{{eqn | r = 1 - x + x^2 - x^3 + x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {1 + x}
      | r = \paren {1 + x}^{-1}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-1} - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 1} } } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = {{Defof|Factorial}} and simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-j} } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = Translation of Index Variable of Product
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k j} {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of Square of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 + x}^2}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k
      | c = 
}}
{{eqn | r = 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {1 + x}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = Power Series Expansion for $\dfrac 1 {1 + x}$
}}
{{eqn | ll= \leadsto
      | l = \frac \d {\d x} \frac 1 {1 + x}
      | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac 1 {\paren {1 + x}^2}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k k x^{k - 1}
      | c = differentiating {{WRT|Differentiation}} $x$
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\paren {1 + x}^2}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} k x^{k - 1}
      | c = taking one of the $-1$s out
}}
{{eqn | r = \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} k x^{k - 1}
      | c = the term in $k = 0$ vanishes
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {1 + x}k x^k
      | c = Translation of Index Variable of Product
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of Square of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 + x}^2}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k
      | c = 
}}
{{eqn | r = 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {1 + x} }
      | r = \paren {1 + x}^{-2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-2}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-2} - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 2} } } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = {{Defof|Factorial}} and simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-\paren {j + 1} } } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = Translation of Index Variable of Product
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k \paren {j + 1} } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Frullani's Integral

:$\ds \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x = \paren {\map f \infty - \map f 0} \ln \frac a b$

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x
	| r = \int_0^\infty \intlimits {\frac {\map f {x t} } x} {t = b} a \rd x
}}
{{eqn	| r = \int_0^\infty \int_b^a \map {f'} {x t} \rd t \rd x
	| c = Fundamental Theorem of Calculus
}} 
{{eqn	| r = \int_b^a \int_0^\infty \map {f'} {x t} \rd x \rd t
	| c = Fubini's Theorem
}}
{{eqn	| r = \int_b^a \intlimits {\frac {\map f {x t} } t} {x = 0} \infty \rd t
	| c = Fundamental Theorem of Calculus
}}
{{eqn	| r = \int_b^a \frac {\map f \infty - \map f 0} t \rd t
}}
{{eqn	| r = \paren {\map f \infty - \map f 0} \paren {\ln a -\ln b}
	| c = Primitive of Reciprocal, Fundamental Theorem of Calculus
}}
{{eqn	| r = \paren {\map f \infty - \map f 0} \ln \frac a b
	| c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Giuliano Frullani|cat = Frullani}}
Theorem: Power Series Expansion for Reciprocal of Cube of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 + x}^3}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k
      | c = 
}}
{{eqn | r = 1 - 3 x + 6 x^2 - 10 x^3 + 15 x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {1 + x}^2}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k
      | c = Power Series Expansion of $\dfrac 1 {\paren {1 + x}^2}$
}}
{{eqn | ll= \leadsto
      | l = \frac \d {\d x} \frac 1 {\paren {1 + x}^2}
      | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac 2 {\paren {1 + x}^3}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} k x^{k - 1}
      | c = differentiating {{WRT|Differentiation}} $x$
}}
{{eqn | ll= \leadsto
      | l = \frac 2 {\paren {1 + x}^3}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \paren {k + 1} k x^{k - 1}
      | c = taking one of the $-1$s out
}}
{{eqn | r = \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} \paren {k + 1} k x^{k - 1}
      | c = the term in $k = 0$ vanishes
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 2} \paren {k + 1} x^k
      | c = Translation of Index Variable of Product
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\paren {1 + x}^3}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of Cube of 1 + x

Let $x \in \R$ such that $-1 < x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 + x}^3}
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k
      | c = 
}}
{{eqn | r = 1 - 3 x + 6 x^2 - 10 x^3 + 15 x^4 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {1 + x}^3 }
      | r = \paren {1 + x}^{-3}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-3}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-3} - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 3} } } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = {{Defof|Factorial}} and simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-\paren {j + 2} } } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = Translation of Index Variable of Product
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k \paren {j + 2} } {\ds \prod_{j \mathop = 1}^k j} x^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of Square Root of 1 + x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\sqrt {1 + x} }
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k
      | c = 
}}
{{eqn | r = 1 - \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 - \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\sqrt {1 + x} }
      | r = \paren {1 + x}^{-\frac 1 2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 2}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 2} - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} } {2^k k!} x^k
      | c = Translation of Index Variable of Product and simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} \prod_{j \mathop = 1}^k \paren {2 j} } {2^k k! \ds \prod_{j \mathop = 1}^k \paren {2 j} } x^k
      | c = multiplying top and bottom by $\ds \prod_{j \mathop = 1}^k \paren {2 j}$
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^{2 k} j} {2^k k! \paren {2^k \ds \prod_{j \mathop = 1}^k j} } x^k
      | c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {2^k k! \paren {2^k k!} } x^k
      | c = {{Defof|Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Square Root of 1 + x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sqrt {1 + x}
      | r = 1 + \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} \frac {\paren {2 \paren {k - 1} }!} {2^{2 k - 1} k! \paren {k - 1}!} x^k
      | c = 
}}
{{eqn | r = 1 + \frac 1 2 x - \frac 1 {2 \times 4} x^2 + \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt {1 + x}
      | r = \paren {1 + x}^{\frac 1 2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 2}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 2 - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 2 j} } {2^k k!} x^k
      | c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {2 j - 1} } {2^k k!} x^k
      | c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \frac {\ds \prod_{j \mathop = 1}^{k - 1} \paren {2 j - 1} } {2^k k!} x^k
      | c = extracting the case where $k = 0$
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \frac {\ds \prod_{j \mathop = 1}^{k - 1} \paren {2 j - 1} \prod_{j \mathop = 1}^{k - 1} \paren {2 j} } {2^k k! \ds \prod_{j \mathop = 1}^{k - 1} \paren {2 j} } x^k
      | c = multiplying top and bottom by $\ds \prod_{j \mathop = 1}^{k - 1} \paren {2 j}$
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \frac {\ds \prod_{j \mathop = 1}^{2 \paren {k - 1} } j} {2^k k! \paren {2^{k - 1} \ds \prod_{j \mathop = 1}^{k - 1} j} } x^k
      | c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \frac {\paren {2 \paren {k - 1} }!} {2^{2 k - 1} k! \paren {k - 1}!} x^k
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Reciprocal of Cube Root of 1 + x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
:$\dfrac 1 {\sqrt [3] {1 + x} } = 1 - \dfrac 1 3 x + \dfrac {1 \times 4} {3 \times 6} x^2 - \dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots$

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\sqrt [3] {1 + x} }
      | r = \paren {1 + x}^{-\frac 1 3}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 3}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 3} - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {3 j - 1} } {3^k k!} x^k
      | c = Translation of Index Variable of Product and simplifying
}}
{{eqn | r = 1 - \frac 1 3 x + \frac {1 \times 4} {3 \times 6} x^2 - \frac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots
      | c = extracting the juice
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Cube Root of 1 + x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
:$\sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots$

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt [3] {1 + x}
      | r = \paren {1 + x}^{\frac 1 3}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 3}^{\underline k} } {k!} x^k
      | c = General Binomial Theorem
}}
{{eqn | r = 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 3 - j} } {k!} x^k
      | c = {{Defof|Falling Factorial}} and extracting $k = 0$
}}
{{eqn | r = 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 3 j} } {3^k k!} x^k
      | c = simplifying
}}
{{eqn | r = \sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for General Exponential Function

Let $a \in \R_{> 0}$ be a (strictly) positive real number.
Then:
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = a^x
      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {x \ln a}^n} {n!}
      | c = 
}}
{{eqn | r = 1 + x \ln a + \frac {\paren {x \ln a}^2} {2!} + \frac {\paren {x \ln a}^3} {3!} + \cdots
      | c = 
}}
{{end-eqn}}

Proof:
By definition of a power to a real number:
:$a^x = \map \exp {x \ln a}$
As $x \ln a$ is itself a real number, we can use Power Series Expansion for Exponential Function:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = \exp x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
      | c = 
}}
{{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots
      | c = 
}}
{{end-eqn}}
substituting $x \ln a$ for $x$.
{{qed}}
Theorem: Power Series Expansion for Half Logarithm of 1 + x over 1 - x

{{begin-eqn}}
{{eqn | l = \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}
}}
{{eqn | r = x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots
}}
{{end-eqn}}
valid for all $x \in \R$ such that $-1 < x < 1$.

Proof:
From Power Series Expansion for $\map \ln {1 + x}$:
:$(1): \quad \ds \map \ln {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$
for $-1 < x \le 1$.
From Power Series Expansion for $\map \ln {1 - x}$:
:$(2): \quad \ds \map \ln {1 - x} = - \sum_{n \mathop = 1}^\infty \frac {x^n} n$
for $-1 < x < 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map \ln {1 + x} - \map \ln {1 - x}
      | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n + \sum_{n \mathop = 1}^\infty \frac {x^n} n
      | c = subtracting $(2)$ from $(1)$
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r - 1} \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}
      | c = separating out into odd and even indices
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} - \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}
      | c = $\paren {-1}^{2 r} = 1$, $\paren {-1}^{2 r - 1} = -1$
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}
      | c = even indices cancel
}}
{{eqn | ll= \leadsto
      | l = \map \ln {\frac {1 + x} {1 - x} }
      | r = 2 \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}
      | c = Difference of Logarithms
}}
{{end-eqn}}
Hence the result.
{{qed}}